The LM7805 voltage regulator has a ripple rejection of 73bB (minimum is 62dB.) My textbook says that this gives a tremendous advantage because we do not have to use any bulky LC filters in the power supply to minimize the ripple.

I don't understand how this IC can reduce the ripple that much.

A sample of the internal circuitry of LM7805:

enter image description here

I don't see any block that will act like a filter capacitor for the input voltage to reduce that ripple that much.

Why do voltage regulator ICs have such a big ripple rejection ratio?

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    \$\begingroup\$ Because they are voltage regulators. The clue is in the name LOL. They have to deal with stuff on Vin and still regulate on the output. \$\endgroup\$
    – Andy aka
    Aug 27, 2020 at 9:47
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    \$\begingroup\$ related question with a diametrically opposite point of view: Are all linear regulators bad at filtering input ripple? \$\endgroup\$ Aug 27, 2020 at 21:42
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    \$\begingroup\$ Are you asking why, or how? The answers so far are to these two different questions. \$\endgroup\$ Aug 28, 2020 at 17:29
  • \$\begingroup\$ More like "how" \$\endgroup\$
    – hontou_
    Aug 29, 2020 at 3:01

5 Answers 5


Let's look at the datasheet to see more details about this ripple rejection:

enter image description here

Note how it says "f = 120 Hz" so that means this ripple rejection is measured at 120 Hz which is quite a low frequency.

The circuit inside the LM7805 (and many other voltage regulators) comes down to this:

enter image description here


A stable reference voltage is generated with zenerdiode \$V_Z\$ (in the LM7805 a "bandgap circuit" is used, it has the same function).

This reference voltage circuit needs to have a very good ripple rejection as well, any ripple on the reference voltage will appear at the output as well. In practice, this is usually not an issue as reference voltage circuits with enough ripple rejection can be made.

An opamp (used as an error amplifier) compares the output voltage (actually a divided-down version of the output voltage, \$R_1\$ and \$R_2\$ are a voltage divider).

The output of the opamp controls a transistor, Q2.

If the opamp is fast enough then it can control the transistor Q2 so well that it will be fast enough to respond to the voltage changes (ripple!) at \$V_i\$. It will respond in such a way that at \$V_o\$ there's as little left of the ripple as the loop can manage. In essence, the loop compensates for the ripple by controlling Q2 such that the ripple is rejected.

If you would do the proper loop analysis you would find that the ripple rejection depends on the excess loop gain inside the loop. For more information, read this.

So with an opamp that is both fast enough and that has a high gain (for 120 Hz, that is not an issue, the gain will be quite high) we can achieve quite high ripple rejection.

At (much) higher frequencies that 120 Hz, like for example at 1 MHz, the opamp will not be fast enough and will have less gain meaning less ripple rejection. Fortunately we can then use capacitors to help us out. For 1 MHz these capacitors can have a relatively small value (a few uF) so size and cost is less of an issue. Also these capacitors are often needed to guarantee stability of the voltage regulator, without the input and output capacitors the voltage regulator might oscillate and generate a new ripple!

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    \$\begingroup\$ Yes, datasheets typically show the schematic at transistor level so that you know what's going on inside - they don't typically abstract things like internal op-amps so it's not always obvious from the datasheet that a bunch of those transistors are effectively building an op-amp. \$\endgroup\$
    – J...
    Aug 27, 2020 at 19:06
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    \$\begingroup\$ What is the use of R1 and R2? \$\endgroup\$
    – Ben
    Aug 28, 2020 at 5:22
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    \$\begingroup\$ R1 and R2 set the output voltage relative to the reference voltage (which is useful because it's easy to make resistors in arbitary values, while it's not so easy to make references with aribitary voltages). \$\endgroup\$ Aug 28, 2020 at 5:47
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    \$\begingroup\$ Something you have to watch for is some regulators can have a bad dip in the rejection. A frequency that is too high to be rejected effectively by the feedback, but too low to be rejected effectively by the input and output capacitors. This is especially likely to be an issue with low-power regulators. \$\endgroup\$ Aug 28, 2020 at 5:48

The regulator has no energy storage (to speak of) internally, what it does is to meter the power from the external storage capacitor to the output. So the ripple voltage "troughs" must never get too close to the dropout voltage of the regulator under worst-case conditions.

Unlike an ideal LC filter, this means that power is burned up in the regulator whenever the voltage is higher than the minimum.

Imagine the city gives you water at 60-70 PSI, but you need 50 PSI +/- 1 PSI. You can always control your valve to get 50 PSI provided the inlet water pressure is enough above 50 PSI to take up the losses in your valve and pipes.


Because filters with capcitors buffer the peaks and dips to smooth the input voltage and create the mean voltage. This need big energy storage.

Regulators just cut of the ripple part and output a smooth voltage, with a value below the lowest value of the dips of the input voltage. enter image description here


That is what regulators do, ideally no matter what the input voltage is (as long as it is withing limits), it will output a fixed output voltage.

They generate an internal reference voltage that does not change much even if input voltage changes, and then there is a feedback mechanism that compares the output voltage with the reference and it keeps the output voltage quite stable.

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    \$\begingroup\$ Yep, it's a control loop. And as long as it's faster than the ripple, you'll be able to compensate ripple to some degree. \$\endgroup\$ Aug 27, 2020 at 9:34

from a conrol loop perspective it's a valid assumtion that most if not all voltage regulators use some sort of PI(D) controller. Also there is no need for our regulator to be able to follow changes of the reference value, as the reference is const. So what the manufactures will do to improve design is to tweak the PI(D) controller for improved error rejection (this normaly will also result in a poorer ability to follow changes of the reference value). There are known methods how to do that.

But as you can see here: https://en.wikipedia.org/wiki/PID_controller#Response_to_disturbances

even simple PI(D)'s involve a lot of knowlede... too much to go deeper here


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