Capacitance and Coulombs Calculation

The SI units for capacitance is C/V (coulomb / volts) and a coulomb is A*s (amps * seconds).
So capacitance = (amps * seconds) / volts

If I want to know how much current a particular capacitor can supply for a determined amount of time, do I calculate it this way:
C: 6800uF V: 75v s: 5-second
therefore amps = 102mA for 5-seconds
Is this correct?
Another question: the voltage will decrease with time. How would I power say an IC chip for some amount of time since the voltage decreases over time? I have an old Sears stereo and the preset stations lose their memory if it's left off for a week or two.
When I was younger, I assumed a capacitor kept a chip charged, but doing the math, it would take a very large capacitor.
Thanks in advance.
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| The SI units for capacitance is C/V (coulomb / volts) and a coulomb is | A*s (amps * seconds). | | | So capacitance = (amps * seconds) / volts | | | | If I want to know how much current a particular capacitor can supply for | a determined amount of time, do I calculate it this way: | | C: 6800uF V: 75v s: 5-second | | | | therefore amps = 102mA for 5-seconds | | | Is this correct? | | | Another question: the voltage will decrease with time. How would I power | say an IC chip for some amount of time since the voltage decreases over | time? I have an old Sears stereo and the preset stations lose their | memory if it's left off for a week or two. | | When I was younger, I assumed a capacitor kept a chip charged, but doing | the math, it would take a very large capacitor.
There are some very large capacitors intended for things just like this. They have ratings in the many whole farads. They are also rated at low voltages and low currents.
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Do research on "time constant". That describes how capacitance charges and discharges over a time period.
Don Young
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Steve wrote:

I'll bet that there is some sort of hold-up circuit, probably involving a battery. If it's a NiCad, it may have lost its capacity. Some systems used a non-rechargable Lithium battery (coin style, like PCs use for RT clocks). Pop the cover off and look for a battery. Replace it. Barring any circuit failures, all will be well.

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----------------------------

No: Both the current and voltage will change with time and how fast this occurs depends on the load resistance. Low resistance and the current will be high and die down quickly If you have a capacitor C charged to some voltage V and you connect it to a load resistor R, the current will be (V/R)* e^-(t/R*C) and the voltage will be I*R Initially the current would be V/R. For example: if your 6800uF capacitor at 75V was connected to a 1K load the initial current would be 75ma and the current would decay exponentially so that at 6.8 seconds it would be would be 27.6ma, at 13.6 seconds it would be 10ma and down to about 0.5 ma in 34seconds. If the load R was larger the time to decay would be more but the current would be proportionally less.
I can see an alternative (my radio has a back up battery just for this purpose) but some knowhow would be needed to add this if there is no provision for it and it may not be worth the cost to have someone do it and I am not recommending that you do it yourself unless you actually know what you are doing.
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I think you can use a sample and hold circuit for your application. It volves components like capacitor, FET and OP AMP.Also it is easy to construct.
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I think you can use a sample and hold circuit for your application. It involves components like capacitor, FET and OP AMP.Also it is easy to construct.
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