Linear regulators - how do they work

Thread Starter

engr_david_ee

Joined Mar 10, 2023
172
I am trying to understand the functionality of linear regulators. I have attached the fundamental circuit of linear regulator.

Can someone please explain how does the circuit work ?

I am also wondering where the V-Ref comes from ? Consider two following two cases.

Vin = 5.0 V
Vout = 4.0 V
Iout = 1 A

Vin = 3.3 V
Vout = 2.5 V
Iout = 1 A

Would V-Ref be the same for both cases or not ?
 

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LowQCab

Joined Nov 6, 2012
4,078
V-Ref is internally generated,
and will be slightly different depending on the design and Part-Number or "Family" of the Regulator.

The way that this is achieved, made to be stable, and insensitive to Temperature changes is quite complex,
and is far too complex to explain in a few sentences in a Forum.

V-Ref is generally the lowest Voltage that the Regulator can Output and still maintain Regulation-Stability.
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Thread Starter

engr_david_ee

Joined Mar 10, 2023
172
Does this means that the V-Ref is fixed for one device. Regardless we change Vin and Vout, the value of V-Ref remain the same per device, right ?
 

Thread Starter

engr_david_ee

Joined Mar 10, 2023
172
Thanks for reply that V-Ref is fixed per device which is at the non-inverting end of the operational amplifier. However, the V_ADJ at the inverting end of the operational amplifier is decided by the voltage divider.

I am familiar with operational amplifier and circuit analysis in general using basic electronics devices like PNP/NPN BJT transistors, operational amplifiers and diodes. Therefore, I am interested to learn how the components inside the linear regulator works to regulate the output voltage.

Can some please explain the internal functionality ? Thanks in advance.
 

Jerry-Hat-Trick

Joined Aug 31, 2022
553
Can some please explain the internal functionality ? Thanks in advance.
The V-Ref is a fixed voltage reference which is probably a "bandgap" diode which maintains a virtually constant voltage across it, typically around 1.25V over a wide temperature range. I believe it's actually two diodes in series, one which is a zener diode, the other an avalanche diode (often also called a zener diode) which have opposite temperature coefficients - quantum physics was never my specialist subject.

The key to understanding op amps, at least from a practical point of view, is that they always do their best to make the +ve and -ve inputs have exactly the same voltage. Your "Vin" by R2 should actually be "Vout" - just a typo. The R2, R1 potential divider will make the voltage at the -ve pin of the op amp equal to Vout x R1/(R1 + R2) which, if it equals V-Ref, will put the op amp in its happy place. Re-written, Vout = Vref x (R1 + R2)/R1 so you can select resistors to achieve your desired value.

Suppose Vout is actually lower than the voltage which makes the -ve input equal to V-Ref. In than case the output of the op amp will go high which will turn the transistor Q1 hard on so that it conducts more current and Vout rises. But as soon as it rises high enough to balance the op amp inputs it will start to turn off. It finds an equilibrium, with the output of the op amp about 0.6V more than the desired Vout.

Really not too complex, and a good example of how op amps function in practice. Hope this makes sense!
 

Thread Starter

engr_david_ee

Joined Mar 10, 2023
172
hi eng,
Look at this very basic regulator.
Choose a R1 resistor to give 6V out.:)

Why do I see 5V with the values of R1 and R2?
E
View attachment 315379

I think I am getting it now.

Referring to the circuit diagram in the fist post.

https://www.digikey.com/en/maker/tutorials/2016/introduction-to-linear-voltage-regulators

1- The linear regulator output voltage v_out is zero in the start.
2- The voltage divider voltage v_adj across R1, will also be zero which is connected to inverting input of the operational amplifier.
3- The v_ref is connected to non-inverting input of the operational amplifier.
4- The output of the operational amplifier will be high as vin+(v_ref) > vin-(v_adj).
5- The NPN transistor Q1 will conduct because the high output of the operational amplifier will turn on the the NPN transistor Q1.
6- Assuming the base to emitter voltage 600 mV (which is usually called drop out voltage in linear regulator).
7- The output voltage of the linear regulator start increasing that will also increase the voltage divider voltage v_adj across R2, which is connected to inverting input of the operational amplifier.
8- As the output of the linear regulator increase further, this will increase vin-(v_adj) compared to vin+(v_ref), causing the output of the operational amplifier low which will turn off the NPN transistor Q1.
9- The capacitor across the output load C2 will stabilize the output voltage through it's dis-charging during the time when NPN transistor Q1 is off.
10- As the output of the linear regulator start decreasing, this will lower the v_adj causing vin+(v_ref) > vin-(v_adj), which turn on the NPN transistor Q1 through the high output of the operational amplifier. This is how the linear regulator stabilize the output.
 

ericgibbs

Joined Jan 29, 2010
18,879
Hi eng.
The current and the allowed heating of the transistor.
And of course the rating of the Vin current available.
E

This is a very basic circuit which can be used for simple, not critical regulation.
EG57_ 1483.png
 

crutschow

Joined Mar 14, 2008
34,471
Your circuit is basically an op amp with a transistor emitter-follower to increase the output current capability, and with negative feedback from the output to the op amp inverting input.
For load currents no greater than the op amp's output current rating, the circuit would work basically the same without the transistor.

A simple way to look at it is, the Ref voltage to the op amp non-inverting input acts like a non-inverting amplifier with gain determined by the value of R1 and R2.
Thus, for example, the circuit in Post #8 has a non-inverting gain of 2, so the output is 2 times 2.5V or 5V.
Then it follows that If you want a 10V output, the required non-inverting gain would be 10 / 2.5 = 4.

Do you understand basic op circuit configurations?
 
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