# Decoupling capacitor

Discussion in 'General Electronics Discussion' started by Darren Booker, Oct 22, 2017.

1. ### Darren Booker

11
0
Oct 22, 2017
Hi I need to keep a constant supply to an alarm module when the voltage switches
I need to fit a decoupling capacitor
Anybody any idea what I exactly need

2. ### Harald KappModeratorModerator

10,071
2,152
Nov 17, 2011
It depends on:
• The switching time (how long does the capacitor have to hold the voltage): dt
• The current consumption of the alarm module (more current means a bigger capacitor): I
• The allowed voltage drop during the switching time (a capacitor cannot hold the voltage completely at the powered level, a certain drop needs to be allowed for):dv

The capacitance can be calculated from C = I*dt/dv.
Chose the next bigger standard value for an added safety margin. The capacitor needs to be rated at the operating voltage plus some margin. For 12 V at least a 16 V capacitor is required, a 25 V capacitor is recommended (taking into account that this is for an alarm circuit where a bit of added reliability is worth every penny).

3. ### Darren Booker

11
0
Oct 22, 2017
The voltage hold would be rather quick lot less than half a second it will be just switching from the on timer relay to the off timer relay
The module consumes 300 milli amp maximum
Can you calculate for me

4. ### twister

172
7
Feb 12, 2012
I think the formula is
time to 60% of charge=CxR
Where C= capacitance in uf
R=resistance
T=time in seconds
Example 100ufx10M=1000seconds

5. ### Darren Booker

11
0
Oct 22, 2017
Im not really good with equations
What capacitor would I need
With the details I've provided
If some one could work it out for me would be much appreciated

6. ### Ratch

1,075
327
Mar 10, 2013
Have you considered a battery? If you can get the right voltage and capacity, your problem will be solved. Otherwise, the use you propose for a capacitor is not decoupling. So you want to supply 300 ma for 0.5 sec? What is the nominal voltage and how low can it decrease and still be acceptable. When I know that, I can calculate the capacitance needed to sustain that voltage range for that length of time while supplying that amount of current.

Ratch

7. ### Darren Booker

11
0
Oct 22, 2017
Nominal voltage would be 9 volts in a 12 volt system
It takes spproximately 0.1 seconds for the timer relays to switch over

8. ### Ratch

1,075
327
Mar 10, 2013
Let me get this correct. The system will operate at 12 volts until it switches to another voltage source. You do not want to go below 9 volts during the switchover. The time for the switchover will be 0.1 seconds, not 0.5 seconds. The current drain will still be 0.3 amps. Do I understand everything correctly?

Ratch

9. ### Darren Booker

11
0
Oct 22, 2017
Yes that's correct
Used 0.5 seconds as a switchover as a maximum but don't think its quite that much according to spec of module
Could you work out for 0.1 second switch over and 0.5 second switch over please

10. ### Ratch

1,075
327
Mar 10, 2013
OK, when the voltage is down to 9 V, and the current is 0.3 amps, the alarm looks like a 30 ohm resistor to the capacitor. We want the RC time constant to be about 5 times 0.1 sec, so that makes the capacitor value 16,667 mf. You may want to double that value because electrolytics are notorious for drying out and losing their capacitance. If the cap is 16.667 mf and the resistance is 30 ohm, the cap will go from 12 volts to 9.82 volts in 0.1 sec. If you want to sustain the same current for 0.5 sec to a drop down to 9.82 volts, you will need a 83.333 mf cap. Those are fairly big units. Are you sure you don't want to use a battery?

Ratch

11. ### Darren Booker

11
0
Oct 22, 2017
A battery is feesable
This would have to be a rechargeable one I presume and would it charge through the normal 12 volt system
What battery would I use?
Also do you have a link to were I can get these capacitors or battery

Last edited: Oct 22, 2017
12. ### Ratch

1,075
327
Mar 10, 2013
That is a question for your local parts supplier or a catalog, but not me, I would think the 12 volt system could keep the right kind of battery fully energized without over energizing it.

Ratch

13. ### motardo

1
0
Oct 22, 2017
Hi,

Assuming a constant current discharge of 300 mA, and C = Q/V = (I * t)/V
- for 0.1 seconds, C = (.3 A * .1 s) / (12 V - 9 V) = 10 mF
- for 0.5 seconds, C = (.3 A * .5 s) / (12 V - 9 V) = 50 mF

14. ### Ratch

1,075
327
Mar 10, 2013
It doesn't work that way. A capacitance that small would never be able to sustain that much current for that length of time. The cap would de-energize in an instant.

Ratch

15. ### Ratch

1,075
327
Mar 10, 2013
The usage determines what kind of cap it is. In your case, what are you bypassing? Caps that big are sure to be electrolytics.

Ratch

16. ### Darren Booker

11
0
Oct 22, 2017
It's just used to keep a supply to the module
But was under the impression that a decoupling capacitor is what was required
Also is the capacitor fitted in parralel on live and earth

17. ### Ratch

1,075
327
Mar 10, 2013
You need to learn about decoupling and bypassing.

Ratch

11
0
Oct 22, 2017
Ok thanks

19. ### duke37

5,364
769
Jan 9, 2011
The capacitors which have been specified are colossal. Do you mean μf rather than mF?

20. ### Ratch

1,075
327
Mar 10, 2013
Yes, microfarads. I never heard of anyone saying millifarads, although the prefix is correct.

Ratch