I want to set up a digital control voltage supply - diyAudio
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Old 29th August 2006, 08:47 AM   #1
xiaonan is offline xiaonan  China
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Angry I want to set up a digital control voltage supply

I want to set up a voltage supply which can be controled by the conputer. Now I hope the output voltage range is from 0 to 30V. Initial plan is to use the LM317 and AD5235. But I can not get the 0V output, because the minimum output voltage of LM317 is 1.2V. I will appreciate to you if you can tell something of suggestion about that. Thank you very much!!
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Old 30th August 2006, 06:06 AM   #2
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LM 317 will go down to 0volts if u lower its reference to -1.2 volts. U will find this approach in LM317 datasheet.

Gajanan Phadte
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Old 30th August 2006, 05:37 PM   #3
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It won't be ideal because the LM317's quiescent current (called Iadj in the data sheet) will get in your way (Iadj*R2 adds to Vo) and the control function is like 1/x:

Taper Vout
0.9 13.6
0.8 7.2
0.7 5.0
...
0.1 1.5
0.0 1.25

Not nice even if you subtract 1.25V by shifting the reference ground to -1.25 ...

Use a circuit with separate reference (can be a cheap one like TL431), opamp and pass transistor. http://www.scopeboy.com/ddragon.html shows something like that, your AD5235 can take the place of R26.
In http://www.circuitcellar.com/library...lier149/2.htm, Brian shows a deign similar to what you want; the only limitation is that the opamp driving the MOSFET needs to handle the full input voltage.

I like the floating topology better (see attached) where the regulator's ground is at the V+ output. TR1/B1/C1 supply output power. The regulator has a small separate supply (TR2/B2/C2) that may also feed your digital circuits. R6 and R7 would be your AD5235 to set voltage and current, respectively (or use a fixed setting for I-limit). Note that for linearity, R10 would have to be large compared to R6, or you'd have to add a buffer inbetween. With single supply as shown the opamps must be able to work at 0V input common, or you give them a little negative supply (a couple of diodes between IC2 GND/C2 with opamp V- on C2 will do the trick).
R5/R4 define the reference voltage (10k each 5V). IC1B tries to hold its +in at V+ level, or at regulator ground, so Vref/R10 = Vout/R9 (D3/D4 protect IC1B). This design scales nicely - only T1 and R9 "see" the full voltage, and you can parallel multiple pass transistors (with emitter Rs) for more current/dissipation, or use MOSFETs.
Note this is a rough schematic - you'd need to work out component values, compensation etc. for your needs.
Oh BTW - you'd need to opto-isolate the computer interface anyway, so regulator ground at V+ is no problem.
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Old 2nd September 2006, 05:41 AM   #4
xiaonan is offline xiaonan  China
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thank gmphadte!!

I see that!!
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Old 2nd September 2006, 05:41 AM   #5
xiaonan is offline xiaonan  China
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thank you wine&dine very much for your so detailed reply!!

My voltage supply is only used for an experiment test system, just need some constant voltage. No AC signal. I don't care the noise and linear of the output. So if that can the last circuit pic be simplified??

Thank you very much.
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Old 2nd September 2006, 07:57 AM   #6
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The obvious simplification is a fixed current limit (trimpot instead of R7) - I wouldn't leave out the limiter itself. And remember to sort out the compensation for your choice of opamp/transistor.
TR1/2 can of course be combined if you find something suitable, but the secondaries have to be separate.
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Old 4th September 2006, 08:28 AM   #7
xiaonan is offline xiaonan  China
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Default another solution

Because the digital potentiometer(resistence) can not afford the high voltage. So I want to change the solution.

Now I want to use the DAC LTC2609(From Linear Technology Company) to transfer the digital signals to analog signals. And then use a Operational Amplifier OPA541(From TI, +-40 power supply) to get the output voltage. The max current of OPA541 is about 1.8A.

Is that solution viable?

Thank you!!
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Old 5th September 2006, 11:20 AM   #8
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>Because the digital potentiometer(resistence) can not afford the high voltage.
>Now I want to use the DAC LTC2609

Neither can deal with "high" voltages, and neither will have to anyway.

Look at the schematic of the floating regulator that I posted - you can use a digital pot, or a DAC, to create a control voltage in the 0-3 or 0-5 V neigborhood to control possibly hundreds of volts - the regulator is run on some +12V, separately supplied, and the pot/DAC have nothing to do with the (high) output voltage, only the pass transistor (MOSFET) and the sense resistor do.
A version of that circuit is found in lots of professional lab/bench supplies by the likes of HP et al. As I said it scales nicely to high voltages/currents as required. Don't let the fact that the regulator's ground potential is at Vout+ scare you too much. You need optoisolation anyway.
See your other thread for the DAC/Power OpAmp story; yes of course you can do that, too.
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Old 6th September 2006, 05:39 PM   #9
xiaonan is offline xiaonan  China
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Look at the schematic of the floating regulator that I posted - you can use a digital pot, or a DAC, to create a control voltage in the 0-3 or 0-5 V neigborhood to control possibly hundreds of volts - the regulator is run on some +12V, separately supplied, and the pot/DAC have nothing to do with the (high) output voltage, only the pass transistor (MOSFET) and the sense resistor do.

You meaning is that the T1 is the amplifier transister that we can only control it's base electrode? But there two inportment things for your design. One the the linearity of the transistor and the other is transistor have to afford a high voltage and high current.
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Old 6th September 2006, 09:06 PM   #10
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> You meaning is that the T1 is the amplifier transister
>that we can only control it's base electrode?

I'm not sure I understand the question. T1 is used as an emitter follower here. What electrode other than the base do you want to control?

> But there two inportment things for your design.
> One the the linearity of the transistor

Nonsense. The transistor is in the opamp's feedback loop.

The trick is the summing node between R9 an R10. The opamp holds that at the same voltage it sees on its inverting input = V+. (The voltage across D3/D4 is always zero, the diodes are only there to protect the opamp while there's a rapid change.) Since the reference is relative to V+ also, the current from R9 is constant (Vref/R9). it has nowhere else to go but R10. Therefore the voltage across R10 (the output voltage) is (Vref/R9)*R10. If it were less, the current through R10 would be less than Vref/R9, causing IC1b (I+) to rise above (I-); IC1B's output would go up; and so would the voltage on T1's emitter (= voltage on its base less ~1.4V), and therefore the ouptut voltage.

Are you sure you know enough basics to tackle this project yourself?

You can believe me this circuit works - my main lab PSU of 20 years is built that way.

>and the other is transistor have to afford a
>high voltage and high current.

That's what I've been telling you all along.
Power transistors for a couple 100V are cheap & easy to come by. (Much easier than opamps for that voltage.)
Power dissipation is inherent in linear regulators and not specific to this circuit. You need appropriate heatsinking (max output current x max input voltage). If necessary, the power dissipation may be shared among a number of transistors. (Also mentioned before).
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