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Lab Power Supply - The Lost Current
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xavier60:

--- Quote from: C on March 02, 2018, 02:17:14 pm ---
I hope that you have found that CV circuit should match CC circuit.

With a CC mode & a CV mode control is Min (current OR Voltage).

With 0.1 R current sense  you have 0 to 3V current sense for a load range of 0 to 3 amps.

With a 0 to 3 volt voltage sense for an output of 0 to 30 volts output you have two matching control circuits.

If the two circuits do not match then you start getting a voltage caused change not equal to a current caused change.

With matched circuits, testing becomes a little easer.

--- End quote ---
Do you mean they should have the same gain and response? Can you tell me where I can find an example? The Current Sense resistor is now close to 0.05 ohms. I'll eventually get some power resistors and mount them on the heat sink.
Anyways, I have corrected my massive blunder with the input circuitry on the CV op-amp.
The PCB is working as well as the breadboard mock up.
I'm still not totally satisfied with the CV to CC transition. Under certain conditions, there is brief dip in current after the initial spike.

Oh yes, also, I ended up adding the extra op-amps to make it digital ready. There are outputs for monitoring voltage and current. And the set inputs are now positive  voltage.
C:

Total system is what counts and is what will power your load.

As I have stated many times you actually have foru states.
CV mode where current is not close to your current setting.
CC mode where voltage is not close to your voltage setting.
CV = CC
The fourth mode actually has two sub modes.
One where you are changing from CV to CC.
One where you are changing from CC to CV


--- Quote from: xavier60 on March 02, 2018, 02:51:05 pm ---
Do you mean they should have the same gain and response? Can you tell me where I can find an example?

--- End quote ---

What happened to using common sense for answer?

Think of a load change.
If the load increases then you have two things happing, Voltage drops & current increases.

If you just look at the voltage drop then control will try to increase the Voltage.

If you just look at the current change then control will try to decrease current.

Only one is correct response.

If times do not match, then you can easily have a increase the voltage followed by a delayed decrease the current.

For a short this could be increase the voltage and output changes to max voltage the 30V+ followed by the decrease the current to x amps at 0V.

I would think that this is common sense.


xavier60:

--- Quote from: C on March 02, 2018, 03:23:55 pm ---

Think of a load change.
If the load increases then you have two things happing, Voltage drops & current increases.

If you just look at the voltage drop then control will try to increase the Voltage.

If you just look at the current change then control will try to decrease current.

Only one is correct response.

If times do not match, then you can easily have a increase the voltage followed by a delayed decrease the current.


--- End quote ---
When the output is overloaded while in CV mode, first the voltage drops, the CV op-amp immediately starts allowing the Gate voltage to rise. The CV op-amp has to react first because it is in control at the time of the overload.
The CV op-amp keeps allowing the Gate voltage to increase for some time after the current has exceeded the set point because the CC op-amp can't react quickly enough.  I have reduced this delay to an acceptable level.
Slowing down the response of the CV op-amp would allow extra time for the CC op-amp to react to the overload before the current rinses too high. I don't want to slow down the response of the CV op-amp.
C:

From you post it is clear to see you are not thinking balanced or matched.

Don't slow the voltage side, speed up the current side.

You would only slow the voltage side as a last choice.

Think about what the test facts are trying to tell you.

IF you have a problem going from CV to CC then you also have a problem going fro CC to CV.

 
xavier60:
This is the corrected schematic without the extra op-amps, in case anyone wants to experiment with it.
The circuitry at the output of the CC op-amp allows it to swing low at its full slew rate to take control of the Gate quickly during an overload. Q1 connects the compensation capacitor, C1, after the CC op-amp is in full control of the Gate.
This quick current control response might not be important to every one.
My Agilent U8002A would output about 40A for 100us in the same test.
The bottom trace shows the current response when the output is short circuited with the regulator set to 10V 3A.
The top trace shows the output voltage upside down because I'm measuring with respect to the + output.
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