I have understood that you can create a linear regulator using a high-current BJT power transistor in the emitter follower configuration and one operational amplifier (and a voltage reference, which can be a voltage source or some kind of Zener diode or similar).

How does such a circuit work?

What is the minimum supported voltage drop of such a circuit? Is it worse or better than dedicated linear regulator ICs?

Are there any reasons to use such a circuit in the modern days where adjustable-voltage linear regulator ICs are widely available at a very cheap price?


4 Answers 4


An emitter follower has a minimum voltage drop below the Vout of the op-amp of about 700mV. If the op-amp can be supplied with a higher supply voltage than input voltage you can reduce that to tens or hundreds of mV, which might be competitive. I have done this in a situation with +/- 8V rails and a 5.6V rail regulated down to 5V using an emitter follower, with the op-amp supplied from the 8V rail. Most commercial regulators do not have a separate supply pin for the internal circuitry so this is not an option.

If you use a PNP or p-channel transistor rather than a follower the drop an be less, but ensuring stability is more difficult, as with commercial LDO regulator chips.

If you design your own you are free to use a better reference, a more accurate op-amp etc so you could get better performance. You might have a free op-amp out of a multiple package.

Generally though, you will be better off using commercial parts in almost all situations.


Yes, you can...


How does such a circuit work?

The opamp acts as a follower (or an amplifier) which takes the reference voltage as input. It drives and controls the pass transistor (the opamp handles the feedback).

What is the minimum supported voltage drop of such a circuit?

If you make it a LDO (with PMOS/PNP pass transistor) you can get low dropout like a good LDO. If it's a NPN pass transistor, then you'll get a few volts.

Since the load is a capacitor, it is essential to understand control theory...

Is it worse or better than dedicated linear regulator ICs?

The main reason to use it is low noise (by using a low noise reference and low noise opamp) and high PSRR. Also it has very low output impedance. But it's much more expensive, complex, and large than an off the shelf regulator.

So before doing such a circuit, you first must be sure that you need the ultra-low noise and other benefits it offers. Otherwise, it's a waste of time and money.


If you look inside any LDO, this is how it is done usually either emitter followers for BJT type or Pch FETs for +ve ultra-low dropout FET type LDO's.

Since they use precision band-gap reference diodes and a circuit with excellent PSRR at low cost, there is no advantage for a discrete circuit these days.

e.g. LM7805 and LM317 both use Darlington NPN emitter followers with the lowest possible Zout.

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These cost only 3 cents at low current and the same topology is used with bigger FETs for 1A and 10A , just lower RdsOn. In some due to Ultralow cap ESR , FET types are less stable to regulate low ripple feedback, so they advise the range of ESR for the load cap. But they perform better with lower I-O voltage drop.

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The is one exception with a PNP LDO rated for 10A from Sharp. enter image description here

There will always be exceptions to the rule that affects cost increases, but these parts define some of the topologies in low cost LDO's , each with tradeoffs for cost and performance. Ultralow Vce(sat) PNP's are very expensive with special processing.

  • 2
    \$\begingroup\$ No. LDOs generally use PNP in common emitter (for positive voltages) so that they can get the dropout voltage below Vbe, down towards Vce(sat) in the limit. \$\endgroup\$
    – user16324
    Sep 9, 2017 at 17:53
  • \$\begingroup\$ @Brian, can you give an example \$\endgroup\$ Sep 9, 2017 at 18:03
  • 1
    \$\begingroup\$ Sure. Any LDO with a dropout around 0.3V or less. Your answer is correct for older linear regulators, but nobody ever called a 7805 or LM317 an LDO. See fig 6 here analog.com/en/analog-dialogue/articles/… or analog.com/en/products/power-management/linear-regulators/… \$\endgroup\$
    – user16324
    Sep 9, 2017 at 18:05
  • \$\begingroup\$ The one I found from Sharp is already obsolete. The one you found is <<1A \$\endgroup\$ Sep 9, 2017 at 18:33

To avoid oscillation, do this


simulate this circuit – Schematic created using CircuitLab


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