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Having trouble in GPIO output, 220 V to 5 V signal
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I'm not an electric guy, I'm just trying to build a small circuit to sense 220 V with my Microcontroller (RPi). But I'm getting either 5 V or 1.3 V on a GPIO wire.
I can't understand the problem. I want 0 V or 5 V on GPIO.
sensor ac gpio 5v
edited 1 hour ago
Michel Keijzers
4,75062149
asked 3 hours ago
Ghazanfar Javed
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Ghazanfar Javed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
2
down vote
favorite
I'm not an electric guy, I'm just trying to build a small circuit to sense 220 V with my Microcontroller (RPi). But I'm getting either 5 V or 1.3 V on a GPIO wire.
I can't understand the problem. I want 0 V or 5 V on GPIO.
sensor ac gpio 5v
edited 1 hour ago
Michel Keijzers
4,75062149
asked 3 hours ago
Ghazanfar Javed
111
New contributor
Ghazanfar Javed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The leakage current out of your GPIO pin in input mode has to be conducted to 0 V by the pull-down resistor. It's often something like 3..10 uA, so the voltage drop it causes across the 370K may well cause your 1.3 V. Change it to 10K and try again.
– TonyM
2 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm not an electric guy, I'm just trying to build a small circuit to sense 220 V with my Microcontroller (RPi). But I'm getting either 5 V or 1.3 V on a GPIO wire.
I can't understand the problem. I want 0 V or 5 V on GPIO.
sensor ac gpio 5v
edited 1 hour ago
Michel Keijzers
4,75062149
asked 3 hours ago
Ghazanfar Javed
111
New contributor
Ghazanfar Javed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm not an electric guy, I'm just trying to build a small circuit to sense 220 V with my Microcontroller (RPi). But I'm getting either 5 V or 1.3 V on a GPIO wire.
I can't understand the problem. I want 0 V or 5 V on GPIO.
sensor ac gpio 5v
sensor ac gpio 5v
edited 1 hour ago
Michel Keijzers
4,75062149
asked 3 hours ago
Ghazanfar Javed
111
New contributor
Ghazanfar Javed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Michel Keijzers
4,75062149
asked 3 hours ago
Ghazanfar Javed
111
New contributor
Ghazanfar Javed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Michel Keijzers
4,75062149
edited 1 hour ago
Michel Keijzers
4,75062149
edited 1 hour ago
Michel Keijzers
4,75062149
4,75062149
asked 3 hours ago
Ghazanfar Javed
111
New contributor
Ghazanfar Javed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Ghazanfar Javed
111
asked 3 hours ago
Ghazanfar Javed
111
111
New contributor
Ghazanfar Javed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ghazanfar Javed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ghazanfar Javed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The leakage current out of your GPIO pin in input mode has to be conducted to 0 V by the pull-down resistor. It's often something like 3..10 uA, so the voltage drop it causes across the 370K may well cause your 1.3 V. Change it to 10K and try again.
– TonyM
2 hours ago
add a comment |Â
The leakage current out of your GPIO pin in input mode has to be conducted to 0 V by the pull-down resistor. It's often something like 3..10 uA, so the voltage drop it causes across the 370K may well cause your 1.3 V. Change it to 10K and try again.
– TonyM
2 hours ago
The leakage current out of your GPIO pin in input mode has to be conducted to 0 V by the pull-down resistor. It's often something like 3..10 uA, so the voltage drop it causes across the 370K may well cause your 1.3 V. Change it to 10K and try again.
– TonyM
2 hours ago
The leakage current out of your GPIO pin in input mode has to be conducted to 0 V by the pull-down resistor. It's often something like 3..10 uA, so the voltage drop it causes across the 370K may well cause your 1.3 V. Change it to 10K and try again.
– TonyM
2 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
5
down vote
Here is using your basic idea, but a circuit that will actually work:
Note the addition of D2. That's important to keep the LED from frying. Yes, D1 does block current in the reverse direction, but it will also have some inevitable leakage. The leakage current should be small (see the datasheet), but that could be enough to build up enough voltage across the LED, also in reverse then, to exceed the LED's reverse voltage spec. D2 provides a safe shunt path for whatever leakage current there may be thru D1 during the negative half-cycle of V1.
Unlike another answer, I recommend keeping D1. With D2 across the LED, D1 can actually leak up to being a short without harm to the LED or D2. You could therefore leave it off. However, it serves a useful purpose in reducing the power dissipation. The diode drops are small compared to 220 V, so you basically have 220 V across a 200 kΩ resistor. That comes out to ¼ W. With D1 in there, that is reduced by half, since current is only flowing during half the V1 voltage cycle. Now you can safely use a ¼ W resistor and still allow for some insulation, packaging, limited air flow, etc, around it. Or you could lower R1 to get more LED current, and thereby require less current transfer ratio of the opto.
On the digital side, all you need is a pullup resistor. But, you need to check the current capability all along the way. The peak voltage of a 220 V sine is 311 V. Let's say 309 V across R1 due to the voltage drop of the LED. That results in 1.5 mA. Of course you don't just want a short blip at the peak of the positive power cycles, so let's say anything over 1.0 mA into the LED should assert the output.
The R2 pullup resistor requires 500 µA for the GPIO line to be held low. That means this circuit would work with a opto-coupler with a CTR (current transfer ratio) of at least ½. Optos with a minimum CTR of 1 are easy to find, especially when fast response is not required, like in this case. You therefore have a fairly comfortable margin of 2x with a 1:1 opto-coupler. That all sounds good.
answered 1 hour ago
Olin Lathrop
276k28329775
add a comment |Â
up vote
2
down vote
Try it like this:
There are two main differences:
There's a diode in antiparallel to the optoisolator's LED. This is better reverse polarity protection than the series diode you used.
The output of the optoisolator pulls the GPIO to ground. This gets you a clear high/low. It will provide a low to the GPIO when AC is present. Because of R2 and C3, it will immediately go low when AC is applied, but will take a bit of time to go high once the AC is gone.
answered 2 hours ago
JRE
18.2k43461
add a comment |Â
up vote
0
down vote
Remove what you have on the output side of U3 at the moment. Connect pin 3 of U3 to ground and pull up pin 4 to VCC with something like a 100k resistor. Add a capacitor between pin 4 and VCC if the output needs smoothing.
You are trying to use npn transistors as high side switches. This is not something you should try to do as a novice.
answered 2 hours ago
RoyC
4,91431431
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Here is using your basic idea, but a circuit that will actually work:
Note the addition of D2. That's important to keep the LED from frying. Yes, D1 does block current in the reverse direction, but it will also have some inevitable leakage. The leakage current should be small (see the datasheet), but that could be enough to build up enough voltage across the LED, also in reverse then, to exceed the LED's reverse voltage spec. D2 provides a safe shunt path for whatever leakage current there may be thru D1 during the negative half-cycle of V1.
Unlike another answer, I recommend keeping D1. With D2 across the LED, D1 can actually leak up to being a short without harm to the LED or D2. You could therefore leave it off. However, it serves a useful purpose in reducing the power dissipation. The diode drops are small compared to 220 V, so you basically have 220 V across a 200 kΩ resistor. That comes out to ¼ W. With D1 in there, that is reduced by half, since current is only flowing during half the V1 voltage cycle. Now you can safely use a ¼ W resistor and still allow for some insulation, packaging, limited air flow, etc, around it. Or you could lower R1 to get more LED current, and thereby require less current transfer ratio of the opto.
On the digital side, all you need is a pullup resistor. But, you need to check the current capability all along the way. The peak voltage of a 220 V sine is 311 V. Let's say 309 V across R1 due to the voltage drop of the LED. That results in 1.5 mA. Of course you don't just want a short blip at the peak of the positive power cycles, so let's say anything over 1.0 mA into the LED should assert the output.
The R2 pullup resistor requires 500 µA for the GPIO line to be held low. That means this circuit would work with a opto-coupler with a CTR (current transfer ratio) of at least ½. Optos with a minimum CTR of 1 are easy to find, especially when fast response is not required, like in this case. You therefore have a fairly comfortable margin of 2x with a 1:1 opto-coupler. That all sounds good.
answered 1 hour ago
Olin Lathrop
276k28329775
add a comment |Â
up vote
5
down vote
Here is using your basic idea, but a circuit that will actually work:
Note the addition of D2. That's important to keep the LED from frying. Yes, D1 does block current in the reverse direction, but it will also have some inevitable leakage. The leakage current should be small (see the datasheet), but that could be enough to build up enough voltage across the LED, also in reverse then, to exceed the LED's reverse voltage spec. D2 provides a safe shunt path for whatever leakage current there may be thru D1 during the negative half-cycle of V1.
Unlike another answer, I recommend keeping D1. With D2 across the LED, D1 can actually leak up to being a short without harm to the LED or D2. You could therefore leave it off. However, it serves a useful purpose in reducing the power dissipation. The diode drops are small compared to 220 V, so you basically have 220 V across a 200 kΩ resistor. That comes out to ¼ W. With D1 in there, that is reduced by half, since current is only flowing during half the V1 voltage cycle. Now you can safely use a ¼ W resistor and still allow for some insulation, packaging, limited air flow, etc, around it. Or you could lower R1 to get more LED current, and thereby require less current transfer ratio of the opto.
On the digital side, all you need is a pullup resistor. But, you need to check the current capability all along the way. The peak voltage of a 220 V sine is 311 V. Let's say 309 V across R1 due to the voltage drop of the LED. That results in 1.5 mA. Of course you don't just want a short blip at the peak of the positive power cycles, so let's say anything over 1.0 mA into the LED should assert the output.
The R2 pullup resistor requires 500 µA for the GPIO line to be held low. That means this circuit would work with a opto-coupler with a CTR (current transfer ratio) of at least ½. Optos with a minimum CTR of 1 are easy to find, especially when fast response is not required, like in this case. You therefore have a fairly comfortable margin of 2x with a 1:1 opto-coupler. That all sounds good.
answered 1 hour ago
Olin Lathrop
276k28329775
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Here is using your basic idea, but a circuit that will actually work:
Note the addition of D2. That's important to keep the LED from frying. Yes, D1 does block current in the reverse direction, but it will also have some inevitable leakage. The leakage current should be small (see the datasheet), but that could be enough to build up enough voltage across the LED, also in reverse then, to exceed the LED's reverse voltage spec. D2 provides a safe shunt path for whatever leakage current there may be thru D1 during the negative half-cycle of V1.
Unlike another answer, I recommend keeping D1. With D2 across the LED, D1 can actually leak up to being a short without harm to the LED or D2. You could therefore leave it off. However, it serves a useful purpose in reducing the power dissipation. The diode drops are small compared to 220 V, so you basically have 220 V across a 200 kΩ resistor. That comes out to ¼ W. With D1 in there, that is reduced by half, since current is only flowing during half the V1 voltage cycle. Now you can safely use a ¼ W resistor and still allow for some insulation, packaging, limited air flow, etc, around it. Or you could lower R1 to get more LED current, and thereby require less current transfer ratio of the opto.
On the digital side, all you need is a pullup resistor. But, you need to check the current capability all along the way. The peak voltage of a 220 V sine is 311 V. Let's say 309 V across R1 due to the voltage drop of the LED. That results in 1.5 mA. Of course you don't just want a short blip at the peak of the positive power cycles, so let's say anything over 1.0 mA into the LED should assert the output.
The R2 pullup resistor requires 500 µA for the GPIO line to be held low. That means this circuit would work with a opto-coupler with a CTR (current transfer ratio) of at least ½. Optos with a minimum CTR of 1 are easy to find, especially when fast response is not required, like in this case. You therefore have a fairly comfortable margin of 2x with a 1:1 opto-coupler. That all sounds good.
answered 1 hour ago
Olin Lathrop
276k28329775
Here is using your basic idea, but a circuit that will actually work:
Note the addition of D2. That's important to keep the LED from frying. Yes, D1 does block current in the reverse direction, but it will also have some inevitable leakage. The leakage current should be small (see the datasheet), but that could be enough to build up enough voltage across the LED, also in reverse then, to exceed the LED's reverse voltage spec. D2 provides a safe shunt path for whatever leakage current there may be thru D1 during the negative half-cycle of V1.
Unlike another answer, I recommend keeping D1. With D2 across the LED, D1 can actually leak up to being a short without harm to the LED or D2. You could therefore leave it off. However, it serves a useful purpose in reducing the power dissipation. The diode drops are small compared to 220 V, so you basically have 220 V across a 200 kΩ resistor. That comes out to ¼ W. With D1 in there, that is reduced by half, since current is only flowing during half the V1 voltage cycle. Now you can safely use a ¼ W resistor and still allow for some insulation, packaging, limited air flow, etc, around it. Or you could lower R1 to get more LED current, and thereby require less current transfer ratio of the opto.
On the digital side, all you need is a pullup resistor. But, you need to check the current capability all along the way. The peak voltage of a 220 V sine is 311 V. Let's say 309 V across R1 due to the voltage drop of the LED. That results in 1.5 mA. Of course you don't just want a short blip at the peak of the positive power cycles, so let's say anything over 1.0 mA into the LED should assert the output.
The R2 pullup resistor requires 500 µA for the GPIO line to be held low. That means this circuit would work with a opto-coupler with a CTR (current transfer ratio) of at least ½. Optos with a minimum CTR of 1 are easy to find, especially when fast response is not required, like in this case. You therefore have a fairly comfortable margin of 2x with a 1:1 opto-coupler. That all sounds good.
answered 1 hour ago
Olin Lathrop
276k28329775
answered 1 hour ago
Olin Lathrop
276k28329775
answered 1 hour ago
Olin Lathrop
276k28329775
answered 1 hour ago
Olin Lathrop
276k28329775
276k28329775
add a comment |Â
add a comment |Â
up vote
2
down vote
Try it like this:
There are two main differences:
There's a diode in antiparallel to the optoisolator's LED. This is better reverse polarity protection than the series diode you used.
The output of the optoisolator pulls the GPIO to ground. This gets you a clear high/low. It will provide a low to the GPIO when AC is present. Because of R2 and C3, it will immediately go low when AC is applied, but will take a bit of time to go high once the AC is gone.
answered 2 hours ago
JRE
18.2k43461
add a comment |Â
up vote
2
down vote
Try it like this:
There are two main differences:
There's a diode in antiparallel to the optoisolator's LED. This is better reverse polarity protection than the series diode you used.
The output of the optoisolator pulls the GPIO to ground. This gets you a clear high/low. It will provide a low to the GPIO when AC is present. Because of R2 and C3, it will immediately go low when AC is applied, but will take a bit of time to go high once the AC is gone.
answered 2 hours ago
JRE
18.2k43461
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Try it like this:
There are two main differences:
There's a diode in antiparallel to the optoisolator's LED. This is better reverse polarity protection than the series diode you used.
The output of the optoisolator pulls the GPIO to ground. This gets you a clear high/low. It will provide a low to the GPIO when AC is present. Because of R2 and C3, it will immediately go low when AC is applied, but will take a bit of time to go high once the AC is gone.
answered 2 hours ago
JRE
18.2k43461
Try it like this:
There are two main differences:
There's a diode in antiparallel to the optoisolator's LED. This is better reverse polarity protection than the series diode you used.
The output of the optoisolator pulls the GPIO to ground. This gets you a clear high/low. It will provide a low to the GPIO when AC is present. Because of R2 and C3, it will immediately go low when AC is applied, but will take a bit of time to go high once the AC is gone.
answered 2 hours ago
JRE
18.2k43461
edited 1 hour ago
answered 2 hours ago
JRE
18.2k43461
answered 2 hours ago
JRE
18.2k43461
answered 2 hours ago
JRE
18.2k43461
18.2k43461
add a comment |Â
add a comment |Â
up vote
0
down vote
Remove what you have on the output side of U3 at the moment. Connect pin 3 of U3 to ground and pull up pin 4 to VCC with something like a 100k resistor. Add a capacitor between pin 4 and VCC if the output needs smoothing.
You are trying to use npn transistors as high side switches. This is not something you should try to do as a novice.
answered 2 hours ago
RoyC
4,91431431
add a comment |Â
up vote
0
down vote
Remove what you have on the output side of U3 at the moment. Connect pin 3 of U3 to ground and pull up pin 4 to VCC with something like a 100k resistor. Add a capacitor between pin 4 and VCC if the output needs smoothing.
You are trying to use npn transistors as high side switches. This is not something you should try to do as a novice.
answered 2 hours ago
RoyC
4,91431431
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Remove what you have on the output side of U3 at the moment. Connect pin 3 of U3 to ground and pull up pin 4 to VCC with something like a 100k resistor. Add a capacitor between pin 4 and VCC if the output needs smoothing.
You are trying to use npn transistors as high side switches. This is not something you should try to do as a novice.
answered 2 hours ago
RoyC
4,91431431
Remove what you have on the output side of U3 at the moment. Connect pin 3 of U3 to ground and pull up pin 4 to VCC with something like a 100k resistor. Add a capacitor between pin 4 and VCC if the output needs smoothing.
You are trying to use npn transistors as high side switches. This is not something you should try to do as a novice.
answered 2 hours ago
RoyC
4,91431431
answered 2 hours ago
RoyC
4,91431431
answered 2 hours ago
RoyC
4,91431431
answered 2 hours ago
RoyC
4,91431431
4,91431431
add a comment |Â
add a comment |Â
Ghazanfar Javed is a new contributor. Be nice, and check out our Code of Conduct.
Ghazanfar Javed is a new contributor. Be nice, and check out our Code of Conduct.
Ghazanfar Javed is a new contributor. Be nice, and check out our Code of Conduct.
Ghazanfar Javed is a new contributor. Be nice, and check out our Code of Conduct.
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The leakage current out of your GPIO pin in input mode has to be conducted to 0 V by the pull-down resistor. It's often something like 3..10 uA, so the voltage drop it causes across the 370K may well cause your 1.3 V. Change it to 10K and try again.
– TonyM
2 hours ago