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I am studying for my EE exam but I couldn't quite get how this circuit works.

Source

Assume R1 = R2, Vout = 10V Vref = 5v. This is stable as it is, and when Vout decreases for a reason below 10V, - input of opamp becomes less than Vref that output of the opamp becomes higher, thus allowing current to escape to ground rather than going to the Load, which decreases Vo more. Where is the thing I am missing?

If I assume an ideal opamp, + and - inputs are equal. Vref = 5V Because of voltage divider rule Vo = 10V. But I am not sure if that's the explanation.

Edit: There is also a reversed one.

Source: Lecture material for SKEE 3263, Electronic Systems, CAMALLIL BIN OMAR, p.49 (http://www.fke.utm.my/new-academics/skem/summary/SKEE3263.php)

I think this is the correct form, when ref is connected to - input, but I am puzzled seeing two different examples.

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  • \$\begingroup\$ @clabacchio - I don't agree, though I would add a resistor to the opamp's output. Opamp output voltage -> \$I_{GND}\$ -> \$I_C\$ -> \$V_{OUT}\$ -> \$V_-\$. \$\endgroup\$ – stevenvh Jun 5 '12 at 6:16
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    \$\begingroup\$ "hihthtiittti"? :-) \$\endgroup\$ – stevenvh Jun 5 '12 at 6:37
  • \$\begingroup\$ @stevenvh: yeah, you're right. I wrote it before realizing what the "PASS XSISTOR" was doing... \$\endgroup\$ – clabacchio Jun 5 '12 at 6:48
  • \$\begingroup\$ @clabacchio - I noted in my answer that it would be much more clear if it showed an actual transistor \$\endgroup\$ – stevenvh Jun 5 '12 at 6:50
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"If I assume an ideal opamp, + and - inputs are equal."

That's not automatically true! Equal inputs are not a property of the ideal opamp! You can have different input voltages even with feedback, think the positive feedback of the Schmitt-trigger. You have to construct the right negative feedback loop to get the inputs equal.

The diagram would probably have been easier to understand if they would have drawn the pass transistor's symbol, instead of that rectangle. Like Olin says, that transistor will be a PNP. \$I_{GND}\$ is the PNP's base current, and the collector current to output is proportional to that. So increasing \$I_{GND}\$ won't draw that away from the output, but instead increase the output current.

What you describe would happen if the pass transistor were an NPN. In that case \$I_{GND}\$ would indeed be taken away from the pass transistor's base current, and an increase in \$I_{GND}\$ would result in a decreased base current, and therefore decreased emitter current to the output. In that case the inputs of the error amp have to be switched.

(Add a resistor to the error amp's output to reduce the amp+transistor's transconductance. Without it, even with real-world components, a 1nV input change may result in a 1A output change, and it may oscillate. (Murphy: your amplifier will only oscillate if you haven't foreseen it.))

edit (shunt regulator)
R4 is an important component. If the input voltage is constant then regulating the output voltage means keeping the current through R4 constant. If the load decreases the transistor will draw more current to keep the total current constant. That's not efficient. Load regulation: if the load current would rise the drop across R4 will increase, and the voltage on the non-inverting input of the opamp will decrease. Then the transistor will conduct less to balance the risen load current.

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Your explanation is basically right:

Assume R1 = R2, Vout = 10V Vref = 5v. This is stable as it is, and when Vout decreases for a reason below 10V, - input of opamp becomes less than Vref that output of the opamp becomes higher, thus allowing current to escape to ground rather than going to the Load, which decreases Vo more.

The thing you seem to be missing is that letting a little current "escape" to ground where the schematic shows Ignd makes PassXsistor pass more current from Vin to Vout. Basically PassXsistor is meant to be a PNP transistor. A real circuit would have a few other details, but the basic scheme is correct. Another way to put this is that Ignd is the control signal to the PNP pass transistor.

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The two pictures in the question show two fundamentally different concepts. It appears that you are confused by these two configurations.

  • The first picture (with the funny box labeled PASS XSISTOR) shows a series regulator. The PASS XSISTOR acts like a variable resistor providing current to the load, dynamically regulated by the OpAmp. Think of PASS XSISTOR as the upper part of a voltage divider and the load (and R1, R2) as the lower part of this voltage divider. (Note that IGND usually is very small compared to the load current IL and does not have anything to do with the main idea behind a series regulator.)

  • In the second picture ("hihthtiittti"), Q1 acts as a load paralleled to the actual load (RL). This is called a shunt regulator. R4 is the upper part of a resistive voltage divider, Q1 and RL (and also R1, R2) are the lower part of this voltage divider.

Both regulators try to maintain a stable voltage across RL. The series regulator does this by putting a variable pass element (usually a transistor) before the load (in series to the load), the shunt regulator uses a variable pass element in parallel to the load (it shunts the load).

This link has a good and simple picture showing only the most basic differences between the two types.

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  • \$\begingroup\$ If the link is just for the picture, why don't you bring it here? (Fair use) \$\endgroup\$ – stevenvh Jun 5 '12 at 15:22

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