0-30 Vdc Stabilized Power Supply

[audioguru2]

1. How to decide the value of R9? Why don't we use 10K or 470R?

R9 feeds the adjusted reference voltage to U2 and limits the loading effect of C4 on the output of U3. If its value is higher then U2 might pickup interference. If its value is lower then U3 might oscillate due to C4 being at its output or R9 won't discharge C4 quick enough during current limiting.

2. What's the purpose of C5? How to calculate the value?

It is a supply bypass capacitor for high frequencies, because C1 is so big it has high inductance. Analog electronic circuits oscillate without it or if its value is too low.
C5 could be 0.1uF to about 1uF and a ceramic disc is best at high frequencies.

3. Is C9 used to improve the stability of U2?

Yes, because Q2 and Q4 are in its negative feedback loop and are much slower than U2 and cause delay, the combination could oscillate without C9.

4. Can I connect output(pin6) of U2 to base of Q2 (bypass R15)? If not, why?

U2 oscillates if its load is capacitive and is directly at its output. R15 isolates the capacitance of Q2 and I recommend reducing its value to 100 ohms.

5. If the system should be grounded, which part of the circuit should connect to ground as well?

Connect the 0V output to ground and the project supplies variable positive voltages above ground.
Connect the Positive output to ground and the 0V connection supplies variable negative voltages below ground.
Don't ground any part of the output and the project can be connected in series with another supply that might be grounded.

 

[audioguru2]

well replacing D7 did nothing new ... so i measure the voltage on r2 where there is 17.7V, and on D5 there's 0V(?!) ... any ideas?

R2 has AC on it. I don't know what DC measurement will be shown on a meter.
Maybe C2 is open.
Maybe D5 is shorted.
Measure these voltages with the negative meter lead connected to the negative terminal of C1.

 

[MonSSter]

i replace d5 and d6, and nothing ... i measure as you said and i got the same measurement's ... that's d5 has 0v, r3, d7 has around 0,7v ... so i realy don't know now ... you think i should start all over again?

 

[indulis]

R9 feeds the adjusted reference voltage to U2 and limits the loading effect of C4 on the output of U3. If its value is higher then U2 might pickup interference. If its value is lower then U3 might oscillate due to C4 being at its output or R9 won't discharge C4 quick enough during current limiting.

C4 has NO affect on U3 until you hit current limit, at which point U3 turns on just hard enough to reduce (clamp) the voltage at U2-pin3 such that the drop across R7 becomes small enough to "release" U3. Having R9 will slow down the discharge of C4, not speed it up. Until you hit current limit, U3's output is at the positive rail... so it will not oscillate. In the end R9 doesn't really do much...

It is a supply bypass capacitor for high frequencies, because C1 is so big it has high inductance. Analog electronic circuits oscillate without it or if its value is too low.

C1 has a large ESR (good for killing Q) but high ESL at line frequencies (60/120 Hz)??

Yes, because Q2 and Q4 are in its negative feedback loop and are much slower than U2 and cause delay, the combination could oscillate without C9.

C9 IS part of the feedback loop.

U2 oscillates if its load is capacitive and is directly at its output. R15 isolates the capacitance of Q2 and I recommend reducing its value to 100 ohms.

U2 can't drive a transistor?? It's junction capacitance is so large that U2 will oscillate? R15 is there to limit the current through Q1 when it has to discharge (via D10) the ouput capacitor (and  load cap, if any).

 

[audioguru2]

i replace d5 and d6, and nothing ... i measure as you said and i got the same measurement's ... that's d5 has 0v, r3, d7 has around 0,7v ... so i realy don't know now ... you think i should start all over again?

Is C2 47uF (maybe it is open) and is it installed with the correct polarity?
Is R2 82 ohms?
Is one wire of the transformer shorted to 0V?

 

[audioguru2]

U2 can't drive a transistor?? It's junction capacitance is so large that U2 will oscillate? R15 is there to limit the current through Q1 when it has to discharge (via D10) the ouput capacitor (and

 

[MonSSter]

r2 is 82ohms, also nothing is shorted ... maybe this will help .. c3 voltage is also 0, and c2 is 17.7V(but as you said, there's AC not DC )

 

[indulis]

The TL081 and TL071 and many other opamps oscillate with the small capacitance of shielded cable connected directly to their output.

Where is there "shielded cable" connected to the output of U2?? QUANTIFY the amount of capacitance it would take to make U2 possibly oscillate, and why that would happen given C9 is in the circuit??  In this configuration it's some sort of gain limited integrator in combination with R15, Q2 b-e, R16, R12/C6 and R11... I don't suppose you have worked out the transfer function??

D10 would be reversed biased if it tried to discharge the output capacitor.

If Q1 turns on, for whatever reason and there had been a voltage at the output, D10 is now forward biased, and C1 is discharged via D10, R15 and Q1.

If the project tried to go quickly to a low voltage then the emitter-base junction of Q2 would break down at a reverse bias of about 6V.

How fast can you turn the knob? Something tells me that transient response of this supply isn't all that good!! besides, R16 would do the same thing.

 

[audioguru2]

r2 is 82ohms, also nothing is shorted ... maybe this will help .. c3 voltage is also 0, and c2 is 17.7V(but as you said, there's AC not DC )

C2 is charged through the rectifiers on half-cycles from the transformer, then discharges a negative current which makes  the negative supply.

 

[audioguru2]

If Q1 turns on, for whatever reason and there had been a voltage at the output, D10 is now forward biased, and C1 is discharged via D10, R15 and Q1.

Yes, I had it backwards, you are correct.

 

[jiangrm]

May I remove R19? Would this cause an problem?

 

[jiangrm]

4. Can I connect output(pin6) of U2 to base of Q2 (bypass R15)? If not, why?

Let me correct my question. My question is:
If I do not need the protection circuit consists of D10, R15 and Q1, can I drive Q2 directly with output of U2?

 

[jiangrm]

If the project tried to go quickly to a low voltage then the emitter-base junction of Q2 would break down at a reverse bias of about 6V. R15 limits its avalanche breakdown current.

Wouldn't D10 clamp the E-B junction and R16 voltage to 0.7V if there is reverse bias?

 

[indulis]

Yes it would.

Will the circuit run without those components in place... yes, but under certain conditions, it might be possible to "flame" some components without it.

R19?? Did you mean R9?

 

[audioguru2]

May I remove R19? Would this cause an problem?

Yes, then Q3 would never turn off. The output of the opamp U3 is a darlington transistor and its max voltage is about 1.5V lower than the positive supply.

Wouldn't D10 clamp the E-B junction and R16 voltage to 0.7V if there is reverse bias?

Yes, I had it backwards.

 

[jiangrm]

Yes it would.

Will the circuit run without those components in place... yes, but under certain conditions, it might be possible to "flame" some components without it.

R19?? Did you mean R9?

I mean R19

 

[jiangrm]

C4 has NO affect on U3 until you hit current limit, at which point U3 turns on just hard enough to reduce (clamp) the voltage at U2-pin3 such that the drop across R7 becomes small enough to "release" U3. Having R9 will slow down the discharge of C4, not speed it up. Until you hit current limit, U3's output is at the positive rail... so it will not oscillate. In the end R9 doesn't really do much...

indulis,

R9 slows down the discharge of C4, but it allows the U2-pin3 to be pulled down faster. Without R9, the circuit has to wait until C4 is fully discharged before entering current limit mode.

 

[jiangrm]

Yes, then Q3 would never turn off. The output of the opamp U3 is a darlington transistor and its max voltage is about 1.5V lower than the positive supply.

This makes sense, thanks

 

[indulis]

R9 slows down the discharge of C4, but it allows the U2-pin3 to be pulled down faster. Without R9, the circuit has to wait until C4 is fully discharged before entering current limit mode.

Yes, but not exactly... true, the U2-p3 node voltage doesn't drop quite as fast (.1

 

[jiangrm]

R8 and C4 are a filter to remove high frequency noise from the D8 reference voltage circuit. R9 isolates C4 so that D9 can reduce the voltage of the project quickly when current regulation needs it to be reduced.

Since the D8 ref voltage is also used in U3, why don't we put a filter before U3-P3?