0
\$\begingroup\$

I am using LTspice trying to find out the extremums of tbmhe output of the following circuit. The diode, op-amp, and PNP transistor are the default components.

What I know at the moment is that \$V_{output}=\frac {R_{3A}+R_{3B}}{R_{3B}} {V_D}\$, but I don't know what determines the extremums of the output.

I also tried to search for examples of similar circuits, but mine is always different from the others in the following three points:

  1. Uses PNP instead of NPN;
  2. Reference connected to the inverting input rather than the non-inverting input;
  3. The direction the diode.

I also hope to know how these differences affect the regulator.

enter image description here

\$\endgroup\$
4
  • 1
    \$\begingroup\$ Welcome to EE.SE! I'm not entierly sure what you are asking or where you are stuck. Is your simulation working as shown above? What do you get? \$\endgroup\$
    – winny
    Commented Oct 14, 2020 at 14:05
  • 2
    \$\begingroup\$ 1) what do you mean by "extremums of the output". Suggestion: read the datasheet of the LM7805 (the most common voltage regulator IC) and see how that describes the circuit's behavior. 2) you have a formula, whoop-tie-doo. But seriously, do you understand how the circuit works. Can you explain it like: when this voltage increases then this voltage...? 3) your circuit is different from another circuit, after you understand how your circuit works, then do the same for other circuits. Then it will become clearer why there are differences. \$\endgroup\$ Commented Oct 14, 2020 at 14:07
  • \$\begingroup\$ With typical unity-gain-stable op-amps, if that circuit is stable at all it'll be barely so, especially if you put any capacitance on the output. Try it with a realistic op-amp model (even LTSpice's "universal" op-amp, but with a GBW set to 1 or 10 MHz), and with at least 10uF on the output. That topology makes a nice regulator, but you have to work to make it stable. \$\endgroup\$
    – TimWescott
    Commented Dec 2, 2021 at 21:21
  • \$\begingroup\$ The voltage of the reference diode varies with current and temperature changes. \$\endgroup\$
    – Audioguru
    Commented Aug 9, 2022 at 1:23

2 Answers 2

1
\$\begingroup\$

What I know at the moment is that Voutput=R3A+R3BR3BVD, but I don't know what determines the extremums of the output.

The dominant factor will be the "reference" diode. You need to be aware that forward-based diodes aren't all that good at setting a constant voltage for varying currents.

As a rough starting point, figure that the forward voltage of the diode is about 1 volt. Since your sine wave has an amplitude of +/- 3 volts, the voltage Tin will vary from 3 to 9 volts. Then the current will be set by R3, and will vary from (about) 20 uA to 80 uA. Note that most silicon diodes such as a 1N4148

https://www.vishay.com/docs/81857/1n4148.pdf

aren't even characterized for such a low current, and the voltage across the diode will be on the order of 0.5 volts. At a guess, the voltage will vary by about 50 mV over the current range. This will give an output variation (for an ideal op amp) of about 0.1 volts.

Other sources of error - the next biggest source would be the non-ideal nature of the op amp. Your equation assumes things like infinitely small input currents and infinite gain at the op amp, and neither of these is true. Most notably, a lot of older op amps have input bias and offset currents in the range of your application.

Finally, you can choose the wrong op amp. Your circuit implies an op amp which can be run essentially as a rail-to-rail op amp with a power supply range of 3 to 9 volts. If you pick an op amp such as a uA741, an LM308, or a TL081/082/084, you will get much larger swings as for at least part of the voltage cycle the op amp simply won't be able to function.

\$\endgroup\$
0
\$\begingroup\$

Unclear what you mean by "extrema" so I'll answer this:

I also hope to know what these differences affects the regulator briefly. the direction the diode

Usually a zener diode would be used as reference, so it would be in the other direction. If you use a standard diode you'll get a 0.6V voltage reference. Since R3A=R3B you'd get 1.2V output.

Use pnp instead of npn, reference connected to the inverting input rather than the non-inverting input.

As shown on the schematic the circuit stability depends on load impedance. Open loop gain is the product of:

  • R3B-R3A voltage divider ratio
  • opamp open loop gain
  • 1/R2
  • transistor current gain
  • load impedance

Open loop phase has a 90° shift due to the opamp, some shift due to the transistor current gain, and of course load impedance. If the load is a cap, that's another 90°.

So the circuit will be unstable on capacitive loads due to 180° phase shift, and conditionally stable on resistive loads depending on what the load value does to the open loop gain, plus transistor behavior vs frequency, etc. Since a voltage regulator is supposed to work on a wide range of loads, it is safe to say this circuit isn't a very useful voltage regulator.

If you want to do a LDO, therefore a PNP-based circuit, then the 90° phase shift of the output cap will always make the 90° phase shift of the opamp a problem. This leads to stability dependent on output capacitor ESR and value, among other things.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.