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Purpose of this oscillator (74HCT14-based) on RS-232 to RS-485 converter?
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I am reverse engineering an RS-232 to RS-485 converter to understand how it works. I have attached the schematic that I reverse-engineered from the PCB (apologies for the hand-drawn schematic). I've checked it several times and quite certain that it is correct. The part that confuses me is the oscillator circled in red.
I recognize it as an oscillator based based around C2 & R8, and on my scope I can see it running at about 10kHz. I also see that the five parallel inverters on it's output are acting as some sort of buffer. However, what the heck is it there for?
This converter will work completely powered by RS-232 port, even if just the RX, TX, and ground lines (pins 2, 3, & 5 of the DE9) are connected. It does NOT require the +5~12V line to be connected to an external power supply.
The only theory that I have is that the oscillator is acting as some sort of charge pump to reduce the source impedances of Vcc & -Vcc (which otherwise would be limited by the source impedances of the DE9 pins, plus 4.7k for -Vcc via RN1d), but that's just a guess and I don't really see how/why that would work in practice.
So, any pointers on why that oscillator is there and how it achieves whatever it does would be appreciated
converter rs232 rs485
edited 1 hour ago
Sparky256
9,64221334
asked 2 hours ago
Chris
162
add a comment |Â
up vote
3
down vote
favorite
I am reverse engineering an RS-232 to RS-485 converter to understand how it works. I have attached the schematic that I reverse-engineered from the PCB (apologies for the hand-drawn schematic). I've checked it several times and quite certain that it is correct. The part that confuses me is the oscillator circled in red.
I recognize it as an oscillator based based around C2 & R8, and on my scope I can see it running at about 10kHz. I also see that the five parallel inverters on it's output are acting as some sort of buffer. However, what the heck is it there for?
This converter will work completely powered by RS-232 port, even if just the RX, TX, and ground lines (pins 2, 3, & 5 of the DE9) are connected. It does NOT require the +5~12V line to be connected to an external power supply.
The only theory that I have is that the oscillator is acting as some sort of charge pump to reduce the source impedances of Vcc & -Vcc (which otherwise would be limited by the source impedances of the DE9 pins, plus 4.7k for -Vcc via RN1d), but that's just a guess and I don't really see how/why that would work in practice.
So, any pointers on why that oscillator is there and how it achieves whatever it does would be appreciated
converter rs232 rs485
edited 1 hour ago
Sparky256
9,64221334
asked 2 hours ago
Chris
162
Are you sure that nothing else is connected to the CT3 / D6 node?
– duskwuff
1 hour ago
This circuit is a wee bit confusing. You have 2 sources of Vcc. You have a zener diode type power supply and the 74HCT is also being used to create another Vcc. It would help if all voltage tags had a voltage value so we could see what is going on.
– Sparky256
1 hour ago
2
See how the 74HCT14 positive supply (pin 14) is ground, and the negative supply (pin 7) is the negative supply rail? The outlined section is indeed a charge pump which will generate a positive supply rail from just the negative supply. Not being familiar with old school rs232 I'm unsure why exactly this is needed, but I do know that the signalling levels swing to both positive and negative voltages.
– jms
1 hour ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I am reverse engineering an RS-232 to RS-485 converter to understand how it works. I have attached the schematic that I reverse-engineered from the PCB (apologies for the hand-drawn schematic). I've checked it several times and quite certain that it is correct. The part that confuses me is the oscillator circled in red.
I recognize it as an oscillator based based around C2 & R8, and on my scope I can see it running at about 10kHz. I also see that the five parallel inverters on it's output are acting as some sort of buffer. However, what the heck is it there for?
This converter will work completely powered by RS-232 port, even if just the RX, TX, and ground lines (pins 2, 3, & 5 of the DE9) are connected. It does NOT require the +5~12V line to be connected to an external power supply.
The only theory that I have is that the oscillator is acting as some sort of charge pump to reduce the source impedances of Vcc & -Vcc (which otherwise would be limited by the source impedances of the DE9 pins, plus 4.7k for -Vcc via RN1d), but that's just a guess and I don't really see how/why that would work in practice.
So, any pointers on why that oscillator is there and how it achieves whatever it does would be appreciated
converter rs232 rs485
edited 1 hour ago
Sparky256
9,64221334
asked 2 hours ago
Chris
162
I am reverse engineering an RS-232 to RS-485 converter to understand how it works. I have attached the schematic that I reverse-engineered from the PCB (apologies for the hand-drawn schematic). I've checked it several times and quite certain that it is correct. The part that confuses me is the oscillator circled in red.
I recognize it as an oscillator based based around C2 & R8, and on my scope I can see it running at about 10kHz. I also see that the five parallel inverters on it's output are acting as some sort of buffer. However, what the heck is it there for?
This converter will work completely powered by RS-232 port, even if just the RX, TX, and ground lines (pins 2, 3, & 5 of the DE9) are connected. It does NOT require the +5~12V line to be connected to an external power supply.
The only theory that I have is that the oscillator is acting as some sort of charge pump to reduce the source impedances of Vcc & -Vcc (which otherwise would be limited by the source impedances of the DE9 pins, plus 4.7k for -Vcc via RN1d), but that's just a guess and I don't really see how/why that would work in practice.
So, any pointers on why that oscillator is there and how it achieves whatever it does would be appreciated
converter rs232 rs485
converter rs232 rs485
edited 1 hour ago
Sparky256
9,64221334
asked 2 hours ago
Chris
162
edited 1 hour ago
Sparky256
9,64221334
asked 2 hours ago
Chris
162
edited 1 hour ago
Sparky256
9,64221334
edited 1 hour ago
Sparky256
9,64221334
edited 1 hour ago
Sparky256
9,64221334
9,64221334
asked 2 hours ago
Chris
162
asked 2 hours ago
Chris
162
asked 2 hours ago
Chris
162
162
Are you sure that nothing else is connected to the CT3 / D6 node?
– duskwuff
1 hour ago
This circuit is a wee bit confusing. You have 2 sources of Vcc. You have a zener diode type power supply and the 74HCT is also being used to create another Vcc. It would help if all voltage tags had a voltage value so we could see what is going on.
– Sparky256
1 hour ago
2
See how the 74HCT14 positive supply (pin 14) is ground, and the negative supply (pin 7) is the negative supply rail? The outlined section is indeed a charge pump which will generate a positive supply rail from just the negative supply. Not being familiar with old school rs232 I'm unsure why exactly this is needed, but I do know that the signalling levels swing to both positive and negative voltages.
– jms
1 hour ago
add a comment |Â
Are you sure that nothing else is connected to the CT3 / D6 node?
– duskwuff
1 hour ago
This circuit is a wee bit confusing. You have 2 sources of Vcc. You have a zener diode type power supply and the 74HCT is also being used to create another Vcc. It would help if all voltage tags had a voltage value so we could see what is going on.
– Sparky256
1 hour ago
2
See how the 74HCT14 positive supply (pin 14) is ground, and the negative supply (pin 7) is the negative supply rail? The outlined section is indeed a charge pump which will generate a positive supply rail from just the negative supply. Not being familiar with old school rs232 I'm unsure why exactly this is needed, but I do know that the signalling levels swing to both positive and negative voltages.
– jms
1 hour ago
Are you sure that nothing else is connected to the CT3 / D6 node?
– duskwuff
1 hour ago
Are you sure that nothing else is connected to the CT3 / D6 node?
– duskwuff
1 hour ago
This circuit is a wee bit confusing. You have 2 sources of Vcc. You have a zener diode type power supply and the 74HCT is also being used to create another Vcc. It would help if all voltage tags had a voltage value so we could see what is going on.
– Sparky256
1 hour ago
This circuit is a wee bit confusing. You have 2 sources of Vcc. You have a zener diode type power supply and the 74HCT is also being used to create another Vcc. It would help if all voltage tags had a voltage value so we could see what is going on.
– Sparky256
1 hour ago
2
2
See how the 74HCT14 positive supply (pin 14) is ground, and the negative supply (pin 7) is the negative supply rail? The outlined section is indeed a charge pump which will generate a positive supply rail from just the negative supply. Not being familiar with old school rs232 I'm unsure why exactly this is needed, but I do know that the signalling levels swing to both positive and negative voltages.
– jms
1 hour ago
See how the 74HCT14 positive supply (pin 14) is ground, and the negative supply (pin 7) is the negative supply rail? The outlined section is indeed a charge pump which will generate a positive supply rail from just the negative supply. Not being familiar with old school rs232 I'm unsure why exactly this is needed, but I do know that the signalling levels swing to both positive and negative voltages.
– jms
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
It's to generate a negative rail.
The oscillator/buffer drives the right side of CT3 between +Vcc and 0V (ignoring diode drops) so the left side goes 0V when high (ground on the buffer is tied to the pin 14 on the chip) and -Vcc when low.
Startup does not look like it's a sure thing with this arrangement especially if power-up is not 'clean', I'd be a lot more comfortable with a 7660 or one of the more modern charge pump chips but this is cheap. That's because the chip essentially generates it's own power supply so it needs to see a couple volts for a bit of time at startup to get going. Maybe I'm missing something on startup, that's from a quick look.
answered 1 hour ago
Spehro Pefhany
194k4139384
add a comment |Â
up vote
1
down vote
Two members claim opposite things, one in a comment and one in an answer. User jms has got it right. The oscillator generates +Vcc if -Vcc happens to be available in the connected RS232 port. The diodes connect possibly available -Vcc to the oscillator IC. The oscillator charges CT2 from -Vcc and the charge is pushed to maintain +Vcc. That pumping is really needed, if -Vcc happens to be the only available voltage via the RS232 port.
If the port happens to output (through a diode) +Vcc, it's used directly and the voltage minus Vcc is not needed altough it does not make any harm.
The design assumes that at least one of voltages -Vcc and +Vcc is available from the RS232 port. RS485 circuit needs only the positive one.
answered 36 mins ago
user287001
8,0051415
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
It's to generate a negative rail.
The oscillator/buffer drives the right side of CT3 between +Vcc and 0V (ignoring diode drops) so the left side goes 0V when high (ground on the buffer is tied to the pin 14 on the chip) and -Vcc when low.
Startup does not look like it's a sure thing with this arrangement especially if power-up is not 'clean', I'd be a lot more comfortable with a 7660 or one of the more modern charge pump chips but this is cheap. That's because the chip essentially generates it's own power supply so it needs to see a couple volts for a bit of time at startup to get going. Maybe I'm missing something on startup, that's from a quick look.
answered 1 hour ago
Spehro Pefhany
194k4139384
add a comment |Â
up vote
2
down vote
It's to generate a negative rail.
The oscillator/buffer drives the right side of CT3 between +Vcc and 0V (ignoring diode drops) so the left side goes 0V when high (ground on the buffer is tied to the pin 14 on the chip) and -Vcc when low.
Startup does not look like it's a sure thing with this arrangement especially if power-up is not 'clean', I'd be a lot more comfortable with a 7660 or one of the more modern charge pump chips but this is cheap. That's because the chip essentially generates it's own power supply so it needs to see a couple volts for a bit of time at startup to get going. Maybe I'm missing something on startup, that's from a quick look.
answered 1 hour ago
Spehro Pefhany
194k4139384
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It's to generate a negative rail.
The oscillator/buffer drives the right side of CT3 between +Vcc and 0V (ignoring diode drops) so the left side goes 0V when high (ground on the buffer is tied to the pin 14 on the chip) and -Vcc when low.
Startup does not look like it's a sure thing with this arrangement especially if power-up is not 'clean', I'd be a lot more comfortable with a 7660 or one of the more modern charge pump chips but this is cheap. That's because the chip essentially generates it's own power supply so it needs to see a couple volts for a bit of time at startup to get going. Maybe I'm missing something on startup, that's from a quick look.
answered 1 hour ago
Spehro Pefhany
194k4139384
It's to generate a negative rail.
The oscillator/buffer drives the right side of CT3 between +Vcc and 0V (ignoring diode drops) so the left side goes 0V when high (ground on the buffer is tied to the pin 14 on the chip) and -Vcc when low.
Startup does not look like it's a sure thing with this arrangement especially if power-up is not 'clean', I'd be a lot more comfortable with a 7660 or one of the more modern charge pump chips but this is cheap. That's because the chip essentially generates it's own power supply so it needs to see a couple volts for a bit of time at startup to get going. Maybe I'm missing something on startup, that's from a quick look.
answered 1 hour ago
Spehro Pefhany
194k4139384
answered 1 hour ago
Spehro Pefhany
194k4139384
answered 1 hour ago
Spehro Pefhany
194k4139384
answered 1 hour ago
Spehro Pefhany
194k4139384
194k4139384
add a comment |Â
add a comment |Â
up vote
1
down vote
Two members claim opposite things, one in a comment and one in an answer. User jms has got it right. The oscillator generates +Vcc if -Vcc happens to be available in the connected RS232 port. The diodes connect possibly available -Vcc to the oscillator IC. The oscillator charges CT2 from -Vcc and the charge is pushed to maintain +Vcc. That pumping is really needed, if -Vcc happens to be the only available voltage via the RS232 port.
If the port happens to output (through a diode) +Vcc, it's used directly and the voltage minus Vcc is not needed altough it does not make any harm.
The design assumes that at least one of voltages -Vcc and +Vcc is available from the RS232 port. RS485 circuit needs only the positive one.
answered 36 mins ago
user287001
8,0051415
add a comment |Â
up vote
1
down vote
Two members claim opposite things, one in a comment and one in an answer. User jms has got it right. The oscillator generates +Vcc if -Vcc happens to be available in the connected RS232 port. The diodes connect possibly available -Vcc to the oscillator IC. The oscillator charges CT2 from -Vcc and the charge is pushed to maintain +Vcc. That pumping is really needed, if -Vcc happens to be the only available voltage via the RS232 port.
If the port happens to output (through a diode) +Vcc, it's used directly and the voltage minus Vcc is not needed altough it does not make any harm.
The design assumes that at least one of voltages -Vcc and +Vcc is available from the RS232 port. RS485 circuit needs only the positive one.
answered 36 mins ago
user287001
8,0051415
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Two members claim opposite things, one in a comment and one in an answer. User jms has got it right. The oscillator generates +Vcc if -Vcc happens to be available in the connected RS232 port. The diodes connect possibly available -Vcc to the oscillator IC. The oscillator charges CT2 from -Vcc and the charge is pushed to maintain +Vcc. That pumping is really needed, if -Vcc happens to be the only available voltage via the RS232 port.
If the port happens to output (through a diode) +Vcc, it's used directly and the voltage minus Vcc is not needed altough it does not make any harm.
The design assumes that at least one of voltages -Vcc and +Vcc is available from the RS232 port. RS485 circuit needs only the positive one.
answered 36 mins ago
user287001
8,0051415
Two members claim opposite things, one in a comment and one in an answer. User jms has got it right. The oscillator generates +Vcc if -Vcc happens to be available in the connected RS232 port. The diodes connect possibly available -Vcc to the oscillator IC. The oscillator charges CT2 from -Vcc and the charge is pushed to maintain +Vcc. That pumping is really needed, if -Vcc happens to be the only available voltage via the RS232 port.
If the port happens to output (through a diode) +Vcc, it's used directly and the voltage minus Vcc is not needed altough it does not make any harm.
The design assumes that at least one of voltages -Vcc and +Vcc is available from the RS232 port. RS485 circuit needs only the positive one.
answered 36 mins ago
user287001
8,0051415
edited 31 mins ago
answered 36 mins ago
user287001
8,0051415
answered 36 mins ago
user287001
8,0051415
answered 36 mins ago
user287001
8,0051415
8,0051415
add a comment |Â
add a comment |Â
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Are you sure that nothing else is connected to the CT3 / D6 node?
– duskwuff
1 hour ago
This circuit is a wee bit confusing. You have 2 sources of Vcc. You have a zener diode type power supply and the 74HCT is also being used to create another Vcc. It would help if all voltage tags had a voltage value so we could see what is going on.
– Sparky256
1 hour ago
2
See how the 74HCT14 positive supply (pin 14) is ground, and the negative supply (pin 7) is the negative supply rail? The outlined section is indeed a charge pump which will generate a positive supply rail from just the negative supply. Not being familiar with old school rs232 I'm unsure why exactly this is needed, but I do know that the signalling levels swing to both positive and negative voltages.
– jms
1 hour ago