# Bypass Capacitors

1. Jul 15, 2012

### mearvk

Why do we branch off and go to ground on bypass capacitors instead of just putting a capacitor inline? Surely the effect would be the same?

2. Jul 15, 2012

### vk6kro

Capacitors conduct AC but do not conduct DC.

So, if we have a DC power line in a circuit and we have some unwanted AC on it we could use a bypass capacitor to conduct the AC to ground.

More importantly, bypass capacitors are an essential part of most circuits and they won't work properly without proper bypassing.

For example in this circuit:

There are bypass capacitors across the power ("+") line to provide a signal path to ground and also across the source resistors to allow the source connection of the FETs to be grounded for signal even though it will have a DC voltage on it for biasing purposes.

Last edited by a moderator: May 6, 2017
3. Jul 15, 2012

### mearvk

No, I don't see how it answers my question.

This is what I'm suggesting may be equivalent:

[+] ------ [ + - ] ------- [-]

Since a capacitor stores energy (let's say choppy DC current from a cheap rectifier) why not just use it like I have above? The choppy energy goes in, is stored, and less choppy current comes out. Hence my original question: why do we see it branching off per your diagram? Is it really necessary to do it that way? Can we run it inline as above and get the same result?

This is the 'normal' way to do it:

[+] ---------------------[-]
|
|
[CAP]
|
----
GRD

4. Jul 15, 2012

### vk6kro

Capacitors conduct AC but do not conduct DC. So, you can't put them in series with the DC current.

5. Jul 15, 2012

### mearvk

Capacitors, I'm fairly certain, exist in DC systems. My audio amplifier has a 220 uF capacitor in its diagram running of a 9v DC battery. It's inline as well.

http://i.imgur.com/JsIvB.jpg

It's small but you can Google 'LM386 diagram' for better images. Check the lower right, you see the big speaker? Right before it is C1, a 1000 uF capacitor. This is a DC circuit.

6. Jul 15, 2012

### vk6kro

No, it isn't.

The speaker runs on AC. The capacitor is there to remove the DC component.

7. Jul 15, 2012

### the_emi_guy

mearvk,
Expanding on vk6kro's input

Have you studied high pass and low pass filters yet?
What you are calling the "normal" connection of the capacitor on the output of the cheap regulator is a low pass filter. Low frequencies, which would include DC, will pass through to the output while high frequencies (the choppyness) is bypassed to ground via the capacitor and does not make it to the output. This is what you want.

With the capacitor arranged inline you have a high pass filter. High frequencies will pass through to the output (through the capacitor), while low frequencies, including DC will be blocked by the capacitor. This is the opposite of what you want.

In the case of you audio circuit, this is not a DC circuit. We do not want DC flowing through the speaker. The capacitor is acting as a high pass filter here blocking DC but allowing audio frequencies to pass.

8. Jul 15, 2012

### mearvk

Thanks for the correction. I meant that it was powered off DC as opposed to an AC fed system. I can see what you guys meant now.

How does the capacitor 'know' to filter high/low? A capacitor, it seems to me, doesn't care if the input is AC or DC, it will store it regardless. And the output, it seems to me, would be basically DC. I'm not saying this is correct but this is how I imagine it.

A capacitor is basically a way to store current. You take a water reservoir and you fill it either with a calm, constant water supply or you fill it with a violent random water supply. Either way the water coming out is basically calm and constant. So you can see what I'm saying is either AC or DC going into a capacitor seems like it would result in a DC-like output.

9. Jul 15, 2012

### PaulS1950

Mearvk,
Capacitors will not store AC current. In AC circuits a capacitor acts more like a filter resistor than a buffer or storage device. In AC circuits the capacitor can be used to change the phase between voltage and current by 90 degrees. (an inductor can be used to bring it back into phase)

10. Jul 15, 2012

### mearvk

I'll play around with my circuit tomorrow. Thanks all.

11. Jul 16, 2012

### sophiecentaur

Perhaps a bit of basic theory wouldn't go amiss here. There are so many sources that there's bound to be one to suit you. Electronics without theory can turn out to be a bit like monkeys and typewriters: a long way round to achieve anything worthwhile.

12. Aug 18, 2012

### mearvk

Hi all.

Quick question. I'm going to make a nice full-bridge rectifier setup for general purpose AC to 'clean' DC applications. This way I can just connect a transformer to it and get going.

My question is, is there a correlation between the size of the decoupling caps and the amount of power the DC device will be using? Smaller caps for less wattage and larger ones for bigger amps? How do I know what size caps to throw on my breadboard and how many for super clean DC out?

I have a prototype setup done which uses two 1000uF caps in parallel but I think that may be overkill since most of the circuits use much less. But again, I was mine to be general purpose, up to say 8 amps at 20-25 volts but probably much lower. The power it produces is cleaner than the pure output for the the rectifier but it's not setting any records - verified it on my scope. Yeah I could do it by guess and check but rather start in the ballpark.

Cheers.

13. Aug 18, 2012

### sophiecentaur

The supply voltage and ripple that you get when smoothing with a capacitor will depend upon three factors. The value of the capacitor, the resistance of the load and the resistance of the source (transformer winding and diodes).
The capacitor will discharge with a time constant of RLC, where RL is the load resistance. Assuming that the source resistance (RS) is substantially lower than RL, the charging time constant will be RSC. You can assume that is instant for a decent transformer.
The most relevant problem is usually the discharge time constant but, in the case of power supplies for high powered audio amps, for instance, when the demand is for high level, low frequency output, the power supply can 'sag' appreciably because the smoothing capacitor doesn't charge fast enough to keep up with demand.

The amount by which the Volts on the Capacitor will drop on each cycle in the simple case will be Exp(-τ/RLC) where τ is the time between charging pulses. This is a simplification which assumes that the C is charged to its peak voltage instantly at the start of each cycle. The value of τ will depend on the frequency of your mains supply and how many charging peaks there are per cycle. For a bridge rectifier this would be twice per cycle.

AS I wrote earlier - it would be a good idea to venture into a bit of theory rather than just suck it and see. It will help in the long run.

14. Aug 18, 2012

### mearvk

I don't want to delve deeply into the theoretical for what is a relatively practical question.

Does ordering matter with capacitors? Does 1x, 2x, 4x, 10x produce the same smoothing output as 10x, 2x, 4x, 1x? Do four 1x caps provide the same smoothing as 1 4x cap?

Do certain capacitances smooth certain frequencies better than others? If so, what capacitances work best with audio applications?

Thanks.

Last edited: Aug 18, 2012
15. Aug 19, 2012

### davenn

That may be so but unless you begin to grasp at least the basic fundamentals of how the capacitor works in a given circuit, you will continue to ask over and over the same old questions and you will drive people, trying to help you, nutz having to repeat themselves.

Yes they do, in general, the smaller the value of capacitance the higher the frequency
as far as audio applications go, it still depends on where in the circuit the capacitor is and what function you want it to perform... eg if its a bypass or say an interstage coupling capacitor.

Often in a regulated power supply you will see bypass capacitors of different values, have a look at this PSU ...

on the input to the voltage regulator IC you have C1 and C2
C1 1000uF --- large value for smoothing the DC out of the bridge rectifier
its bypassing the low frequency ( ~ 120Hz) AC ripple voltage to the 0V rail (negative rail)
C2 100nF (0.1uF) ----a small value... its there to inhibit any high frequency oscillations that may cause the regulator IC to become unstable, C4 of the same value does the same job

cheers
Dave

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16. Aug 19, 2012

### davenn

as a general rule of thumb, you use ~ 1000uF per 1 Amp of current that the PSU is designed to supply. So if your PSU that you state above was capable of supplying 8 Amps then you should be using ~ 10,000uF to provide really good smoothing. It doesnt hurt to have a bit more than needed. Also if your PSU as stated was capable of 25VDC output then those smoothing caps should be rated at twice that voltage ...ie 50VDC

Dave

17. Aug 19, 2012

### sophiecentaur

That's a bit like saying you don't want to get involved with arithmetic when you go shopping. Calculations go with the territory when you are designing and building electronics circuits. 'Big' capacitors are not cheap and, if you don't want to get involved with simple calculations, you could spend much more than you need or end up needing to add more capacity later on.

At mains frequency, you can connect capacitors in parallel, in any old way you want. The result is just the sum of the values. Capacitors for use at Audio frequencies (and that includes aspects of the power supply) are often used in pairs - one with a large C value, which will have significant inductance, used in parallel with a small C value and very low inductance. Look at circuit diagrams and you will often find a 100uF electrolytic and a 0.1uF ceramic, both connected to ground. There's a lot to be said for just copying such practices even if you're not sure why.