You have no discharge path for 100uF cap. Once it is charged it will remain a loooong time charged since op amp input is high impedance. Is that ok?

Yes, in doing so I have somewhat of a sample and hold circuit so that if I need to pump the brakes at some point during the rise time I can without having to restart from 0. Funny enough, this part also is not working in the real circuit, it discharges just as fast as it charges.

 

hi 55,
This is your original asc sim, plus a sim with the OPA.

E

Added: This could explain why your original sim looked OK, but real circuit not.

That could be it, any theory on why it has such a big impact? My understanding was that the high impedance input of the circuit would prevent any impact on the RC circuit output. If that's not the case, I'll need to find a better way to tune the output and isolate it from the next part of the circuit.

 

Yes, in doing so I have somewhat of a sample and hold circuit so that if I need to pump the brakes at some point during the rise time I can without having to restart from 0. Funny enough, this part also is not working in the real circuit, it discharges just as fast as it charges.

I think there are some parts and circuits involved which are not mentioned or visible here, or some part(s) in your circuit is broken/leaking/misfunctioning. From our perspective it is impossible to tell..

 

If you were to isolate and consider R1, C1 and U1 alone, I don't see how you can get ms charge time and not RC = 10s, unless R1 and/or C1 values are not correct or C1 is not properly connected in the circuit.

 

I think you’re making it complicated. Won’t this do it?View attachment 248214
100Ω is to limit the current into the comparator when it switches on.

It works for when the comparator is off as I'll have even closer to 0 volts on the output than I do now. However, when the comparator turns on I will have the rail voltage applied to the output rather than my setpoint voltage which is meant to be variable. The nmos switching adds that needed isolation between the control and the output.

 

how did you measure the time it takes to get fully charged?, Oscilloscope picture where you have gate voltage and RC node?
What discharges the 100uF cap ?
should be something like... View attachment 248215

Yes, I used an oscilloscope on the output. I didn't really "measure" the charging time as much as I just observed and saw that it was faster than a second and obviously not instantaneous.

 

hi 55,
There are different current levels for I5 and I6
A small charge coming from the OPA input.???

What is the OPA type.?
E

 

hi 55,
There are different current levels for I5 and I6
A small charge coming from the OPA input.???

What is the OPA type.?
E

I am using a TLV272IDR

 

You have no discharge path for 100uF cap. Once it is charged it will remain a loooong time charged since op amp input is high impedance. Is that ok?

Through the 1k resistor, the 100k resistor and the diode - so it's about the same as the charging time constant.

 

I am using a TLV272IDR

Problem solved, I ended up replacing the opamp with a new one of the same kind and it ended up working. I didn't suspect any issues with that before because it was still accurately amplifying my output. Not sure what kind of issue could cause that but I was able to find it thanks to our discussion pointing to it. Thank you so much for the help!

 

Thanks everyone for taking the time on this post! It was very helpful!

 

Not sure what kind of issue could cause that

For whatever reason, it seems that it had a high input leakage/bias current.

 

For whatever reason, it seems that it had a high input leakage/bias current.

I don't see how high input leakage current would change RC time-constant from 10s to 10ms.
C would just not be able to hold a charge.

 

I don't see how high input leakage current would change RC time-constant from 10s to 10ms.
C would just not be able to hold a charge.

Edited: But isn't that the result (short charge time) if a bad OP amp input became low impedence?

 

I don't see how high input leakage current would change RC time-constant from 10s to 10ms.

Well, he said he changed the opamp and the problem went away, so I think it's reasonable to assume the problem was a seriously leaky input on the opamp.

 

Sure, I acknowledge the evidence but the math does not compute.
Replace the opamp with a resistor across C1. Yes, the time constant is reduced but also C1 does not charge.

 

Replace the opamp with a resistor across C1. Yes, the time constant is reduced but also C1 does not charge.

He has the opposite problem,
It charges too fast (below).
So the leakage is to V+, not ground.

on the PCB. Instead, it charges up to the setpoint within milliseconds

 

He has the opposite problem,
It charges too fast (below).
So the leakage is to V+, not ground.

Which means that the capacitor would already be fully charged independent of the state of the MOSFET.

 

Which means that the capacitor would already be fully charged independent of the state of the MOSFET.

So than what's your solution to why changing the opamp solved the problem?

 

Which means that the capacitor would already be fully charged independent of the state of the MOSFET.

I think crutschow is right. Only I think the opamp input became low impedence to ground.....say 1k.
The cap would charged normally but would discharge very quickly...which is what the TS was describing...