Mixing
Opinions on what I've come up with for a 12 V-to-5 V automotive circuit?
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
Is this a solid solution for an automotive 12 V-to-5 V circuit to power an ATtiny? I've been trying to come up with a solution for my needs that can handle the automotive environment that is so scary some places won't even talk about it.
From recommendations, scouring datasheets and lots of googling this is what I've come up with so far. It will be placed inside the cabin of a vehicle (not the engine bay) and everything selected is AEC qualified, but I don't know if it will actually work as intended or if I don't have enough protection (the ESD/TVS should kill most of the automotive nightmare right?). The other part of my circuit (not shown) is an ATtiny that will operate some auxiliary lighting.
Please don't say just use a USB 12 V adapter, because I'm not trying to couple multiple boards together hence trying to nail this circuit so I can include it on my device's PCB, but other than that please school me if I have anything incorrect.
**Also note the resistor for the LED is incorrectly rated in the schematic. It's supposed to be 150Â ohm like the component list.
Components:
- D1: SMA6J5.0CA-TR
- IC1: ZLDO1117QK50TC
- C1: C1210X106M3RACAUTO 10 ÂµF 25 V
- C2: TPSC107K010R0150 100 ÂµF 10 V
-- Optional, for testing (not really important) --
- R1: ERJ-U06J151V (150 ohm)
- LED1: 150080GS75000
voltage-regulator automotive circuit-protection
edited 9 mins ago
Peter Mortensen
1,58131422
asked 11 hours ago
dawm
195
 |Â
show 2 more comments
up vote
3
down vote
favorite
Is this a solid solution for an automotive 12 V-to-5 V circuit to power an ATtiny? I've been trying to come up with a solution for my needs that can handle the automotive environment that is so scary some places won't even talk about it.
From recommendations, scouring datasheets and lots of googling this is what I've come up with so far. It will be placed inside the cabin of a vehicle (not the engine bay) and everything selected is AEC qualified, but I don't know if it will actually work as intended or if I don't have enough protection (the ESD/TVS should kill most of the automotive nightmare right?). The other part of my circuit (not shown) is an ATtiny that will operate some auxiliary lighting.
Please don't say just use a USB 12 V adapter, because I'm not trying to couple multiple boards together hence trying to nail this circuit so I can include it on my device's PCB, but other than that please school me if I have anything incorrect.
**Also note the resistor for the LED is incorrectly rated in the schematic. It's supposed to be 150Â ohm like the component list.
Components:
- D1: SMA6J5.0CA-TR
- IC1: ZLDO1117QK50TC
- C1: C1210X106M3RACAUTO 10 ÂµF 25 V
- C2: TPSC107K010R0150 100 ÂµF 10 V
-- Optional, for testing (not really important) --
- R1: ERJ-U06J151V (150 ohm)
- LED1: 150080GS75000
voltage-regulator automotive circuit-protection
edited 9 mins ago
Peter Mortensen
1,58131422
asked 11 hours ago
dawm
195
1
It would be better if you specified the expected current draw. From your text circuit get about (5-1.2)/95 = ~40mA. (B.T.W Your text says 150 ohm, your diagram says 95 ohm...)
– Oldfart
10 hours ago
2
There are "cigarette lighter" adapters available with selectable voltage outputs...
– Solar Mike
10 hours ago
1
@Oldfart eek! I changed LEDs and guess I missed updating the schematic, the schematic should be 150ohm just as the component list. The circuit on the other end of the regulated 5v 1A shouldn't be more than 0.8A
– dawm
10 hours ago
3
They come in a complete package - not "multiple boards" and tend to be very compact and efficient providing a clean supply, and I found the one I bought has been perfect for the multiple uses I have given it.
– Solar Mike
10 hours ago
2
@dawm I will let you do it your way - however, as I was a vehicle electrician in a previous life I would be looking at various options personally. You will need some cabling somewhere...
– Solar Mike
10 hours ago
 |Â
show 2 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Is this a solid solution for an automotive 12 V-to-5 V circuit to power an ATtiny? I've been trying to come up with a solution for my needs that can handle the automotive environment that is so scary some places won't even talk about it.
From recommendations, scouring datasheets and lots of googling this is what I've come up with so far. It will be placed inside the cabin of a vehicle (not the engine bay) and everything selected is AEC qualified, but I don't know if it will actually work as intended or if I don't have enough protection (the ESD/TVS should kill most of the automotive nightmare right?). The other part of my circuit (not shown) is an ATtiny that will operate some auxiliary lighting.
Please don't say just use a USB 12 V adapter, because I'm not trying to couple multiple boards together hence trying to nail this circuit so I can include it on my device's PCB, but other than that please school me if I have anything incorrect.
**Also note the resistor for the LED is incorrectly rated in the schematic. It's supposed to be 150Â ohm like the component list.
Components:
- D1: SMA6J5.0CA-TR
- IC1: ZLDO1117QK50TC
- C1: C1210X106M3RACAUTO 10 ÂµF 25 V
- C2: TPSC107K010R0150 100 ÂµF 10 V
-- Optional, for testing (not really important) --
- R1: ERJ-U06J151V (150 ohm)
- LED1: 150080GS75000
voltage-regulator automotive circuit-protection
edited 9 mins ago
Peter Mortensen
1,58131422
asked 11 hours ago
dawm
195
Is this a solid solution for an automotive 12 V-to-5 V circuit to power an ATtiny? I've been trying to come up with a solution for my needs that can handle the automotive environment that is so scary some places won't even talk about it.
From recommendations, scouring datasheets and lots of googling this is what I've come up with so far. It will be placed inside the cabin of a vehicle (not the engine bay) and everything selected is AEC qualified, but I don't know if it will actually work as intended or if I don't have enough protection (the ESD/TVS should kill most of the automotive nightmare right?). The other part of my circuit (not shown) is an ATtiny that will operate some auxiliary lighting.
Please don't say just use a USB 12 V adapter, because I'm not trying to couple multiple boards together hence trying to nail this circuit so I can include it on my device's PCB, but other than that please school me if I have anything incorrect.
**Also note the resistor for the LED is incorrectly rated in the schematic. It's supposed to be 150Â ohm like the component list.
Components:
- D1: SMA6J5.0CA-TR
- IC1: ZLDO1117QK50TC
- C1: C1210X106M3RACAUTO 10 ÂµF 25 V
- C2: TPSC107K010R0150 100 ÂµF 10 V
-- Optional, for testing (not really important) --
- R1: ERJ-U06J151V (150 ohm)
- LED1: 150080GS75000
voltage-regulator automotive circuit-protection
voltage-regulator automotive circuit-protection
edited 9 mins ago
Peter Mortensen
1,58131422
asked 11 hours ago
dawm
195
edited 9 mins ago
Peter Mortensen
1,58131422
asked 11 hours ago
dawm
195
edited 9 mins ago
Peter Mortensen
1,58131422
edited 9 mins ago
Peter Mortensen
1,58131422
edited 9 mins ago
Peter Mortensen
1,58131422
1,58131422
asked 11 hours ago
dawm
195
asked 11 hours ago
dawm
195
asked 11 hours ago
dawm
195
195
1
It would be better if you specified the expected current draw. From your text circuit get about (5-1.2)/95 = ~40mA. (B.T.W Your text says 150 ohm, your diagram says 95 ohm...)
– Oldfart
10 hours ago
2
There are "cigarette lighter" adapters available with selectable voltage outputs...
– Solar Mike
10 hours ago
1
@Oldfart eek! I changed LEDs and guess I missed updating the schematic, the schematic should be 150ohm just as the component list. The circuit on the other end of the regulated 5v 1A shouldn't be more than 0.8A
– dawm
10 hours ago
3
They come in a complete package - not "multiple boards" and tend to be very compact and efficient providing a clean supply, and I found the one I bought has been perfect for the multiple uses I have given it.
– Solar Mike
10 hours ago
2
@dawm I will let you do it your way - however, as I was a vehicle electrician in a previous life I would be looking at various options personally. You will need some cabling somewhere...
– Solar Mike
10 hours ago
 |Â
show 2 more comments
1
It would be better if you specified the expected current draw. From your text circuit get about (5-1.2)/95 = ~40mA. (B.T.W Your text says 150 ohm, your diagram says 95 ohm...)
– Oldfart
10 hours ago
2
There are "cigarette lighter" adapters available with selectable voltage outputs...
– Solar Mike
10 hours ago
1
@Oldfart eek! I changed LEDs and guess I missed updating the schematic, the schematic should be 150ohm just as the component list. The circuit on the other end of the regulated 5v 1A shouldn't be more than 0.8A
– dawm
10 hours ago
3
They come in a complete package - not "multiple boards" and tend to be very compact and efficient providing a clean supply, and I found the one I bought has been perfect for the multiple uses I have given it.
– Solar Mike
10 hours ago
2
@dawm I will let you do it your way - however, as I was a vehicle electrician in a previous life I would be looking at various options personally. You will need some cabling somewhere...
– Solar Mike
10 hours ago
1
1
It would be better if you specified the expected current draw. From your text circuit get about (5-1.2)/95 = ~40mA. (B.T.W Your text says 150 ohm, your diagram says 95 ohm...)
– Oldfart
10 hours ago
It would be better if you specified the expected current draw. From your text circuit get about (5-1.2)/95 = ~40mA. (B.T.W Your text says 150 ohm, your diagram says 95 ohm...)
– Oldfart
10 hours ago
2
2
There are "cigarette lighter" adapters available with selectable voltage outputs...
– Solar Mike
10 hours ago
There are "cigarette lighter" adapters available with selectable voltage outputs...
– Solar Mike
10 hours ago
1
1
@Oldfart eek! I changed LEDs and guess I missed updating the schematic, the schematic should be 150ohm just as the component list. The circuit on the other end of the regulated 5v 1A shouldn't be more than 0.8A
– dawm
10 hours ago
@Oldfart eek! I changed LEDs and guess I missed updating the schematic, the schematic should be 150ohm just as the component list. The circuit on the other end of the regulated 5v 1A shouldn't be more than 0.8A
– dawm
10 hours ago
3
3
They come in a complete package - not "multiple boards" and tend to be very compact and efficient providing a clean supply, and I found the one I bought has been perfect for the multiple uses I have given it.
– Solar Mike
10 hours ago
They come in a complete package - not "multiple boards" and tend to be very compact and efficient providing a clean supply, and I found the one I bought has been perfect for the multiple uses I have given it.
– Solar Mike
10 hours ago
2
2
@dawm I will let you do it your way - however, as I was a vehicle electrician in a previous life I would be looking at various options personally. You will need some cabling somewhere...
– Solar Mike
10 hours ago
@dawm I will let you do it your way - however, as I was a vehicle electrician in a previous life I would be looking at various options personally. You will need some cabling somewhere...
– Solar Mike
10 hours ago
 |Â
show 2 more comments
5 Answers
5
active
oldest
votes
up vote
9
down vote
You specified a current draw of 800mA. The voltage drop is 12-5=7V thus the regulator will consume 0.8*7=5.6 Watts. That will likely burn your regulator from the board.
Even if you take a different type and a heat sink, you don't want a big heat source in your car.
I strongly suggest you use a ready-made 5V switching regulator. They come in a bit larger TO220 housing.
answered 10 hours ago
Oldfart
7,1732723
1
One of my earlier ideas was with a switching reg but the extra components just to drive it properly took up a lot of real estate on the board design (unless I never came across the one pictured that is all in one?), a friend of mine has been using the AZ1117CH-5.0TRG1 which is the non-automotive version of the ZLDO1117 in a similar application (used on ATVs and circle track cars) with no heat issues (claims it doesn't even get warm) but his circuit lacks the protection i'm trying to figure out.
– dawm
9 hours ago
1
All you do is add the protection to the regulator, just as you did with the ZLDO1117QK50TC. See it as a drop-in replacement for the ZLDO1117QK50TC
– Oldfart
9 hours ago
2
@dawm Try this digikey category: digikey.com/products/en/power-supplies-board-mount/… There's a parametric search option for "linear regulator replacement", too (under the "type" field), which tends to be devices like the one Oldfart suggested; drop-in replacements for linear regulators.
– Felthry
7 hours ago
@dawm Don't follow the red herring. Safely dissipating 5W takes way more real estate.
– Agent_L
1 hour ago
add a comment |Â
up vote
5
down vote
If you must use a linear regulator (Oldfart is right about the unwanted dissipation), then you don't need a LDO. LDO's withstand maybe 20V input voltage; the standard 7805 is rated at 35V, which for automotive applications is desirable. I would also suggest a fuse on the input; if this is undesirable, then possibly a 1-ohm 1W resistor to take the edge off a spike. Finally, a solid ground connection is essential.
answered 9 hours ago
henros
45023
1
a fuse will be added, just haven't looked into what size/type..etc I'm going to require. I thought that using the TVS would help reduce the chance of higher voltages reaching the LDO.
– dawm
9 hours ago
add a comment |Â
up vote
3
down vote
The most common solution is to use a switching regulator. Switching regulators dissipate a lot less heat than linear regulators, especially when there is a large voltage drop. The down side is that they produce more noise, but this can be mitigated by following them with a linear regulator to produce a fairly clean supply.
Texas make some good switching regulators and have a free design tool on their site which will produce a schematic, select components and recommend a PCB layout. Alternatively you can buy a module from vendors like Recom and Meanwell, which is literally a black box that sits on the PCB taking 12V in and generating a selectable output voltage.
If your system does not need a low noise supply you could use a 5V output switch more supply. If it does then you could select a 6V output and use a 5V LDO to further regulate and clean the supply. A popular technique is to use the output of the switching regulator to drive high power devices that aren't sensitive to noise, such as lights or battery chargers, and have only the sensitive parts run from the LDO to keep heat generation down.
answered 7 hours ago
user
1,084614
add a comment |Â
up vote
2
down vote
Your major problems are:-
- The +12V can actually be close to +30V. Car batteries are charged at around +14V, and tow trucks use two batteries in series to deliver a nominal +24V to really crank that starter motor, even if they need to use long jump leads. Your regulator needs to survive continuous overvoltage, or you need to add protection for it.
- The +12V can actually be -12V if the inputs are reversed. (Or -30V, see above.) Can your regulator and electrolytic capacitor handle this? If not, add a diode.
- The +12V can actually be shorted together, or during cranking can drop to substantially less than +12V, discharging any capacitors in front of the regulator, and discharging any capacitors behind the regulator back through the regulator. You may need a diode to stop this, and/or remove your 10uF cap goes.
- The +5V supply may draw more current than the regulator can supply. You may need a low-side fuse.
- The whole circuit may draw more current than the wiring from the +12V can supply. (A car battery can supply a lot, but remember this typically goes through a fuse box and relays.) You may want a high-side fuse, or give an explicit warning that you will rely on the car's own fuse box.
- Your +5V supply will have instantaneous current spikes, and the regulator and 100uF will not be able to keep up. (Electrolytic caps are pretty slow.) Normal practise for this is to parallel an electrolytic with a faster, smaller capacitor (typically the same you'd use for decoupling) to handle the transients.
- The capacitors will give you a huge inrush current. A resistor in the +12V supply is a good idea, and also helps solve some of your shorting-out issues.
Some of your circuit protection issues could be taken care of by using an automotive-spec regulator like an LM2940. Others are still an issue. Similarly, changing to a switch-mode regulator will help you with some problems but not others, and can introduce different problems too (not least noise/ripple on the supply line).
Figuring out solutions to these issues is left as an exercise for the OP, since it may well be coursework. :)
answered 1 hour ago
Graham
2,377411
add a comment |Â
up vote
1
down vote
I would use unidirectional TVS at the input so that negative spike goes through with minimal residual voltage.
Also note that TVS has to protect 12V input which can be also 15V in normal conditions. You have choosen 5V TVS, which is wrong. Use a SMA6J18A 18V.
answered 11 hours ago
Marko BurÅ¡iÄÂ
9,4042812
2
Sorry I'm still confused on how TVS diodes operate, using the SMA6J18A with 18V working and clamping at 28.3V to 33.2V that seems to exceed protecting the regulator from higher than needed voltage. Would the TPD1E10B09QDPYRQ1 be better suited?, its working voltage is 9V and clamping is 13V, with the ZLDO1117's SOA curve peak being 10V for 1A that would put me in the SOA during clamping and normal operation right? Datasheet says Vin -Vout on the SOA curve chart, 9-5 = 4V, 13-5 = 8V for a safe 5V 1A output.
– dawm
10 hours ago
1
Or do I completely misunderstand TVS diodes completely? Does current flow past a TVS untouched at the working voltage, then starts getting clamped when it goes over the breakdown voltage? That's how I understood it, that is why I picked the SMA6J5.0CA-TR, I thought it would allow at least 5V through it and start clamping it to 9V-13.4V after it reached the breakdown.
– dawm
10 hours ago
1
@dawm Look at the diagram on page 2 of the datasheet. Because the diode is normally used in reverse direction only the left half of the diagram is of interest. If you're going to use e.g. an SMA6J15A you get V[WM]=15V, where at 15V only 1µA will leak through the diode, and beyond the breakdown voltage V[BR]=18.5V (max.) the device will act like a resistor in the order of 0.1...0.2Ohms.
– JimmyB
6 hours ago
2
You will want to include a fuse in the supply line before the TVS which blows in case the overvoltage is not just spurious or the supply gets reversed to prevent the TVS from creating a short curcuit and burning out.
– JimmyB
6 hours ago
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
You specified a current draw of 800mA. The voltage drop is 12-5=7V thus the regulator will consume 0.8*7=5.6 Watts. That will likely burn your regulator from the board.
Even if you take a different type and a heat sink, you don't want a big heat source in your car.
I strongly suggest you use a ready-made 5V switching regulator. They come in a bit larger TO220 housing.
answered 10 hours ago
Oldfart
7,1732723
1
One of my earlier ideas was with a switching reg but the extra components just to drive it properly took up a lot of real estate on the board design (unless I never came across the one pictured that is all in one?), a friend of mine has been using the AZ1117CH-5.0TRG1 which is the non-automotive version of the ZLDO1117 in a similar application (used on ATVs and circle track cars) with no heat issues (claims it doesn't even get warm) but his circuit lacks the protection i'm trying to figure out.
– dawm
9 hours ago
1
All you do is add the protection to the regulator, just as you did with the ZLDO1117QK50TC. See it as a drop-in replacement for the ZLDO1117QK50TC
– Oldfart
9 hours ago
2
@dawm Try this digikey category: digikey.com/products/en/power-supplies-board-mount/… There's a parametric search option for "linear regulator replacement", too (under the "type" field), which tends to be devices like the one Oldfart suggested; drop-in replacements for linear regulators.
– Felthry
7 hours ago
@dawm Don't follow the red herring. Safely dissipating 5W takes way more real estate.
– Agent_L
1 hour ago
add a comment |Â
up vote
9
down vote
You specified a current draw of 800mA. The voltage drop is 12-5=7V thus the regulator will consume 0.8*7=5.6 Watts. That will likely burn your regulator from the board.
Even if you take a different type and a heat sink, you don't want a big heat source in your car.
I strongly suggest you use a ready-made 5V switching regulator. They come in a bit larger TO220 housing.
answered 10 hours ago
Oldfart
7,1732723
1
One of my earlier ideas was with a switching reg but the extra components just to drive it properly took up a lot of real estate on the board design (unless I never came across the one pictured that is all in one?), a friend of mine has been using the AZ1117CH-5.0TRG1 which is the non-automotive version of the ZLDO1117 in a similar application (used on ATVs and circle track cars) with no heat issues (claims it doesn't even get warm) but his circuit lacks the protection i'm trying to figure out.
– dawm
9 hours ago
1
All you do is add the protection to the regulator, just as you did with the ZLDO1117QK50TC. See it as a drop-in replacement for the ZLDO1117QK50TC
– Oldfart
9 hours ago
2
@dawm Try this digikey category: digikey.com/products/en/power-supplies-board-mount/… There's a parametric search option for "linear regulator replacement", too (under the "type" field), which tends to be devices like the one Oldfart suggested; drop-in replacements for linear regulators.
– Felthry
7 hours ago
@dawm Don't follow the red herring. Safely dissipating 5W takes way more real estate.
– Agent_L
1 hour ago
add a comment |Â
up vote
9
down vote
up vote
9
down vote
You specified a current draw of 800mA. The voltage drop is 12-5=7V thus the regulator will consume 0.8*7=5.6 Watts. That will likely burn your regulator from the board.
Even if you take a different type and a heat sink, you don't want a big heat source in your car.
I strongly suggest you use a ready-made 5V switching regulator. They come in a bit larger TO220 housing.
answered 10 hours ago
Oldfart
7,1732723
You specified a current draw of 800mA. The voltage drop is 12-5=7V thus the regulator will consume 0.8*7=5.6 Watts. That will likely burn your regulator from the board.
Even if you take a different type and a heat sink, you don't want a big heat source in your car.
I strongly suggest you use a ready-made 5V switching regulator. They come in a bit larger TO220 housing.
answered 10 hours ago
Oldfart
7,1732723
answered 10 hours ago
Oldfart
7,1732723
answered 10 hours ago
Oldfart
7,1732723
answered 10 hours ago
Oldfart
7,1732723
7,1732723
1
One of my earlier ideas was with a switching reg but the extra components just to drive it properly took up a lot of real estate on the board design (unless I never came across the one pictured that is all in one?), a friend of mine has been using the AZ1117CH-5.0TRG1 which is the non-automotive version of the ZLDO1117 in a similar application (used on ATVs and circle track cars) with no heat issues (claims it doesn't even get warm) but his circuit lacks the protection i'm trying to figure out.
– dawm
9 hours ago
1
All you do is add the protection to the regulator, just as you did with the ZLDO1117QK50TC. See it as a drop-in replacement for the ZLDO1117QK50TC
– Oldfart
9 hours ago
2
@dawm Try this digikey category: digikey.com/products/en/power-supplies-board-mount/… There's a parametric search option for "linear regulator replacement", too (under the "type" field), which tends to be devices like the one Oldfart suggested; drop-in replacements for linear regulators.
– Felthry
7 hours ago
@dawm Don't follow the red herring. Safely dissipating 5W takes way more real estate.
– Agent_L
1 hour ago
add a comment |Â
1
One of my earlier ideas was with a switching reg but the extra components just to drive it properly took up a lot of real estate on the board design (unless I never came across the one pictured that is all in one?), a friend of mine has been using the AZ1117CH-5.0TRG1 which is the non-automotive version of the ZLDO1117 in a similar application (used on ATVs and circle track cars) with no heat issues (claims it doesn't even get warm) but his circuit lacks the protection i'm trying to figure out.
– dawm
9 hours ago
1
All you do is add the protection to the regulator, just as you did with the ZLDO1117QK50TC. See it as a drop-in replacement for the ZLDO1117QK50TC
– Oldfart
9 hours ago
2
@dawm Try this digikey category: digikey.com/products/en/power-supplies-board-mount/… There's a parametric search option for "linear regulator replacement", too (under the "type" field), which tends to be devices like the one Oldfart suggested; drop-in replacements for linear regulators.
– Felthry
7 hours ago
@dawm Don't follow the red herring. Safely dissipating 5W takes way more real estate.
– Agent_L
1 hour ago
1
1
One of my earlier ideas was with a switching reg but the extra components just to drive it properly took up a lot of real estate on the board design (unless I never came across the one pictured that is all in one?), a friend of mine has been using the AZ1117CH-5.0TRG1 which is the non-automotive version of the ZLDO1117 in a similar application (used on ATVs and circle track cars) with no heat issues (claims it doesn't even get warm) but his circuit lacks the protection i'm trying to figure out.
– dawm
9 hours ago
One of my earlier ideas was with a switching reg but the extra components just to drive it properly took up a lot of real estate on the board design (unless I never came across the one pictured that is all in one?), a friend of mine has been using the AZ1117CH-5.0TRG1 which is the non-automotive version of the ZLDO1117 in a similar application (used on ATVs and circle track cars) with no heat issues (claims it doesn't even get warm) but his circuit lacks the protection i'm trying to figure out.
– dawm
9 hours ago
1
1
All you do is add the protection to the regulator, just as you did with the ZLDO1117QK50TC. See it as a drop-in replacement for the ZLDO1117QK50TC
– Oldfart
9 hours ago
All you do is add the protection to the regulator, just as you did with the ZLDO1117QK50TC. See it as a drop-in replacement for the ZLDO1117QK50TC
– Oldfart
9 hours ago
2
2
@dawm Try this digikey category: digikey.com/products/en/power-supplies-board-mount/… There's a parametric search option for "linear regulator replacement", too (under the "type" field), which tends to be devices like the one Oldfart suggested; drop-in replacements for linear regulators.
– Felthry
7 hours ago
@dawm Try this digikey category: digikey.com/products/en/power-supplies-board-mount/… There's a parametric search option for "linear regulator replacement", too (under the "type" field), which tends to be devices like the one Oldfart suggested; drop-in replacements for linear regulators.
– Felthry
7 hours ago
@dawm Don't follow the red herring. Safely dissipating 5W takes way more real estate.
– Agent_L
1 hour ago
@dawm Don't follow the red herring. Safely dissipating 5W takes way more real estate.
– Agent_L
1 hour ago
add a comment |Â
up vote
5
down vote
If you must use a linear regulator (Oldfart is right about the unwanted dissipation), then you don't need a LDO. LDO's withstand maybe 20V input voltage; the standard 7805 is rated at 35V, which for automotive applications is desirable. I would also suggest a fuse on the input; if this is undesirable, then possibly a 1-ohm 1W resistor to take the edge off a spike. Finally, a solid ground connection is essential.
answered 9 hours ago
henros
45023
1
a fuse will be added, just haven't looked into what size/type..etc I'm going to require. I thought that using the TVS would help reduce the chance of higher voltages reaching the LDO.
– dawm
9 hours ago
add a comment |Â
up vote
5
down vote
If you must use a linear regulator (Oldfart is right about the unwanted dissipation), then you don't need a LDO. LDO's withstand maybe 20V input voltage; the standard 7805 is rated at 35V, which for automotive applications is desirable. I would also suggest a fuse on the input; if this is undesirable, then possibly a 1-ohm 1W resistor to take the edge off a spike. Finally, a solid ground connection is essential.
answered 9 hours ago
henros
45023
1
a fuse will be added, just haven't looked into what size/type..etc I'm going to require. I thought that using the TVS would help reduce the chance of higher voltages reaching the LDO.
– dawm
9 hours ago
add a comment |Â
up vote
5
down vote
up vote
5
down vote
If you must use a linear regulator (Oldfart is right about the unwanted dissipation), then you don't need a LDO. LDO's withstand maybe 20V input voltage; the standard 7805 is rated at 35V, which for automotive applications is desirable. I would also suggest a fuse on the input; if this is undesirable, then possibly a 1-ohm 1W resistor to take the edge off a spike. Finally, a solid ground connection is essential.
answered 9 hours ago
henros
45023
If you must use a linear regulator (Oldfart is right about the unwanted dissipation), then you don't need a LDO. LDO's withstand maybe 20V input voltage; the standard 7805 is rated at 35V, which for automotive applications is desirable. I would also suggest a fuse on the input; if this is undesirable, then possibly a 1-ohm 1W resistor to take the edge off a spike. Finally, a solid ground connection is essential.
answered 9 hours ago
henros
45023
answered 9 hours ago
henros
45023
answered 9 hours ago
henros
45023
answered 9 hours ago
henros
45023
45023
1
a fuse will be added, just haven't looked into what size/type..etc I'm going to require. I thought that using the TVS would help reduce the chance of higher voltages reaching the LDO.
– dawm
9 hours ago
add a comment |Â
1
a fuse will be added, just haven't looked into what size/type..etc I'm going to require. I thought that using the TVS would help reduce the chance of higher voltages reaching the LDO.
– dawm
9 hours ago
1
1
a fuse will be added, just haven't looked into what size/type..etc I'm going to require. I thought that using the TVS would help reduce the chance of higher voltages reaching the LDO.
– dawm
9 hours ago
a fuse will be added, just haven't looked into what size/type..etc I'm going to require. I thought that using the TVS would help reduce the chance of higher voltages reaching the LDO.
– dawm
9 hours ago
add a comment |Â
up vote
3
down vote
The most common solution is to use a switching regulator. Switching regulators dissipate a lot less heat than linear regulators, especially when there is a large voltage drop. The down side is that they produce more noise, but this can be mitigated by following them with a linear regulator to produce a fairly clean supply.
Texas make some good switching regulators and have a free design tool on their site which will produce a schematic, select components and recommend a PCB layout. Alternatively you can buy a module from vendors like Recom and Meanwell, which is literally a black box that sits on the PCB taking 12V in and generating a selectable output voltage.
If your system does not need a low noise supply you could use a 5V output switch more supply. If it does then you could select a 6V output and use a 5V LDO to further regulate and clean the supply. A popular technique is to use the output of the switching regulator to drive high power devices that aren't sensitive to noise, such as lights or battery chargers, and have only the sensitive parts run from the LDO to keep heat generation down.
answered 7 hours ago
user
1,084614
add a comment |Â
up vote
3
down vote
The most common solution is to use a switching regulator. Switching regulators dissipate a lot less heat than linear regulators, especially when there is a large voltage drop. The down side is that they produce more noise, but this can be mitigated by following them with a linear regulator to produce a fairly clean supply.
Texas make some good switching regulators and have a free design tool on their site which will produce a schematic, select components and recommend a PCB layout. Alternatively you can buy a module from vendors like Recom and Meanwell, which is literally a black box that sits on the PCB taking 12V in and generating a selectable output voltage.
If your system does not need a low noise supply you could use a 5V output switch more supply. If it does then you could select a 6V output and use a 5V LDO to further regulate and clean the supply. A popular technique is to use the output of the switching regulator to drive high power devices that aren't sensitive to noise, such as lights or battery chargers, and have only the sensitive parts run from the LDO to keep heat generation down.
answered 7 hours ago
user
1,084614
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The most common solution is to use a switching regulator. Switching regulators dissipate a lot less heat than linear regulators, especially when there is a large voltage drop. The down side is that they produce more noise, but this can be mitigated by following them with a linear regulator to produce a fairly clean supply.
Texas make some good switching regulators and have a free design tool on their site which will produce a schematic, select components and recommend a PCB layout. Alternatively you can buy a module from vendors like Recom and Meanwell, which is literally a black box that sits on the PCB taking 12V in and generating a selectable output voltage.
If your system does not need a low noise supply you could use a 5V output switch more supply. If it does then you could select a 6V output and use a 5V LDO to further regulate and clean the supply. A popular technique is to use the output of the switching regulator to drive high power devices that aren't sensitive to noise, such as lights or battery chargers, and have only the sensitive parts run from the LDO to keep heat generation down.
answered 7 hours ago
user
1,084614
The most common solution is to use a switching regulator. Switching regulators dissipate a lot less heat than linear regulators, especially when there is a large voltage drop. The down side is that they produce more noise, but this can be mitigated by following them with a linear regulator to produce a fairly clean supply.
Texas make some good switching regulators and have a free design tool on their site which will produce a schematic, select components and recommend a PCB layout. Alternatively you can buy a module from vendors like Recom and Meanwell, which is literally a black box that sits on the PCB taking 12V in and generating a selectable output voltage.
If your system does not need a low noise supply you could use a 5V output switch more supply. If it does then you could select a 6V output and use a 5V LDO to further regulate and clean the supply. A popular technique is to use the output of the switching regulator to drive high power devices that aren't sensitive to noise, such as lights or battery chargers, and have only the sensitive parts run from the LDO to keep heat generation down.
answered 7 hours ago
user
1,084614
answered 7 hours ago
user
1,084614
answered 7 hours ago
user
1,084614
answered 7 hours ago
user
1,084614
1,084614
add a comment |Â
add a comment |Â
up vote
2
down vote
Your major problems are:-
- The +12V can actually be close to +30V. Car batteries are charged at around +14V, and tow trucks use two batteries in series to deliver a nominal +24V to really crank that starter motor, even if they need to use long jump leads. Your regulator needs to survive continuous overvoltage, or you need to add protection for it.
- The +12V can actually be -12V if the inputs are reversed. (Or -30V, see above.) Can your regulator and electrolytic capacitor handle this? If not, add a diode.
- The +12V can actually be shorted together, or during cranking can drop to substantially less than +12V, discharging any capacitors in front of the regulator, and discharging any capacitors behind the regulator back through the regulator. You may need a diode to stop this, and/or remove your 10uF cap goes.
- The +5V supply may draw more current than the regulator can supply. You may need a low-side fuse.
- The whole circuit may draw more current than the wiring from the +12V can supply. (A car battery can supply a lot, but remember this typically goes through a fuse box and relays.) You may want a high-side fuse, or give an explicit warning that you will rely on the car's own fuse box.
- Your +5V supply will have instantaneous current spikes, and the regulator and 100uF will not be able to keep up. (Electrolytic caps are pretty slow.) Normal practise for this is to parallel an electrolytic with a faster, smaller capacitor (typically the same you'd use for decoupling) to handle the transients.
- The capacitors will give you a huge inrush current. A resistor in the +12V supply is a good idea, and also helps solve some of your shorting-out issues.
Some of your circuit protection issues could be taken care of by using an automotive-spec regulator like an LM2940. Others are still an issue. Similarly, changing to a switch-mode regulator will help you with some problems but not others, and can introduce different problems too (not least noise/ripple on the supply line).
Figuring out solutions to these issues is left as an exercise for the OP, since it may well be coursework. :)
answered 1 hour ago
Graham
2,377411
add a comment |Â
up vote
2
down vote
Your major problems are:-
- The +12V can actually be close to +30V. Car batteries are charged at around +14V, and tow trucks use two batteries in series to deliver a nominal +24V to really crank that starter motor, even if they need to use long jump leads. Your regulator needs to survive continuous overvoltage, or you need to add protection for it.
- The +12V can actually be -12V if the inputs are reversed. (Or -30V, see above.) Can your regulator and electrolytic capacitor handle this? If not, add a diode.
- The +12V can actually be shorted together, or during cranking can drop to substantially less than +12V, discharging any capacitors in front of the regulator, and discharging any capacitors behind the regulator back through the regulator. You may need a diode to stop this, and/or remove your 10uF cap goes.
- The +5V supply may draw more current than the regulator can supply. You may need a low-side fuse.
- The whole circuit may draw more current than the wiring from the +12V can supply. (A car battery can supply a lot, but remember this typically goes through a fuse box and relays.) You may want a high-side fuse, or give an explicit warning that you will rely on the car's own fuse box.
- Your +5V supply will have instantaneous current spikes, and the regulator and 100uF will not be able to keep up. (Electrolytic caps are pretty slow.) Normal practise for this is to parallel an electrolytic with a faster, smaller capacitor (typically the same you'd use for decoupling) to handle the transients.
- The capacitors will give you a huge inrush current. A resistor in the +12V supply is a good idea, and also helps solve some of your shorting-out issues.
Some of your circuit protection issues could be taken care of by using an automotive-spec regulator like an LM2940. Others are still an issue. Similarly, changing to a switch-mode regulator will help you with some problems but not others, and can introduce different problems too (not least noise/ripple on the supply line).
Figuring out solutions to these issues is left as an exercise for the OP, since it may well be coursework. :)
answered 1 hour ago
Graham
2,377411
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Your major problems are:-
- The +12V can actually be close to +30V. Car batteries are charged at around +14V, and tow trucks use two batteries in series to deliver a nominal +24V to really crank that starter motor, even if they need to use long jump leads. Your regulator needs to survive continuous overvoltage, or you need to add protection for it.
- The +12V can actually be -12V if the inputs are reversed. (Or -30V, see above.) Can your regulator and electrolytic capacitor handle this? If not, add a diode.
- The +12V can actually be shorted together, or during cranking can drop to substantially less than +12V, discharging any capacitors in front of the regulator, and discharging any capacitors behind the regulator back through the regulator. You may need a diode to stop this, and/or remove your 10uF cap goes.
- The +5V supply may draw more current than the regulator can supply. You may need a low-side fuse.
- The whole circuit may draw more current than the wiring from the +12V can supply. (A car battery can supply a lot, but remember this typically goes through a fuse box and relays.) You may want a high-side fuse, or give an explicit warning that you will rely on the car's own fuse box.
- Your +5V supply will have instantaneous current spikes, and the regulator and 100uF will not be able to keep up. (Electrolytic caps are pretty slow.) Normal practise for this is to parallel an electrolytic with a faster, smaller capacitor (typically the same you'd use for decoupling) to handle the transients.
- The capacitors will give you a huge inrush current. A resistor in the +12V supply is a good idea, and also helps solve some of your shorting-out issues.
Some of your circuit protection issues could be taken care of by using an automotive-spec regulator like an LM2940. Others are still an issue. Similarly, changing to a switch-mode regulator will help you with some problems but not others, and can introduce different problems too (not least noise/ripple on the supply line).
Figuring out solutions to these issues is left as an exercise for the OP, since it may well be coursework. :)
answered 1 hour ago
Graham
2,377411
Your major problems are:-
- The +12V can actually be close to +30V. Car batteries are charged at around +14V, and tow trucks use two batteries in series to deliver a nominal +24V to really crank that starter motor, even if they need to use long jump leads. Your regulator needs to survive continuous overvoltage, or you need to add protection for it.
- The +12V can actually be -12V if the inputs are reversed. (Or -30V, see above.) Can your regulator and electrolytic capacitor handle this? If not, add a diode.
- The +12V can actually be shorted together, or during cranking can drop to substantially less than +12V, discharging any capacitors in front of the regulator, and discharging any capacitors behind the regulator back through the regulator. You may need a diode to stop this, and/or remove your 10uF cap goes.
- The +5V supply may draw more current than the regulator can supply. You may need a low-side fuse.
- The whole circuit may draw more current than the wiring from the +12V can supply. (A car battery can supply a lot, but remember this typically goes through a fuse box and relays.) You may want a high-side fuse, or give an explicit warning that you will rely on the car's own fuse box.
- Your +5V supply will have instantaneous current spikes, and the regulator and 100uF will not be able to keep up. (Electrolytic caps are pretty slow.) Normal practise for this is to parallel an electrolytic with a faster, smaller capacitor (typically the same you'd use for decoupling) to handle the transients.
- The capacitors will give you a huge inrush current. A resistor in the +12V supply is a good idea, and also helps solve some of your shorting-out issues.
Some of your circuit protection issues could be taken care of by using an automotive-spec regulator like an LM2940. Others are still an issue. Similarly, changing to a switch-mode regulator will help you with some problems but not others, and can introduce different problems too (not least noise/ripple on the supply line).
Figuring out solutions to these issues is left as an exercise for the OP, since it may well be coursework. :)
answered 1 hour ago
Graham
2,377411
answered 1 hour ago
Graham
2,377411
answered 1 hour ago
Graham
2,377411
answered 1 hour ago
Graham
2,377411
2,377411
add a comment |Â
add a comment |Â
up vote
1
down vote
I would use unidirectional TVS at the input so that negative spike goes through with minimal residual voltage.
Also note that TVS has to protect 12V input which can be also 15V in normal conditions. You have choosen 5V TVS, which is wrong. Use a SMA6J18A 18V.
answered 11 hours ago
Marko BurÅ¡iÄÂ
9,4042812
2
Sorry I'm still confused on how TVS diodes operate, using the SMA6J18A with 18V working and clamping at 28.3V to 33.2V that seems to exceed protecting the regulator from higher than needed voltage. Would the TPD1E10B09QDPYRQ1 be better suited?, its working voltage is 9V and clamping is 13V, with the ZLDO1117's SOA curve peak being 10V for 1A that would put me in the SOA during clamping and normal operation right? Datasheet says Vin -Vout on the SOA curve chart, 9-5 = 4V, 13-5 = 8V for a safe 5V 1A output.
– dawm
10 hours ago
1
Or do I completely misunderstand TVS diodes completely? Does current flow past a TVS untouched at the working voltage, then starts getting clamped when it goes over the breakdown voltage? That's how I understood it, that is why I picked the SMA6J5.0CA-TR, I thought it would allow at least 5V through it and start clamping it to 9V-13.4V after it reached the breakdown.
– dawm
10 hours ago
1
@dawm Look at the diagram on page 2 of the datasheet. Because the diode is normally used in reverse direction only the left half of the diagram is of interest. If you're going to use e.g. an SMA6J15A you get V[WM]=15V, where at 15V only 1µA will leak through the diode, and beyond the breakdown voltage V[BR]=18.5V (max.) the device will act like a resistor in the order of 0.1...0.2Ohms.
– JimmyB
6 hours ago
2
You will want to include a fuse in the supply line before the TVS which blows in case the overvoltage is not just spurious or the supply gets reversed to prevent the TVS from creating a short curcuit and burning out.
– JimmyB
6 hours ago
add a comment |Â
up vote
1
down vote
I would use unidirectional TVS at the input so that negative spike goes through with minimal residual voltage.
Also note that TVS has to protect 12V input which can be also 15V in normal conditions. You have choosen 5V TVS, which is wrong. Use a SMA6J18A 18V.
answered 11 hours ago
Marko BurÅ¡iÄÂ
9,4042812
2
Sorry I'm still confused on how TVS diodes operate, using the SMA6J18A with 18V working and clamping at 28.3V to 33.2V that seems to exceed protecting the regulator from higher than needed voltage. Would the TPD1E10B09QDPYRQ1 be better suited?, its working voltage is 9V and clamping is 13V, with the ZLDO1117's SOA curve peak being 10V for 1A that would put me in the SOA during clamping and normal operation right? Datasheet says Vin -Vout on the SOA curve chart, 9-5 = 4V, 13-5 = 8V for a safe 5V 1A output.
– dawm
10 hours ago
1
Or do I completely misunderstand TVS diodes completely? Does current flow past a TVS untouched at the working voltage, then starts getting clamped when it goes over the breakdown voltage? That's how I understood it, that is why I picked the SMA6J5.0CA-TR, I thought it would allow at least 5V through it and start clamping it to 9V-13.4V after it reached the breakdown.
– dawm
10 hours ago
1
@dawm Look at the diagram on page 2 of the datasheet. Because the diode is normally used in reverse direction only the left half of the diagram is of interest. If you're going to use e.g. an SMA6J15A you get V[WM]=15V, where at 15V only 1µA will leak through the diode, and beyond the breakdown voltage V[BR]=18.5V (max.) the device will act like a resistor in the order of 0.1...0.2Ohms.
– JimmyB
6 hours ago
2
You will want to include a fuse in the supply line before the TVS which blows in case the overvoltage is not just spurious or the supply gets reversed to prevent the TVS from creating a short curcuit and burning out.
– JimmyB
6 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I would use unidirectional TVS at the input so that negative spike goes through with minimal residual voltage.
Also note that TVS has to protect 12V input which can be also 15V in normal conditions. You have choosen 5V TVS, which is wrong. Use a SMA6J18A 18V.
answered 11 hours ago
Marko BurÅ¡iÄÂ
9,4042812
I would use unidirectional TVS at the input so that negative spike goes through with minimal residual voltage.
Also note that TVS has to protect 12V input which can be also 15V in normal conditions. You have choosen 5V TVS, which is wrong. Use a SMA6J18A 18V.
answered 11 hours ago
Marko BurÅ¡iÄÂ
9,4042812
answered 11 hours ago
Marko BurÅ¡iÄÂ
9,4042812
answered 11 hours ago
Marko BurÅ¡iÄÂ
9,4042812
answered 11 hours ago
Marko BurÅ¡iÄÂ
9,4042812
9,4042812
2
Sorry I'm still confused on how TVS diodes operate, using the SMA6J18A with 18V working and clamping at 28.3V to 33.2V that seems to exceed protecting the regulator from higher than needed voltage. Would the TPD1E10B09QDPYRQ1 be better suited?, its working voltage is 9V and clamping is 13V, with the ZLDO1117's SOA curve peak being 10V for 1A that would put me in the SOA during clamping and normal operation right? Datasheet says Vin -Vout on the SOA curve chart, 9-5 = 4V, 13-5 = 8V for a safe 5V 1A output.
– dawm
10 hours ago
1
Or do I completely misunderstand TVS diodes completely? Does current flow past a TVS untouched at the working voltage, then starts getting clamped when it goes over the breakdown voltage? That's how I understood it, that is why I picked the SMA6J5.0CA-TR, I thought it would allow at least 5V through it and start clamping it to 9V-13.4V after it reached the breakdown.
– dawm
10 hours ago
1
@dawm Look at the diagram on page 2 of the datasheet. Because the diode is normally used in reverse direction only the left half of the diagram is of interest. If you're going to use e.g. an SMA6J15A you get V[WM]=15V, where at 15V only 1µA will leak through the diode, and beyond the breakdown voltage V[BR]=18.5V (max.) the device will act like a resistor in the order of 0.1...0.2Ohms.
– JimmyB
6 hours ago
2
You will want to include a fuse in the supply line before the TVS which blows in case the overvoltage is not just spurious or the supply gets reversed to prevent the TVS from creating a short curcuit and burning out.
– JimmyB
6 hours ago
add a comment |Â
2
Sorry I'm still confused on how TVS diodes operate, using the SMA6J18A with 18V working and clamping at 28.3V to 33.2V that seems to exceed protecting the regulator from higher than needed voltage. Would the TPD1E10B09QDPYRQ1 be better suited?, its working voltage is 9V and clamping is 13V, with the ZLDO1117's SOA curve peak being 10V for 1A that would put me in the SOA during clamping and normal operation right? Datasheet says Vin -Vout on the SOA curve chart, 9-5 = 4V, 13-5 = 8V for a safe 5V 1A output.
– dawm
10 hours ago
1
Or do I completely misunderstand TVS diodes completely? Does current flow past a TVS untouched at the working voltage, then starts getting clamped when it goes over the breakdown voltage? That's how I understood it, that is why I picked the SMA6J5.0CA-TR, I thought it would allow at least 5V through it and start clamping it to 9V-13.4V after it reached the breakdown.
– dawm
10 hours ago
1
@dawm Look at the diagram on page 2 of the datasheet. Because the diode is normally used in reverse direction only the left half of the diagram is of interest. If you're going to use e.g. an SMA6J15A you get V[WM]=15V, where at 15V only 1µA will leak through the diode, and beyond the breakdown voltage V[BR]=18.5V (max.) the device will act like a resistor in the order of 0.1...0.2Ohms.
– JimmyB
6 hours ago
2
You will want to include a fuse in the supply line before the TVS which blows in case the overvoltage is not just spurious or the supply gets reversed to prevent the TVS from creating a short curcuit and burning out.
– JimmyB
6 hours ago
2
2
Sorry I'm still confused on how TVS diodes operate, using the SMA6J18A with 18V working and clamping at 28.3V to 33.2V that seems to exceed protecting the regulator from higher than needed voltage. Would the TPD1E10B09QDPYRQ1 be better suited?, its working voltage is 9V and clamping is 13V, with the ZLDO1117's SOA curve peak being 10V for 1A that would put me in the SOA during clamping and normal operation right? Datasheet says Vin -Vout on the SOA curve chart, 9-5 = 4V, 13-5 = 8V for a safe 5V 1A output.
– dawm
10 hours ago
Sorry I'm still confused on how TVS diodes operate, using the SMA6J18A with 18V working and clamping at 28.3V to 33.2V that seems to exceed protecting the regulator from higher than needed voltage. Would the TPD1E10B09QDPYRQ1 be better suited?, its working voltage is 9V and clamping is 13V, with the ZLDO1117's SOA curve peak being 10V for 1A that would put me in the SOA during clamping and normal operation right? Datasheet says Vin -Vout on the SOA curve chart, 9-5 = 4V, 13-5 = 8V for a safe 5V 1A output.
– dawm
10 hours ago
1
1
Or do I completely misunderstand TVS diodes completely? Does current flow past a TVS untouched at the working voltage, then starts getting clamped when it goes over the breakdown voltage? That's how I understood it, that is why I picked the SMA6J5.0CA-TR, I thought it would allow at least 5V through it and start clamping it to 9V-13.4V after it reached the breakdown.
– dawm
10 hours ago
Or do I completely misunderstand TVS diodes completely? Does current flow past a TVS untouched at the working voltage, then starts getting clamped when it goes over the breakdown voltage? That's how I understood it, that is why I picked the SMA6J5.0CA-TR, I thought it would allow at least 5V through it and start clamping it to 9V-13.4V after it reached the breakdown.
– dawm
10 hours ago
1
1
@dawm Look at the diagram on page 2 of the datasheet. Because the diode is normally used in reverse direction only the left half of the diagram is of interest. If you're going to use e.g. an SMA6J15A you get V[WM]=15V, where at 15V only 1µA will leak through the diode, and beyond the breakdown voltage V[BR]=18.5V (max.) the device will act like a resistor in the order of 0.1...0.2Ohms.
– JimmyB
6 hours ago
@dawm Look at the diagram on page 2 of the datasheet. Because the diode is normally used in reverse direction only the left half of the diagram is of interest. If you're going to use e.g. an SMA6J15A you get V[WM]=15V, where at 15V only 1µA will leak through the diode, and beyond the breakdown voltage V[BR]=18.5V (max.) the device will act like a resistor in the order of 0.1...0.2Ohms.
– JimmyB
6 hours ago
2
2
You will want to include a fuse in the supply line before the TVS which blows in case the overvoltage is not just spurious or the supply gets reversed to prevent the TVS from creating a short curcuit and burning out.
– JimmyB
6 hours ago
You will want to include a fuse in the supply line before the TVS which blows in case the overvoltage is not just spurious or the supply gets reversed to prevent the TVS from creating a short curcuit and burning out.
– JimmyB
6 hours ago
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f399929%2fopinions-on-what-ive-come-up-with-for-a-12-v-to-5-v-automotive-circuit%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
It would be better if you specified the expected current draw. From your text circuit get about (5-1.2)/95 = ~40mA. (B.T.W Your text says 150 ohm, your diagram says 95 ohm...)
– Oldfart
10 hours ago
2
There are "cigarette lighter" adapters available with selectable voltage outputs...
– Solar Mike
10 hours ago
1
@Oldfart eek! I changed LEDs and guess I missed updating the schematic, the schematic should be 150ohm just as the component list. The circuit on the other end of the regulated 5v 1A shouldn't be more than 0.8A
– dawm
10 hours ago
3
They come in a complete package - not "multiple boards" and tend to be very compact and efficient providing a clean supply, and I found the one I bought has been perfect for the multiple uses I have given it.
– Solar Mike
10 hours ago
2
@dawm I will let you do it your way - however, as I was a vehicle electrician in a previous life I would be looking at various options personally. You will need some cabling somewhere...
– Solar Mike
10 hours ago