Why do we use two parallel capacitors in a voltage regulator circuit? [duplicate]

C1 and C2 are parallel capacitors and their total capacitance is 1000.1 uF. I think C1 is large enough and I can remove C2 from the circuit. The result will be an open circuit.

Let's assume that I can buy one capacitor that has the value of 1000.1 uF. I know I can't do it in the real world because a 1000.1uf capacitor does not exist in the shops but the question will help me understand the circuit well.

Can I replace C1 and C2 with the 1000.1 uF cap?
Does this 0.1 uF really matter? Is it important?

marked as duplicate by The Photon, PeterJ, Robherc KV5ROB, Matt Young, jippieMar 6 '16 at 16:50

• I am not going to attempt to answer this question, as it's too complicated. 7805 is too slow to illustrate the principles. The 'correct' answer has to take account of the hazards of paralleling a small C2 with the residual inductance of C1, potentially creating a tuned circuit that defeats the very object of using the extra, small cap. Warning, any answer you get that says 'yes you must' or 'no you must not' is not a sufficient answer. – Neil_UK Mar 5 '16 at 11:33
• I'm interested in what D5 is for. – abligh Mar 5 '16 at 16:44
• This question has been asked before: What's the use of a decoupling capacitor near a reservoir capacitor? – The Photon Mar 5 '16 at 16:58
• – The Photon Mar 5 '16 at 17:02
• Also see Murata's application guide. – The Photon Mar 5 '16 at 17:03

Take a look at this graph: -

http://electronicdesign.com/files/29/1478/figure_01.gif

Capacitors have a resonant frequency due to the inherent small series inductance they have. The "generalized" capacitor showing the various parasitic components is shown below: -

It's the L$_{ESL}$ that causes this series resonance.

For a typical 10 uF tantalum capacitor this might occur at about 1 MHz. A 1000 uF electrolytic will basically look like an inductor above several tens of kHz: -

Notice that the 100 nF (MLCC) ceramic is good as a capacitor all the way to over 10 MHz therefore, putting two caps together gives you the best of both worlds. For a 7805 this might not make much difference but on different types of regulators not having the 100 nF could turn a power supply into a power oscillator.

• A high quality answer as ever Andy. What would you have to add about causing a parallel resonance between a small C, and the ESL of the adjacent big C? This tends not to be a problem with alli as the big C, the ESR is enough to drop the Q. A big and a small ceramic or tant however might need additional loss. This loss was often in the form of series ferrite 'stoppers' when I was engineering microwave circuits. – Neil_UK Mar 5 '16 at 12:27
• @Neil_UK theoretically this could be a problem but i suspect the ESR of the big cap would dampen the parallel resonant impedance down to a few tens of ohms max but definitely a good point for the possibility of problems between the resonant points of the 1000 uF and 100 nF. – Andy aka Mar 5 '16 at 12:41
• @Neil_UK I meant to add at the time that if you calculate the series L of the 1000 uF (from the graph above) it comes out at about 63 nH (based on resonance being ~20 kHz. If you work out the Q based on ESR being about 0.02 ohms (again from the graph above) you get Q = 0.4 i.e. very low and therefore a very poor resonant circuit. – Andy aka Mar 5 '16 at 21:19
• No but it's polite (and often required) to reference the work of others. It only took me a only a moment to find the author of the second image and he (looks like an Elcap employee) does request attribution: commons.wikimedia.org/wiki/… – PeterJ Mar 6 '16 at 12:50
• @Andyaka You need to say where you got the images, links are good. Posting them without attribution implies that you made them. – W5VO Mar 6 '16 at 15:37

If all capacitors were ideal your 1000.1 uF capacitor idea would work. They're not ideal though and real capacitors have non-ideal behaviour.

• C1 is there to hold the voltage up between pulses from the rectifier. It needs to be large in value and electrolytic capacitors are the most practical solution to this. Unfortunately they can have some internal resistance and, worse, some inductance which reduces their ability to react at high frequencies.
• C2 is typically 0.1 uF and will usually be mylar type or similar. These have very little inductance and work really well at high frequencies, shunting the noise to ground.
• C3 and 4 have a similar relationship.

Many novices try to skimp on the capacitors only to find that the voltage regulators go unstable and the output voltage starts to oscillate. Obey the datasheet recommendations!

In short: "high" capacitors (like the 1000 µF) are used to smoothen the voltage signal to a straight DC voltage, "low" capacitors (like the 0.1 µF) are used to suppress interference voltages. So the two capacitors have two different "jobs" to do and can not be replaced by one with the same capacitance.

0.1uF is low ESR, for stability, ensure non oscillation under certain 'bad' load conditions.

Higher cap are electrolytic with interval coiled structure with higher ESR

• The issue is not the ESR, but the ESL, of the big cap. – Neil_UK Mar 5 '16 at 11:37
• Thanks Neil. Yes, it is ESL and ESR combined. Quoted Murata information,"there is some resistance (ESR) from loss due to dielectric substances, electrodes or other components in addition to the capacity component C and some parasitic inductance (ESL) due to electrodes,leads and other components.As a result,the frequency characteristics of |Z| form a V-shaped curve(or U-shaped curve depending on the type of capacitor)as shown in Figure 4,and ESR also shows frequency characteristics for values equivalent to loss" murata.com/en-us/products/emiconfun/capacitor/2013/02/14/… – EEd Mar 5 '16 at 18:29
• Some info from Sun technical paper: google.com/… – EEd Mar 5 '16 at 18:43