# Cpacitor discharge

Discussion in 'Electronic Design' started by Joe G \(Home\), Feb 11, 2007.

1. ### Joe G \(Home\)Guest

Hi All,

I am hope some can confirm statements below.

For a capacitor

Q = C * V = I * t

I am trying to work out how long an electrolytic will hold up the Vdd on
circuit.

nominal Vdd = 3.3v and lowest Vdd = 2.4v

If the current drawn by the circuit is constant 10mA = I

C * V = I * t

Thus t (time) = (C * V) / I Where V = 3.3v - 2.4v =
0.9 v

Hence for a 10,000uF the time the Vdd rail is held up is

t(time) = (10,000 E-6 * 0.9) / 0.01 = 900mS

QUESTIONS

Q1
Are my statements above correct?

Q2
Should I use the natural e discharge model V = 3.3v * (1 - e(t))
style of equation?
any ideas on this equation would be appreciated.

Joe

2. ### John PopelishGuest

I think they are.
Not if the current is constant. That form applies to a
capacitor discharged by a resistor, so that as the voltage
falls, so does the current.
Another handy form of the capacitor equation that relates
voltage and current is I=C*(dv/dt). The current is
proportional to the time rate of change of voltage. If the
current is not a constant (where the C * V = I * T solution
fits) or proportional to the total capacitor voltage
(where the exponential solution fits) this completely
general form may allow a solution.
Note that there are non electrolytic capacitors for low
voltages that have capacitances very much higher than 10,000
uF for low current supply hold up applications. e.g.:
http://www.elna-america.com/products/dynacap/dynacap_frame.htm

3. ### mpmGuest

Should I use the natural e  discharge model      V = 3.3v  * (1 - e(t))
Yes. For a constant current charge or drain, the charge/discharge
rates will vary with instantaneous voltage. See equations here: (Do I
don't have to type them in....)

http://www.rowan.edu/colleges/lasold/physicsandastronomy/LabManual/labs/Capacitor.pdf

I actually just researched this exact same question two weeks ago...
Keep in mind electrolytics can easily vary +/- 20% or more in value,
and also age.
This will very likely affect your timing calculations.

Also, I happened upon a neat little online simulator where you could
(for free) "virtually" build your circuit online and run some simple
simulations. This might help you verify your assumptions / math.

Visit: http://www.falstad.com/circuit/index.html

-mpm

4. ### EeyoreGuest

Not if the load is indeed constant current.

Graham

Er NO !

Graham

6. ### mpmGuest

Er NO !
Did I screw this up??
Maybe I didn't understand his question.

Take a cap, charge it, disconnect it and let is sit there static.
Now, let's say it's charged to 3.3 votls.

He wants to run a uPC (or some such) that will operate from 3.3 Volts
down to about 2.4 Volts.
He wants to know how long it will take for the cap to reach 2.4 Volts.

And he's going to draw a constant 10mA current.
(And the resistance of the uPC is not changing.)

Therefore, the discharge function of the capacitor must be a function
of the instantaneous voltage on the capacitor. As the capacitor
drains, less voltage is applied to the load, so the current through
the load is likewise reduced.

7. ### Tom BruhnsGuest

As stated, the current is constant, not dependent on voltage. In that
case, the capacitor voltage will be a linear ramp. However, if indeed
he's running some cmos circuit whose power supply current is
practically all used to charge gate capacitances, the current will
almost certainly be even more strongly dependent on power supply
voltage than it would if the load were a resistor.

Cheers,
Tom

8. ### EeyoreGuest

You said -A. " he's going to draw a constant 10mA current "
then B. " As the capacitor drains, less voltage is applied to the load, so the
current through
the load is likewise reduced. "

Either it's constant current or not.

The exponential discharge is with a resistive load. Most ICs tend to demand a
near constant current regardless of supply voltage btw.

Graham

10. ### Joe G \(Home\)Guest

Thanks all who replied.

My assumption is the load will draw a constant 10mA.

My load is a microcontrolller circuit.... and I have made a big assumption -
constant 10mA.

I will have to either build or model the microcontroller circuit to see
f - over the range 3.3v to 2.4v the current is constant.

Regards
Joe

11. ### Jan PanteltjeGuest

Sounds correct.
No, because you discharge wit ha _constant_ current, not in a resistor.

13. ### EeyoreGuest

Introducing a strawman is so totally predictable for you.

Do you have any other tricks you'd like to perform for us ?

Graham

14. ### Fred BloggsGuest

Right, if it's a CMOS digital then I=Cp*F*Vdd where Cp is the so-called
equivalent power dissipation capacitance. This makes CpF the load
conductance or an equivalent discharge resistance of Re=1/(CpF). So the
hold up time is derived from 2.4=3.3*exp(-t/CRe) or t=ReC*Ln(3.3/2.4) or
t=C*Ln(3.3/2.4)/CpF, which means it will hold up longer than the
constant current drain model, so he should stay with his original method
if he doesn't mind a bit oversized component.

15. ### Tom BruhnsGuest

If it's op amps, the current probably will be very nearly constant,
assuming the signal levels stay the same. If it's the input to a
regulator, the output current will presumably be constant so long as
the input doesn't drop below what's required to hold up the output
voltage. But if it's a whole mess of cmos gates operating directly
off the supply at a constant frequency, the current certainly will
depend on the supply voltage. HOWEVER, the OP DID STATE that the
current is (assumed) constant.

Assuming the current is constant will adequetely cover the case where
current decreases with voltage, of course, and unless space is very
tight or the production volumes are huge, it's cheap insurance to just
assume the current is constant. But there's one case where you could
get burned: if that input voltage goes to a switching regulated
supply, the current will INCREASE as the input voltage drops. You can
get fancy and look at the efficiency of the particular supply as a
function of input voltage for the particular output, or just assume
constant input power and probably be OK.

Cheers,
Tom

16. ### Andrew EdgeGuest

The only way way you can get a current out of a capacitor is by using
a resistor.It may have a small value but a resistor it is. All
currents and voltages in an RC circuit
have the form
i= io*exp -t/Req*C
v=voexp-t/Req*C

Andy

17. ### John PopelishGuest

(snip)

You might want to rethink that generality a bit. The
universe is more complicated than that.

18. ### Jan PanteltjeGuest

No way, 'current' equals electrons.
There are many ways other then 'resistors'.
Look up vacuum tube, transistor, semiconductor, and in this context
especially 'current source'.

To make it a bit more clear, in a _resistor_ the current decreases linear
with the increase in resistance.

In something that allows control of the current, say a transistor, or tube,
the current does _not_ rise proportionally to the applied voltage.

In a (junction) transistor the Ic becomes greatly independent above some minimum
voltage, and then only depends on Ib.
Same for a 'penthode' tube.
Many integrated circuits use current sources and current mirrors, and as such
Isupply may depend little on Usupply.

This is one for [sci].electronics.basics.

19. ### Andrew EdgeGuest

Ouch ... That one has a damn large resistance ... Infinity!
..

You said it.

Andy

20. ### Andrew EdgeGuest

It sure is. I am always willing to learn ... so give me an example.

Andy

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