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Tech note: voltage regulators-1

Posted 7th January 2010 at 06:10 PM by jan.didden
Updated 11th February 2010 at 04:31 AM by Jason

There are lots of types of voltage regulators, but in this installment I’ll talk about series regulators.

What’s a regulator? It’s all in the name: it REGULATES the voltage to the circuit to be powered to keep it constant and as free of noise and ripple as practical. The ‘regulation’ means that there is some circuitry that compares a reference voltage, like from a zener diode, to the regulated output voltage, and then uses the difference between the two to adjust another element to null that difference. The ‘compare-and-correct’ is crucial for a regulator, and is done by negative feedback….

Look at Fig 1: is there a regulator in there? No, they are all circuits that try to give a constant, ripple free voltage, but if you start to draw varying currents from them, the output will vary with that current and there is no mechanism that somehow tries to null out that variation. Fig 1c is better than 1b, because Q1 buffers the voltage from the zener reference, so the Vo will vary less with load current, but there still is no ‘compare-and-correct’ mechanism.

But, building on Fig 1c, we can add that ‘compare-and-correct’ element as in Fig 2. Let’s say we want a 14VDC regulated supply. We use a 7V zener diode to give us the reference; since it is only very little loaded, it will be pretty constant and ripple free. The opamp compares the ref to half of Vo (from the two 1k resistor divider) and if that is smaller than 7V, the opamp output will rise and open the transistor further, to increase Vo until it is 14VDC again so half of that will be equal to that 7V reference.

A few observations. The correction element (the opamp) needs an error to work; it needs a (very small) difference between its inputs. So the error is never 100% nulled out, but the DC difference is probably much less than the difference caused by the tolerance of the two Vo divider resistors. If you draw varying currents from this circuit, the opamp will immediately correct the output voltage to keep up with the Vo changes from the load current, again to within a few mV. That means, from the load point of view, it is almost an ideal voltage source that has very low output impedance, just a few milli-ohms for a good regulator.
Why is the output impedance Zo important? Load current will cause a voltage drop across that Zo. It doesn’t matter that Zo is not a physical resistor, Ohms’ law is unrelenting! Therefore, with a varying signal load current, this will cause signal ripple on Vo which may cause unwanted output signals in the amp that is supplied by this thing. Anyway, since the opamp will do its best to keep Vo at twice the zener voltage, any input ripple and hum coming from the rectifier is also greatly reduced.

If that opamp is very fast, the opamp will try to open that transistor quickly to correct any errors it senses, but if the pass transistor is much slower, the opamp actually overshoots the correction. At worst case, the system may start to oscillate: when the transistor finally reacts, it'll keep going and will be slow to turn off, so Vo will become too high, the opamp immediately wants to shut off the transistor, Vo gets too low, the opamp immediately wants to open the transistor, etc, you get the point.

To break that vicious cycle, most regulators have a capacitor at their output to prevent Vo to change too fast which prevents the opamp seeing a large and fast jump in Vo. This output cap is NOT responsible for supplying current to the load; it is in parallel with the few milli-ohms output impedance so it would probably need to be a farad or more to make any difference at all. No, that cap is there for stability, and it actually works better when it is a bit lossy so it can damp out any onset of oscillations better. So, don’t put a boutique foil cap here, as it is not only a waste of money but it actually works less well.

Stay tuned, and happy soldering!

Jan Didden
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  1. Old Comment
    Superbly written!

    I'm listening.
    Posted 8th January 2010 at 04:14 PM by ra7 ra7 is offline
  2. Old Comment
    Yes, very good explained
    Posted 23rd February 2010 at 12:15 PM by Charles Charles is offline
  3. Old Comment
    jan.didden's Avatar
    Originally Posted by fjpompeo View Comment
    Very good information. I have a question, what is the difference between regulate the positive rail, usually using a NPN power transistor or regulate the negative rail? I know that the most common is positive rail but some HAM told me that negative the power loss is small due to less difference between input and output voltage and then dissipating less heat. Let me explain 20V input 13,8V output with 10A = 20V-13,8V=6,2VX10A=62W (positive rail) and 15V-13,8V=1,2Vx10A=12W (negative rail).
    Fernando J. Pompeo - Brazil
    I don't think there is a principle difference. In your example, if you reduce the pos input to the same as the neg input, 15V, the dissipation is the same.

    jan didden
    Posted 23rd February 2010 at 07:08 PM by jan.didden jan.didden is offline
  4. Old Comment
    jan.didden's Avatar
    Originally Posted by pteron View Comment
    Good article. The output cap will also provide current during transients - remember the Zo for the regulator is frequency dependent and will not be milli-ohms at high frequencies.
    True, the Zo generally rises with frequency, although with a good reg it can be quite low. But for the purpose you mention, the cap doesn't have to be so large in value. You would like to have a small ESR, and in particular a small ESL for that purpose, but such a cap might also compromise stability. A tough call.

    jan didden
    Posted 23rd February 2010 at 11:20 PM by jan.didden jan.didden is offline
  5. Old Comment
    Irakli's Avatar
    Hi Jan,

    Excellent synopsis on regulators.

    I have a quick question regarding C1.
    Will it hurt much if I put C1 in 10-100uF range instead of 1-10uF ?

    My understanding is that C1 reduces A1-Q2 loop gain at high frequencies and increasing C1 value will result in increased output impedance at high frequencies.

    I really appreciate your comments.
    Posted 15th September 2010 at 11:49 PM by Irakli Irakli is offline

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