Discharge a Goldcap Capacitor

Thread Starter

nicmon

Joined Mar 18, 2011
30
I'm using a 1 uF, 5.5V Goldcap in my circuit which will be charged with a 3,3V current, Sometimes I would like to disconnect it from whole circuit, therefore I put a SPDT switch between 3,3V Source (comes from a LM1086) and the positive of Goldcap and a LED as indicator.

the problem is once I change the switch to OFF position, the LED remains on for a while (I know, this is the normal behaviour of a Cap :D)

referring to this article: http://www.allaboutcircuits.com/vol_6/chpt_3/17.html, can somebody help me and say which resistor can use there to discharge the Goldcap?

Thank you.
 

SgtWookie

Joined Jul 17, 2007
22,230
If the 1uF cap is the only one in the circuit, then a 33k resistor connected across its' terminals would bring the cap voltage below 1.5v in about 30mS. To you, it would appear that the LED went off nearly instantly.

While the circuit is powered, 3.3v across the 33k resistor is only a 0.1mA (100uA) load, so it won't be a significant power drain.
 

Thread Starter

nicmon

Joined Mar 18, 2011
30
Thank you very much, as I said, I'm using a SPDT to switch between with and without Capacitor modes in my circuit.
Do I harm the Capacitor if I put its legs be connected to the resistor whole time that I'm in "without cap" mode?

Sorry, for my noob questions.
 

SgtWookie

Joined Jul 17, 2007
22,230
The 33k resistor is so large that the current flow through it will be very small at 100uA, even at the full 3.3v voltage applied while the circuit is operating. That is 330 microwatts of power, or 0.00033 Watts; a very small amount of power.

You will not damage the capacitor.
 

strantor

Joined Oct 3, 2010
6,782
If the 1uF cap is the only one in the circuit, then a 33k resistor connected across its' terminals would bring the cap voltage below 1.5v in about 30mS. To you, it would appear that the LED went off nearly instantly.

While the circuit is powered, 3.3v across the 33k resistor is only a 0.1mA (100uA) load, so it won't be a significant power drain.
SgtWookie, how did you get this answer? Is it a formula using RC time constant or from a chart in the datasheet or what? I would like to know for the future how to calculate how long it will take a capacitor to discharge for a given load.
thanks
 

#12

Joined Nov 30, 2010
18,224
The formula is Ecap = dE (e^(-t/RC)) (I have it written on the wall behind my work bench.)

In English,
New voltage on the cap is the original voltage on the cap +/- [the change in voltage applied times e (2.7182818) to the (negative time over Resistance Capacitance)]

Theoretically, the voltage never gets to zero, but like Bernard says...close enough for government work.

People almost never use this formula bc it's easier to remember things like "10 time constants and it's empty" or, "one RC and its down two thirds". That's why the formula is written on the wall. I can never remember it.
 
Last edited:

Adjuster

Joined Dec 26, 2010
2,148
If the LED is remaining on for a significant time at present, is the capacitor really only 1 microfarad? Perhaps it's really one of those 1F memory backup caps
 

SgtWookie

Joined Jul 17, 2007
22,230
You would be surprised how long an LED remains illuminated once lit, albeit dimly, by even a small cap across it, and how large a resistance you can use to cause the cap to discharge rapidly enough.
 

strantor

Joined Oct 3, 2010
6,782
The formula is Ecap = dE (e^(-t/RC)) (I have it written on the wall behind my work bench.)

In English,
New voltage on the cap is the original voltage on the cap +/- [the change in voltage applied times e (2.7182818) to the (negative time over Resistance Capacitance)]

Theoretically, the voltage never gets to zero, but like Bernard says...close enough for government work.

People almost never use this formula bc it's easier to remember things like "10 time constants and it's empty" or, "one RC and its down two thirds". That's why the formula is written on the wall. I can never remember it.
There's one for my wall. thanks.
 

#12

Joined Nov 30, 2010
18,224
I never know which bits will be the most helpful, or who needs them, but it's always nice to get a "thank you".

#12
 

Adjuster

Joined Dec 26, 2010
2,148
You would be surprised how long an LED remains illuminated once lit, albeit dimly, by even a small cap across it, and how large a resistance you can use to cause the cap to discharge rapidly enough.
I believe that "Goldcap" normally refers to a Panasonic double-layer "supercap" product - they are not made in sizes as small as 1μF.

Of course, the OP may have something else in mind, and in these days of Chinese copies tradenames can't be relied upon so much.

http://panasonic.com/industrial/ele...arch.aspx?src=/www-ctlg/ctlg/qABC0000_AM.html
 

SgtWookie

Joined Jul 17, 2007
22,230
OK, our OP did say 1uF in his first post. If they really meant 1F instead, then it'll take about 8-1/3 hours for that same 1 time constant... :rolleyes:

In that case, our OP could instead use a SPDT switch, the common being the input to charge the cap, and the two other terminals switching either to power or ground. Using a light bulb in series with the common to the cap (say like an automotive brake lamp) would limit the maximum current while charging or discharging the cap, but would have a very low resistance once the cap was charged or discharged.
 

Thread Starter

nicmon

Joined Mar 18, 2011
30
upps, they are 1.0F indeed (and not 1uF), so I must ask you again about the correct value of resistor.

Can I use another value for resistor, e.g. 470K as I don't have a 330K in the hand?
 

#12

Joined Nov 30, 2010
18,224
I'd say you can use any value from 47 ohms 1/2 watt up.
I'd also say I gave you the equasion to figure it out yourself in post #9.
 

SgtWookie

Joined Jul 17, 2007
22,230
I'd still suggest that you use an incandescent lamp in the charge and discharge path for the cap. Incandescent bulbs have a non-linear resistance; when they are cold, the resistance is very low. As they heat up, the resistance increases a great deal.

If your cap is completely discharged, the regulator on your input will be very stressed due to the high current required to charge the capacitor.

Here's one way to connect it up:


I don't know why you need such a large capacitor, but perhaps you'll explain that.
 

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