How to use capacitors in regulator circuit

Thread Starter

CoderJohn

Joined May 22, 2014
16
Ok this question I have now is not much to do with the pins/resistors, but how are capacitors used to stabilize voltage peaks/drops? After they are charged up the capacitor doesn't allow current flow anymore.... so are they just stuck to ground in parallel to the vcc of the integrated circuit?

How is it that they stabilize voltage? Is it because if the voltage drops quickly the capacitor will discharge to provide equilibrium voltage potential while the battery has time to catch up or what? If I was to use one with a voltage regulator do you typically want to place them after the regulator? How do you know what size capacitor to use in terms of Farads....
 

#12

Joined Nov 30, 2010
18,224
Use the capacitor named in the datasheet for the regulator. There are instability ranges you try to get above. I recently saw one that was unstable at 1 nf but stable for all frequencies with 1 uf...so that's what you use.
Is it because if the voltage drops quickly the capacitor will discharge to provide equilibrium voltage
Yep. Pretty much, but its about nanohenries in the circuit board traces. You can't do microsecond switching with a nanohenry in series with the power to the chip. The local capacitor supplies the tiny amount of current for a tiny amount of time.

This is what I use for datasheets:http://datasheetcatalog.com/
 

#12

Joined Nov 30, 2010
18,224
Note to Moderators. I accidentally answered a necropost and then reported the hijacking to you. Don't tell me I hijacked the thread.:mad:
 

wayneh

Joined Sep 9, 2010
17,496
After they are charged up the capacitor doesn't allow current flow anymore.... so are they just stuck to ground in parallel to the vcc of the integrated circuit?

How is it that they stabilize voltage?
It's true that a capacitor can't pass DC current at steady state, but it can absorb or source current during transients, until its capacity is expended.
 

hp1729

Joined Nov 23, 2015
2,304
Ok this question I have now is not much to do with the pins/resistors, but how are capacitors used to stabilize voltage peaks/drops? After they are charged up the capacitor doesn't allow current flow anymore.... so are they just stuck to ground in parallel to the vcc of the integrated circuit?

How is it that they stabilize voltage? Is it because if the voltage drops quickly the capacitor will discharge to provide equilibrium voltage potential while the battery has time to catch up or what? If I was to use one with a voltage regulator do you typically want to place them after the regulator? How do you know what size capacitor to use in terms of Farads....
Reservoir capacitors
Those high microfarad caps in power supplies. Be mindful that the line power does not drive the load at the end. The rectified AC charges up the reservoir cap in high current spurts and the load draws the charge off of the capacitor. Think reservoir lakes. Rains fill up the lake in spurts and the river draws from the lake at a lower consistent rate.
 

Thread Starter

CoderJohn

Joined May 22, 2014
16
What if I wanted to use a switch to manually operate the clock signal for testing purposes? I know that in the tiny amount of time of the switch making contact there is usually a dip/spike. What size capacitor is generally recommended for this? I wouldn't think more than 1 uF would be needed to fix that but I want your guys opinions.
 

dl324

Joined Mar 30, 2015
16,846
I know that in the tiny amount of time of the switch making contact there is usually a dip/spike. What size capacitor is generally recommended for this? I wouldn't think more than 1 uF would be needed to fix that but I want your guys opinions.
Depends on the value of the series resistance. You want the time constant to be long enough for the switch contacts to stop bouncing.
 

Thread Starter

CoderJohn

Joined May 22, 2014
16
I'm not sure of the time the switch bounces.... but if I were to assume a millisecond then would I just use

t = RC
.001 = R(1uF)
R = 1000 ohms

Is that how I would calculate such a thing?
 

dl324

Joined Mar 30, 2015
16,846
I'm not sure of the time the switch bounces.... but if I were to assume a millisecond then would I just use

t = RC
.001 = R(1uF)
R = 1000 ohms

Is that how I would calculate such a thing?
You got it. If it were me, I'd make it 100mS or so to have some safety margin. How long it can be depends the application.
 

Thread Starter

CoderJohn

Joined May 22, 2014
16
Alrighty, I just finished this image so it doesn't replicate the 100ms but is this how it would be wired....
Assume the black box is the device that it would lead into.
 

MrChips

Joined Oct 2, 2009
30,712
Ok this question I have now is not much to do with the pins/resistors, but how are capacitors used to stabilize voltage peaks/drops? After they are charged up the capacitor doesn't allow current flow anymore.... so are they just stuck to ground in parallel to the vcc of the integrated circuit?

How is it that they stabilize voltage? Is it because if the voltage drops quickly the capacitor will discharge to provide equilibrium voltage potential while the battery has time to catch up or what? If I was to use one with a voltage regulator do you typically want to place them after the regulator? How do you know what size capacitor to use in terms of Farads....
A capacitor is one of the three basic passive components - resistor, capacitor and inductor.

The properties of a capacitor can be stated in three simple statements:

1. A capacitor stores charge.
2. A capacitor impedes low frequency signals (including DC).
3. A capacitor admits high frequency signals (AC).

There is a lot more than meets the eye. There is a lot more about capacitors that depend on the specific application.

In an unregulated DC power supply, the function of the capacitor is to store charge, hence the term "reservoir capacitor". Capacitance values can be anywhere from 100μF to 10000μF (10mF) depending on the load current and how much ripple one can tolerate.

The ripple voltage can be estimated using the formula:

ripple voltage = load current /(2 x frequency x capacitance)

Hence to determine the capacitance required, we work backwards:

capacitance in Farads = load current in Amps /( 2 x frequency in Hertz x ripple voltage in volts)

For example, let us assume,

load current = 1.2A
frequency = 60 Hz
ripple voltage = 1V

capacitance = 1.2/120 F = 10,000μF or 10mF

http://www.zen22142.zen.co.uk/Design/dcpsu.htm

When using an IC voltage regulator, the rule of thumb is to use the reservoir capacitor as calculated above PLUS two capacitors as recommended in the data sheet of the IC regulator, one capacitor on the input side and another capacitor on the output side of the regulator.

Typical values are 10μF on the input and 0.1 to 1μF on the output.

What you cannot see is the inductance in your physical circuit depending on your wiring practice and PCB layout.

In some application notes you might see two capacitors placed in parallel, 10μF and 0.1μF. Sometimes you might see three capacitors, 10μF, 10nF and 100nF.

Wouldn't 100nF added to 10μF make 10.1μF?

The straight answer is NO.

First of all, the tolerance of a 10μF electrolytic capacitor could be 10% or 20%. This vastly nullifies placing any additional capacitance lower than 1μF in parallel with the first. What you don't see is the effect of internal and external resistance and inductance on the performance of the capacitor. The additional low value capacitors will have different frequency responses, i.e. their role is to absorb voltage fluctuations occurring at higher frequencies, even in MHz and tens of MHz ranges. Board layout plays a crucial role in such applications.

You can learn more about this here:

https://www.intersil.com/content/dam/Intersil/documents/an13/an1325.pdf

Bottom line is - There is much more to it than meets the eye.
 

MrChips

Joined Oct 2, 2009
30,712
What if I wanted to use a switch to manually operate the clock signal for testing purposes? I know that in the tiny amount of time of the switch making contact there is usually a dip/spike. What size capacitor is generally recommended for this? I wouldn't think more than 1 uF would be needed to fix that but I want your guys opinions.
The straight answer is it doesn't work.

Switch bounce is a well known phenomenon and a nuisance. A simple hardware solution is to use a SPDT push-button and an R-S flip-flop using two NAND gates.

For more in-depth information on debounce techniques, see this:

http://www.eng.utah.edu/~cs5780/debouncing.pdf
 
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