I am designing a current pulse generator, and need to come up with a way to indicate to the CPU when a 280uF capacitor bank is charged to 200V. The design involves switching the capacitor bank between the 200V supply (charging) and the output circuit (isolated). I am not aware of too many op-amps that allow that high of a differential voltage, and using a voltage divider will drain the charge I am trying to measure. I considered using a current sense resistor in the charging circuit and indicating "ready" when the current approaches zero, but this would not be fool-proof in the case of a power supply failure, and it would not indicate when the capacitor bank has leaked charge when in the output mode. Now I am trying to think of a clever way to use a FET without exceeding Vgs max. I would appreciate any suggestions.
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capacitor charge indicator 3
- Thread starter DJam
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- #3
If by accuracy you mean the threshold of the indicator, I would specify it at about 95% (190V). The signal will simply be a logic "high" or "low" to the CPU. The purpose of the design is to provide a pulse of specified current and pulse width, in this case 40A for up to 500us, to a low (0.5 - 5)ohm load. The capacitor bank is charged through the 200V supply, then switched via a relay to the isolated output circuit, which controls the current and pulse width.
I'm not sure what 'VTG' means. The 200V comes from a 300Vdc nominal AC-DC front end module. I use a crude zener diode regulator to get the 200V.
I'm not sure what 'VTG' means. The 200V comes from a 300Vdc nominal AC-DC front end module. I use a crude zener diode regulator to get the 200V.
check the time constant on your r/c discharge. you are hardly achieving 40 A specified for the full 500us.
your voltage divider 101 class will help you monitor the cap charge with just about any op circuit you have.
by the way using a "crude zener" typically will require such a large series resistor, why are you switching the capacity and not the load?
are your contacts rated for 40 A. you may have designed a nice spot welder for your relay.
your voltage divider 101 class will help you monitor the cap charge with just about any op circuit you have.
by the way using a "crude zener" typically will require such a large series resistor, why are you switching the capacity and not the load?
are your contacts rated for 40 A. you may have designed a nice spot welder for your relay.
- Thread starter
- #5
Hacksaw, that was a useless post. I might be able to forgive you since you're "mechanical". The bottom line is, I was asking for a suggestion to monitor capacitor charge, not a critique of my whole design. If I wanted that, I would have to describe the entire project in detail, which would take several pages. But instead I will respond to your charges and hopefully enlighten you and any others who are monitoring this thread:
1. This is a current controlled device, not simply a RC discharge. You have to look at the energy required for the pulse. If I switched 200v into a 0.5 ohm load then the 400 amps would fry the circuit. That is why the current is controlled at 40A. To provide 40A into a maximum load of 5 ohms I need 200V of compliance. The maximum energy required is then (200V)*(40A)*(500us)= 4 Joules. Using E=0.5CV^^2. That works out to C=200uF.
2. Your grade 3 reading and comprehension class will help you understand as I wrote above that a voltage divider will drain charge from the capacitor.
3. The crude zener is simple, cheap, and it works.
4. Are you familiar with something called a power relay? They can handle the current with no problem.
Please consider posting something of value next time. Thanks.
1. This is a current controlled device, not simply a RC discharge. You have to look at the energy required for the pulse. If I switched 200v into a 0.5 ohm load then the 400 amps would fry the circuit. That is why the current is controlled at 40A. To provide 40A into a maximum load of 5 ohms I need 200V of compliance. The maximum energy required is then (200V)*(40A)*(500us)= 4 Joules. Using E=0.5CV^^2. That works out to C=200uF.
2. Your grade 3 reading and comprehension class will help you understand as I wrote above that a voltage divider will drain charge from the capacitor.
3. The crude zener is simple, cheap, and it works.
4. Are you familiar with something called a power relay? They can handle the current with no problem.
Please consider posting something of value next time. Thanks.
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- #6
Skogsgurra
Electrical
DJam,
Why is it so important not to load your circuit? A voltage divider with 10 megohms resistance will not leak more than 20 uA away. And if that is too much, you can go much higher. No problem there. I say this because you zener regulator shunts a lot more current away than your voltage divider will. If your concern is about loosing charge in the transfer period (when switching from charging to discharging) I see no reason for that. The voltage divider discharge time constant will be at least 2000 seconds and that means that you will loose less than 100 mV in one second. The transfer probably takes less than 100 ms, which makes the problem even smaller. If that still is a problem, you can always keep the voltage divider on the power supply side when you discharge the circuit.
And DJam. I did not at all like the way you answered Hacksaw. He has a very valid point when asking about the time constant. You mentioned nowhere that you have a current controlled discharge. So how could he know?
Why is it so important not to load your circuit? A voltage divider with 10 megohms resistance will not leak more than 20 uA away. And if that is too much, you can go much higher. No problem there. I say this because you zener regulator shunts a lot more current away than your voltage divider will. If your concern is about loosing charge in the transfer period (when switching from charging to discharging) I see no reason for that. The voltage divider discharge time constant will be at least 2000 seconds and that means that you will loose less than 100 mV in one second. The transfer probably takes less than 100 ms, which makes the problem even smaller. If that still is a problem, you can always keep the voltage divider on the power supply side when you discharge the circuit.
And DJam. I did not at all like the way you answered Hacksaw. He has a very valid point when asking about the time constant. You mentioned nowhere that you have a current controlled discharge. So how could he know?
- Thread starter
- #7
Thanks for your input Skogsgurra. Using a 10M-ohm divider means the capacitor voltage would drop 5% after about 2 minutes. This would probably be acceptable but I will need input from our intended user first. I just wanted to know if anyone knew of another way to do it.
Just for clarification, the zener is on the power supply side of the switch, so it does not shunt any current from the capacitors. Once the capacitor is switched to the output circuit, it is totally isolated from the zener. Also if I put the voltage divider on the power supply side then it will always indicate a charge as long as the power supply is working. What I would like to monitor is the charge on the capacitor when isolated. If it drops I would then signal the CPU to flip the switch and recharge it.
I'm sorry if my reply to Hacksaw offended you, but I did mention in my reply to Nbucska that it was a current controlled device.
Just for clarification, the zener is on the power supply side of the switch, so it does not shunt any current from the capacitors. Once the capacitor is switched to the output circuit, it is totally isolated from the zener. Also if I put the voltage divider on the power supply side then it will always indicate a charge as long as the power supply is working. What I would like to monitor is the charge on the capacitor when isolated. If it drops I would then signal the CPU to flip the switch and recharge it.
I'm sorry if my reply to Hacksaw offended you, but I did mention in my reply to Nbucska that it was a current controlled device.
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- #8
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- #9
Hello DJam.
My handle says Automotive, but actually I am a retired electronic engineer, but don't tell anyone that.
I can appreciate what you are trying to achieve, my first thoughts are that the self discharge of your capacitor and associated circuitry due to leakage is probably going to be more troublesome than a resistor chain to monitor voltage.
If it really worries you, how about using the resistor from a Fluke 20Kv EHT probe ?
Are you using a solid dialectric pulse rated capacitor, or the electrolytic type ?
My handle says Automotive, but actually I am a retired electronic engineer, but don't tell anyone that.
I can appreciate what you are trying to achieve, my first thoughts are that the self discharge of your capacitor and associated circuitry due to leakage is probably going to be more troublesome than a resistor chain to monitor voltage.
If it really worries you, how about using the resistor from a Fluke 20Kv EHT probe ?
Are you using a solid dialectric pulse rated capacitor, or the electrolytic type ?
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cbarn24050
Industrial
Hi, a high voltage zener diode (or series string), resistor into an opto led. The circuit only drains cap if over the threshold voltage.
no hard feelings here, not knowing the full details of your design reqm't i'm still wondering why you have to disconnect (isolate) the capacitor from the charging circuit.
sounds like you are using electrolytics (280 uf is a common size)and the leakage rates can be significant, often more than a voltage divider used for monitoring.
sounds like you are using electrolytics (280 uf is a common size)and the leakage rates can be significant, often more than a voltage divider used for monitoring.
- Thread starter
- #13
Thanks for all of your responses. Cornell Dublier and United Chemi-Con both make an electrolytic cap designed for repetetive high energy pulses. However, the leakage is spec'd at 1XC in uA after 5 minutes. So I am looking at 300uA leakage, which means the dV/dt will be about 1V/sec after the 5 minute stabilization. The output driver is a P-channel power mosfet with an off-state leakage of 1uA, so its not a big factor. Indeed, the voltage divider will not have much of an impact either. I do like the zener diode into an opto idea though. If anyone knows of a better rated capacitor (that will fit in your palm), please let me know.
Thank you again, everyone.
Thank you again, everyone.
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- #14
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- #16
That particlar double break is just to isolate the power supply after charging. The output circuit is also isolated by a double-break which is enabled by the CPU only after certain conditions are met. Finally, the output is also enabled or disabled by a manual keylock switch.
The rep rate is not specified by the user, but the current design has the capacitor charging from 0 to 200V in about 25 seconds. Of course, this would be worst case. Smaller amplitude pulses would require less recharge time.
The rep rate is not specified by the user, but the current design has the capacitor charging from 0 to 200V in about 25 seconds. Of course, this would be worst case. Smaller amplitude pulses would require less recharge time.
Hi all,
Two words, series R... Very small series R, Diff amp with good common mode rejection and decoupling as the spikes that will be generated on charge and discharge will be fairly large you will also have to watch the board layout, bandwidth and biasing current (obviously)...
Regards,
Carl Norman
Two words, series R... Very small series R, Diff amp with good common mode rejection and decoupling as the spikes that will be generated on charge and discharge will be fairly large you will also have to watch the board layout, bandwidth and biasing current (obviously)...
Regards,
Carl Norman
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