Looking at the datasheet I can see that the voltage regulators are not just a zenner diode inside, they are complex devices. I have noticed that there is always a capacitor at the input and another one at the output. An example is the uA7800 series fixed voltage regulators.

I have read that one of them is to "stabilize the circuit operation" while the other is to "reduce ripple on the output". Looking at the datasheet, why do they have this fixed value? And if they do have a fixed value then why not just fabricate them into the voltage regulator itself? e.g for the uA7800 series it is 0.33uF at the input and 0.1uF at the output. It is not explained why they have these values.

  • \$\begingroup\$ plz provide the circuit diagram \$\endgroup\$
    – user107801
    Commented Jun 2, 2016 at 7:39
  • \$\begingroup\$ Datasheet of uA7800 will do \$\endgroup\$
    – quantum231
    Commented Jun 2, 2016 at 7:46

3 Answers 3


Most voltage regulators (especially LDO types) require a capacitor on the output for stability, and it will usually improve transient response even for regulators like the 7800 that may not strictly require it. An input capacitor is usually required to reduce source impedance.

It is impractical to make capacitors more than tens of pF (or so) on an inexpensive chip- they take up too much expensive silicon area, and external ceramic or electrolytic capacitors are very cheap in quantity. That is not in the cards. And the capacitors actually provide energy storage so it's not something that clever circuitry can substitute for.

The values are compromises that make sense based on the chip stability behavior at different load currents, and also what caps were common when the datasheet was composed (that might be 35 or 40 years ago for the 7800 series). It is almost always acceptable to use a larger capacitance on the input, and usually acceptable on the output, however there may be minimum/maximum values on the capacitor ESR- the equivalent series resistance. In some cases a capacitor that is too ideal may cause the regulator to oscillate.

Most modern regulators will indicate what values and types of capacitor are acceptable, so reading and understanding the datasheet is all you need to do.

  • \$\begingroup\$ How can an input capacitor reduce source impedance, by source you mean output of the voltage regulator? \$\endgroup\$
    – quantum231
    Commented May 8, 2016 at 19:15
  • 3
    \$\begingroup\$ Input to the voltage regulator. The capacitor is in parallel with the source to the regulator. At high frequencies the capacitor will have a low impedance. Xc = 1/(jwC) \$\endgroup\$ Commented May 8, 2016 at 20:27
  • \$\begingroup\$ Input to a voltage regulator is DC but a higher voltage than the output. The input to it is not AC. So how does this low impedance help? Or, are you talking about output from a bridge rectifier being connected to a capacitor shunted to ground and the output from that going into voltage regulator, since that would be a smoothing capacitor just like the one at the output of the voltage regulator (Vout). By input capacitor I assume you mean capacitor at Vin? \$\endgroup\$
    – quantum231
    Commented May 8, 2016 at 21:11
  • 2
    \$\begingroup\$ Imagine you load the output with a few MHz square wave- the input current will have an AC component. \$\endgroup\$ Commented May 8, 2016 at 21:36
  • 1
    \$\begingroup\$ @quantum231 "Input to a voltage regulator is DC but a higher voltage than the output." This is only true in an ideal situation. Real power sources have a finite resistance, so any change in the in current taken from the regulator will change the input voltage. There may be other sources of noise in the input voltage, such as the power line acting as an antenna and picking up radio frequency interference, or voltage spikes when other equipment switches on and off - or even a lightning strike hitting the mains power distribution system. \$\endgroup\$
    – alephzero
    Commented May 8, 2016 at 23:43

I thought I would build my latest circuit one component at a time so that I could "debug" and learn what was actually happening at each step. So I hooked up my 12V supply to my Input and Ground and eagerly waited for the 5V regulated output. What I got was 10 seconds of looking for a 5V output followed by smoke! That is what happens if you don't have the capacitors in the circuit. Self-oscillation, huge heat and smoke. Please take my word for it and don't try this at home. Teaches you to read the datasheet properly from cover to cover.


The answer to this question lies in practical experience.

Omit the input capacitor and, sooner or later, the stabilizer goes into self-oscillation, over-heats and (literally) explodes.

These chips are, in fact, series-stabilized shunt-controlled arrangements with enormous gain in the side-chain. That input capacitor controls phase-shifts.

As stated the output capacitor is normal with any regulator/supplier circuit.

Full details see: http://www.clovellydonkeys.co.uk/kengreen_website/index.htm


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