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0-30 Vdc Stabilized Power Supply


Sallala
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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.
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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).
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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?
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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.
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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.
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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?
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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.
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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.
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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
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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?
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Prior to the current limit threshold, U3 is "acting" like a comparator, sitting at the positive rail waiting for the voltage across R7 to increase. Unless someone wanted to use the supply as a "regulated current source" whiles it's in current limit, does current regulation matter?? Not that it isn't a good idea, but I can't think of a reason why a "typical hobbyist" would care.

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Prior to the current limit threshold, U3 is "acting" like a comparator, sitting at the positive rail waiting for the voltage across R7 to increase. Unless someone wanted to use the supply as a "regulated current source" whiles it's in current limit, does current regulation matter?? Not that it isn't a good idea, but I can't think of a reason why a "typical hobbyist" would care.


Agree. I want to make it a CV/CC PSU.
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  • 2 weeks later...

If you use a 30VAC transformer to allow the output to provide a well regulated 30VDC at 3A, then without a load the unregulated positive voltage will be about +43.8V. Then the total supply voltage for two of the opamps is 43.8V + 5.6V (negative supply)= 49.4V which is too high for 36V and 44V opamps. The OPA445AP is the only high voltage opamp I could find.

Where in the world are you? The OPA445AP is made by Texas Instruments and is very common in North America.

Here is my recommended parts list again:

post-1706-14279143016351_thumb.gif

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