lab power supply design

[xReM1x]

Schematic:
https://i.gyazo.com/5a083c121a1b25fbf66f0b0f7ac5e045.png
is there anything I should change?
I'm going to use MAX472 instead of the differential amplifier.
and OP275 op amp instead of the LT1013.

[dom0]

There are several flaws in this circuit, but it might still just work. Transient response will be sluggish at best and current regulation will be very poor due to U4 (a simple op amp current divider will have no errors beyond basic resistor inaccuracy and the alpha error of the "downstream" pass transistor).


If you want to know how a proper lab-grade PSU is done, look at the schematic of commercially available units (e.g. HP/Agisight).

[c4757p]

Q9-Q8 has extremely high voltage gain and so will be hard to control - you may find it's unstable without a load on the output. A resistor in series with Q8's emitter could solve that.

What is the purpose of M1?

[xReM1x]

There are several flaws in this circuit, but it might still just work. Transient response will be sluggish at best and current regulation will be very poor due to U4 (a simple op amp current divider will have no errors beyond basic resistor inaccuracy and the alpha error of the "downstream" pass transistor).


If you want to know how a proper lab-grade PSU is done, look at the schematic of commercially available units (e.g. HP/Agisight).

I'm going to use MAX472 and not U4..

Q9-Q8 has extremely high voltage gain and so will be hard to control - you may find it's unstable without a load on the output. A resistor in series with Q8's emitter could solve that.

What is the purpose of M1?

can you explain more? just a resistor in Q8 doesnt work..
the purpose of M1 is that I added it because without it, the current regulation will go max to ~3A and not 5A.

[c4757p]

How is M1 supposed to provide current regulation?

Anyway - yes, I can explain more, but what do you mean by "just a resistor in Q8 doesnt work"? I promise you, it'll work for what I'm saying.

The voltage at Q8's emitter is constant, so it can be considered to be a non-emitter-degenerated common-emitter amplifier. The gain of such an amplifier is - assuming low output current, say 10V output with no load except the 10mA drawn by R21:

A_V = -R_OUT g_m = -R_OUT I_C/V_T = -1k (10mA / 26mV) = -385

Where:
A_V = gain of the amplifier
R_OUT = output impedance, here R21 = 1k
g_m = transconductance of Q8
I_C = collector current of Q8
V_T = thermal voltage = about 26mV

That is further multiplied by the gain of the amplifier formed by Q9 and its resistors, which is -330/14 = -23.6.

So the total Q9/Q8 gain is -385 * -23.6 = 9066.

Waaaaaay too much. U5 isn't meant to control something so fussy. It's like trying to drive a car that has a very, very sensitive accelerator while having a heavy weight tied to your foot...

Adding an emitter resistor to Q8 will decrease its gain significantly from 385.

[xReM1x]

How is M1 supposed to provide current regulation?

Anyway - yes, I can explain more, but what do you mean by "just a resistor in Q8 doesnt work"? I promise you, it'll work for what I'm saying.

The voltage at Q8's emitter is constant, so it can be considered to be a non-emitter-degenerated common-emitter amplifier. The gain of such an amplifier is - assuming low output current, say 10V output with no load except the 10mA drawn by R21:

A_V = -R_OUT g_m = -R_OUT I_C/V_T = -1k (10mA / 26mV) = -385

Where:
A_V = gain of the amplifier
R_OUT = output impedance, here R21 = 1k
g_m = transconductance of Q8
I_C = collector current of Q8
V_T = thermal voltage = about 26mV

That is further multiplied by the gain of the amplifier formed by Q9 and its resistors, which is -330/14 = -23.6.

So the total Q9/Q8 gain is -385 * -23.6 = 9066.

Waaaaaay too much. U5 isn't meant to control something so fussy. It's like trying to drive a car that has a very, very sensitive accelerator while having a heavy weight tied to your foot...

Adding an emitter resistor to Q8 will decrease its gain significantly from 385.

M1 isnt providing current regulation, but without it the current is limited for unknown reason to 3A and not 5A.
it does some magics that make this psu work correctly.

what is the resistence value for the resistor R1?
is it supposd to be connected like this?

[c4757p]

Yes, it should be connected like that. Rule of thumb - choose the resistor to drop about a tenth of V_BE (so 65mV) at the full load current. It can be chosen more 'properly' via loop stability analysis but I think that's a bit beyond our scope right now