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Bench CC/CV PSU Based on Daves uSupply (Not Anymore)
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KC0PPH:

--- Quote from: Kleinstein on March 18, 2019, 07:04:06 am ---The small caps at the OPs are usually needed to prevent the regulator from oscillating. So they are not just for a smoother output, but really essential parts.  A main purpose of the simulations is to find the right values for the caps. This circuit is type is not that critical, but the cap value should still be about right.

How far one can go down depends on the OPs. The constant current load needs a negative supply to work - so some 0.6 V sound like a reasonable lower limit for the constant current load.

--- End quote ---

Any oscillating should be minimal if the load is fixed correct? I should have my cap assortment in 2 days. Next research project is going to be looking into negative supplies for the OpAmp.
Kleinstein:
For a correct linear regulator circuit there should be not oscillation for the steady state. For a good supply this should be true for any practical load, including the relatively difficult case of a large (up to a few 1000 µF) low ESR capacitor at the output.  One may not be able to avoid some ringing on transients for all load cases: so when the current is doing a step change, there can be a decaying oscillations, but this should be decaying reasonably fast. With an easy load like just a resistor there should be essentially no ringing  there could be still some odd wiggles when coming out of saturation, but ideally there should be no or only minimal overshoot.
jaycee:

--- Quote from: KC0PPH on March 18, 2019, 03:03:48 am ---
--- Quote from: jaycee on March 17, 2019, 05:53:20 pm ---With no error amps, it wont regulate at all

--- End quote ---

I dont understand how this circuit works, so I am trying to decompose it and test each "system" independently to learn about it. My current understanding is that I should get a constant voltage/current out of the transistor with what I have currently breadboarded up. (The 4mA current source and the TIP120 with 100 ohm to ground as a load). I understand that with varying loads and such I am going to get inconsistent results,


--- End quote ---

OK a fair question, so I'll try and explain it :) I will refer to my original circuit here.

Q3, Q4 and associated resistors form a constant current source. There's a few ways to do this, and I just happened to pick the two transistor version. Some will say the version with the transistor + LED as the voltage reference is lower noise. I dont think it matters much here. Anyway, this current source is then used to feed the Darlington output stage formed by Q1 and Q2. This allows those transistors to conduct power accordingly.

The regulation loop is set up by sinking away this current accordingly in order to maintain regulation. Lets look at the voltage control. U1 is set up as an error amplifier, and compares your set voltage with the output voltage. R18 + R8 and R19 form a potential divider so that only a small portion of the output voltage is sensed - this "scales" it down to the same range as the set voltage from the DAC. C6 and R7 act as a "speed up" compensation network to improve transient response. U4 simply acts as a buffer.

So, what happens is that U1 compares the set and output voltage, and due to opamp action it adjusts its output accordingly. The diode D1 is there so that the opamp can only "sink" current away from the output stage. This setup provides voltage regulation.

The current regulation works in a similar manner, this time opamp U3 and it's associated resistors form a differential amplifier which monitors the voltage drop across the shunt resistor. As you should know V=IR, and since we can measure V, and know R, we can work out the current. The amplifier is configured for gain because the voltage drop is very small. So what we end up with is a voltage that is proportional to the current flowing through the shunt resistor. This is then used to regulate the current flow via U2, which is set up as an error amp in the same way as U1 is. Again, diode D2 means that this error amp can only sink current.

The two diodes D1 and D2 effectively form an "OR" gate.. normally the voltage error amp is in control, and regulating the voltage at the output.. however if the current being drawn exceeds the set value, the current error amp takes over the regulation loop.

Q5 and Q6 form a constant current sink. This works exactly like the constant current source, except it sinks current instead of providing it. This works to provide the minimum load which helps keep the regulator stable.

(edit: added the schematic I am referring to for clarity)
iMo:
The stability analysis with the above schematics, with 100ohm load.
40deg phase margin.
KC0PPH:

--- Quote from: jaycee on March 18, 2019, 05:45:26 pm ---
--- Quote from: KC0PPH on March 18, 2019, 03:03:48 am ---
--- Quote from: jaycee on March 17, 2019, 05:53:20 pm ---With no error amps, it wont regulate at all

--- End quote ---

I dont understand how this circuit works, so I am trying to decompose it and test each "system" independently to learn about it. My current understanding is that I should get a constant voltage/current out of the transistor with what I have currently breadboarded up. (The 4mA current source and the TIP120 with 100 ohm to ground as a load). I understand that with varying loads and such I am going to get inconsistent results,


--- End quote ---

OK a fair question, so I'll try and explain it :) I will refer to my original circuit here.

Q3, Q4 and associated resistors form a constant current source. There's a few ways to do this, and I just happened to pick the two transistor version. Some will say the version with the transistor + LED as the voltage reference is lower noise. I dont think it matters much here. Anyway, this current source is then used to feed the Darlington output stage formed by Q1 and Q2. This allows those transistors to conduct power accordingly.

The regulation loop is set up by sinking away this current accordingly in order to maintain regulation. Lets look at the voltage control. U1 is set up as an error amplifier, and compares your set voltage with the output voltage. R18 + R8 and R19 form a potential divider so that only a small portion of the output voltage is sensed - this "scales" it down to the same range as the set voltage from the DAC. C6 and R7 act as a "speed up" compensation network to improve transient response. U4 simply acts as a buffer.

So, what happens is that U1 compares the set and output voltage, and due to opamp action it adjusts its output accordingly. The diode D1 is there so that the opamp can only "sink" current away from the output stage. This setup provides voltage regulation.

The current regulation works in a similar manner, this time opamp U3 and it's associated resistors form a differential amplifier which monitors the voltage drop across the shunt resistor. As you should know V=IR, and since we can measure V, and know R, we can work out the current. The amplifier is configured for gain because the voltage drop is very small. So what we end up with is a voltage that is proportional to the current flowing through the shunt resistor. This is then used to regulate the current flow via U2, which is set up as an error amp in the same way as U1 is. Again, diode D2 means that this error amp can only sink current.

The two diodes D1 and D2 effectively form an "OR" gate.. normally the voltage error amp is in control, and regulating the voltage at the output.. however if the current being drawn exceeds the set value, the current error amp takes over the regulation loop.

Q5 and Q6 form a constant current sink. This works exactly like the constant current source, except it sinks current instead of providing it. This works to provide the minimum load which helps keep the regulator stable.

(edit: added the schematic I am referring to for clarity)

--- End quote ---

Thank you for the explanation.

So if I am getting the theory right C6 begins to look like a "short" as the frequency goes up. Even though this is a DC circuit the transients are AC and thus this AC "noise" gets pushed right through C6 into the OA.

The thing I am confused about is why the feedback Resistors and Capacitors are required in a steady state system? I would think that oscillation would not happen with a fixed load, and without oscillation there would be no "AC" transients to go through the feedback loops.

I promise I am not trying to be arrogant here, I am trying to understand in depth what is going on.
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