# PSPICE rectified AC to voltage regulator

1. Nov 15, 2013

### Zeuss1220

I'm using a LM[Three-One-Seven] to regulate dc voltage output from mains power(240V ac). For PSPICE simulation, I was instructed to use an ideal OpAmp and Darlington BJT to model the IC in PSPICE. I came up with the following model, which satisfies the equation Vo=Vref(1 + RL/RH), where Vref = 1.25V(nominal voltage for the IC). Function of the circuit is to supply power to another device.

Refer to attachment regulatorDC.jpg

[For OPAMP, VPOS = 50V, VNEG= -50V]
I tried using a dc input to the circuit and it works as expected i.e. Output voltage = 1.25(1+RL/RH) for any input voltage and load resistance as long as the input is higher than the desired output. For example, having RL=1320, RH = 120 and Vin = 10, I only obtain 10.2V instead of 15V.

To obtain power from the mains, a transformer and a rectifier is required. So firstly, I attempted to simulate the circuit without the transformer.
I have tried connecting a AC source(VSIN) to a full-wave diode rectifier(w/ smoothing cap) to the circuit. However, the circuit still gives a output of 15V(same resistor values as example above), for any VAMPL of the VSIN, even VAMPL = 0. I used two VSIN with ground in the middle because PSPICE requires me to ground that loop else I get an convergence error.

Refer to regulatorAC.jpg

What is the mistake I'm making? I tried the function of the rectifier seperately in a simple test circuit, and it works fine. Is it something to do with the OPAMP's VPOS and VNEG? It didn't seem to affect during the test with dc input. Or is the current being drawn for the source(although the voltage isn't sufficient) to give the desired voltage at the output? If that's the case, do I need to use the transformer to isolate the regulator from the source? I have tried using XFRM_Linear as a transformer and still have the same undesired results.

Please tell me if any clarification is needed. Thank you in advance.

EDIT: The circuit still gives 15V output WITHOUT any input source connected. Only sources remaining are the 1.25V VDC for the reference voltage and the supply to the OpAmp

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Last edited: Nov 15, 2013
2. Nov 15, 2013

### CWatters

I haven't used spice for a long long time but...

When the input voltage is set to zero is it possible the circuit continues to power itself from C8? Perhaps monitor a few more nodes in the input side. See what the voltage is on C8 when the AC input is zero.

3. Nov 15, 2013

### Zeuss1220

There's 15.68V DC at the input side, with or without the smoothing cap (in fact with or without the input bypass cap too).

4. Nov 16, 2013

### CWatters

Weird. Has to be a bug somewhere in the model.

5. Nov 16, 2013

### Zeuss1220

It has something to do with the diode and the op amp. The power to the output comes from the OpAmp supply. During the test without the rectifier, the input(Vin) of the regulator is connected directly to the source(VSIN). So Vin must = VSIN. So the source is the one limiting the output voltage. Using the rectifier, Vin is not forced to = VSIN. So Vin goes up to any desired value to give the output voltage as 15V, again the power coming from the OpAmp. The difference between VSIN and Vin, is dropped across the diode (forward bias voltage), which in practice may break the diode.

The voltage across the diode is not forced to be = forward bias voltage (e.g 0.6V). Either something is wrong with my regulator model or I should use a different diode, if it helps?

6. Nov 16, 2013

### Staff: Mentor

Yes. When you use a high-powered OP-AMP it can supply all the power you need, there is no requirement for the transformer and rectifier! The regulated output is all sourced from the OP-AMP; being ideal it has no trouble compensating for the Darlington Pair's gain being very low (read, 0).

7. Nov 16, 2013

### Zeuss1220

The ideal Op amp should be an feedback error amplifier. So should I set the supply voltage lower? or is there a problem with regulator model design?

8. Nov 16, 2013

### Staff: Mentor

You could try restricting the OP-AMP's output capability.

9. Nov 16, 2013

### Zeuss1220

The only parameters of the OpAmp I can change is its supply. I tried setting it at 5V, but it is limiting the output voltage.

Source: 30V dc
Input without rectifier.
Op amp supply: 5V

Output votlage: 4.05V

The BJT is not "*drawing current" from the source. The BJT supposed to be a current regulator, so my understanding is that it will increase current flow from collecter to emitter when there's voltage at the base. Is this correct?

10. Nov 16, 2013

### Staff: Mentor

In a "real" device the power for the op-amp would be derived from the mains. No mains, no power for the op-amp and so no "phantom" output.

Why not make the op-amp power supply and the "regulator" reference supply controlled voltage sources? Choose a voltage or current that would only exist when the AC mains are present to activate these supplies.

11. Nov 16, 2013

### Zeuss1220

@gneill
I understand what you're trying to say, and I see how that's sensible in practice. But I was specifically instructed to use an Ideal Op Amp for simulation.

As for the reference voltage, I have no other idea besides a DC source. I'm not sure whether using another "source" in the model affects the function of the circuit.

The idea is to simulate the a voltage regulator circuit using an OpAmp(for feedback error amplification) and a Darlingtion BJT(for current regulation).

12. Nov 16, 2013

### Zeuss1220

13. Nov 16, 2013

### Staff: Mentor

For starters, you could add a resistance in series with the base of the Darlington Pair to limit the drive current from the OP-AMP.

14. Nov 16, 2013

### Zeuss1220

Adding a high value resistor at the base of the darlington pair seemed to do the trick. Even at 50V op amp supply, sufficient voltage is required from the source to provide an output voltage of 15V. I used an 8k resistor, are there any recommended value I should use instead?

From what I understand, the most of the voltage from the opamp output falls across the resistor, whereas only the *forward-bias voltage appears across the base-emitter. Earlier, all of the voltage appears across BE, hence the OpAmp supplies the output voltage.

Please correct my understanding if necessary. Its very important for me that I understand completely anything I work on or learn.

15. Nov 16, 2013

### Staff: Mentor

Whoa! I didn't have in mind kΩ! I thought a few Ohms might do, enough to limit the base current to about 5% of the load current. Or pretend the Darlington Pair has a gain of, say, 40, and work backwards.

A practical OP-AMP would have some output resistance and drive limitation.

16. Nov 16, 2013

### Zeuss1220

Low ohms give me convergence errors. Not sure what it actually means though, came across it numerous times working with this circuit.

17. Nov 16, 2013

### Staff: Mentor

You don't seem to have indicated over what power range you wish to test this model. The required value of added resistance is load dependent. With no load, it may indeed need multiple kΩ. (A resistor may not be best representative of a real OP-AMP, perhaps a current limiter. But I think a single well-chosen resistor may well suffice provided your needs are modest.)

18. Nov 16, 2013

### Zeuss1220

For this simulation, load current absorption is to be assumed as 0.5A, hence 30 Ohms load resistor.
I get no difference in outpt for op amp resistor value varied to different kOhm values. Like I mentioned earlier, values in the tens or hundred ohms give me a convergence error. I'll assume the value as 8k for now.

I'm having some doubts on the temperature dependency of the circuit. I simulated the circuit for multiple temperatures, no difference were observed. However, does the real 317 performance gets affected by temperature besides component damage?

19. Nov 17, 2013

### Staff: Mentor

Using an "ideal OP-AMP" and that ideal 1.25V voltage source leaves little else to show temp dependencies, I guess.

You can find datasheets online. http://imageshack.us/a/img811/5412/thgooglefriend1.gif [Broken]

Last edited by a moderator: May 6, 2017