PSPICE rectified AC to voltage regulator

[Zeuss1220]

I'm using a LM317 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

 

[duke37]

Looks as if you have positive feedback not negative.

 

[Harald Kapp]

I think the circuit is o.k. The feedback is negative.

Your OpAmp has no supply, however. So the model gets supply from who knows where. If it has supply, the Opmp will drive base current into the output transistor until the output satisfies the 15V regulation it is designed for.

Do you have another OpAmp model that you can tie to the unregulated supply (after the rectifier and filter capacitor)?

 

[duke37]

An ideal op-amp doesn't need a supply !

The op-amp and Darlington have a gain of one, just a complcated emitter follower.
Rh and Rl attenuates the output (X) to 0.916 *X.
The input to the op-amp is the same as the output.
X = 0.916 *X + 1.25
X (1 - 0.916) = 1.25
X = 1.25/0.084 = 14.9

QED

 

[Harald Kapp]

An ideal op-amp doesn't need a supply !

That's an artefact of the simulation. A real OpAmp does. And a good model of a real OpAmp shouldn't give an output without supply. Any simulation is only as good aas the models used.

I don't doubt your calculation, It gives the output voltage correctly (to within rounding errors). But it doesn't tell anythiong about positive or negative feedback. Positive feedback would make the regulator run amok against one of the supply voltage rails.
As it is, the "-" input of the OpAmp sees the full output voltage whereas the "+" input sees the lower value 0.916*Vout. Therefore any change in Vout is negatively corrected by the OpAmp (rising Vout -> lower output of the OpAmp and vice versa). This is negative feedback.

 

[duke37]

I am a simple soul and I often get things wrong but not here I think.

The negative feedback is only around the op-amp/Darlington, giving a gain of one.
To get oscillation the loop gain has to be more than one, but in this case there is attenuation so oscillation does not occur.

A much better circuit would compare a portion of the output to the 1.25V reference.