Calculate capacitor replacement value?

Time will be proportional to C.

Leakage could be a problem out to 150 minutes. Good electrolytics could be OK.

I like nice mechanical twist timers. Mo uses her iPhone.

Thanks John, appreciate the fast reply. I?ll get out to the shed th is evening and try that simple formula. I was expecting something with expo nents or antilogs...

The voltage on the cap never gets high, so the bottom of the charge curve is almost linear. So you can assume that the cap would charge linearly to Vcc in R*C seconds. Good enough for cooking.

If you needed to charge to a higher voltage, it would get nonlinear and need more math. I'd just Spice it.

o, I found the culprit was the capacitor, nominally 1500uF but now lifeless . Its actual original value is unknown. There are 8 markings on the knob of the timer?s linear pot (1 to 8 minutes) and 11 on the knob of a 1P x12W rotary switch (from 10 to 150 minutes) . As a first (arguably optimist ic) first step I want to calculate its value, for the replacement. It? ?s not practical to replacie the pot and all 11 resistances, some of them with more than one resistor and/or reposition all 19 markings individually . Nor does my temperament allow me to bin it.

s 1.45V. If I stop-watch the time (in seconds) taken to trigger the alarm w ith a known value of C (uF) and R (megohms), what formula will then give m e the new value of C needed to change that time to its original value, as m arked please?

many from the longer duration rotary switch values as patience allows.

ver value is calculated. Will that have a detrimental effect on leakage?

It should be 1.45 = 9 x (1 - exp(-T/RC)) from which you solve T= - RC x LN (1 - (1.45/9)), from which you find T1/T2= C1/C2 or

C1= T1 x C2/T2, meaning if you measure T2 with known C2, then you need a new C1 for desired T1 in the ratio C1 = T1 x (C2/T2)

That looks better, thanks.

John: I did not get a result agreeing with your formula on the few tests I ?ve done so far.

For example, with C = 1090 uF and R = 790 K the alarm triggered after 4

Terry, East Grinstead, UK

That is a gross difference which indicates something else is influencing your measurements. Make sure your 9V supply is accurate. Check for parallel leakage paths. Measure your capacitor leakage (is it installed with correct polarity?)

You can find the discrepancy if you look hard enough.

I think you must be right. Fred's formula gave 146 s, close to John Larkin's simpler estimate based on linearity of the lower part of the curve.

The supply is 9.0 V, from a professional bench unit. Not sure about "parallel leakage paths" and if I drew a circuit 20 years ago I cannot find it, so testing that will be difficult. And as for capacitor leakage (presumably the 'ESR' my reading covers?) I have no equipment to measure that. Given that this is a DC circuit, are there simpler rough tests for leakage?

That particular electrolytic (unused but maybe 5-10 years old) looked OK, but my next step tomorrow will be to try a few others of similar value.

Terry, East Grinstead, UK

ts I?e done so far.

ter 409 s. But the formula predicts 138s. I measured the voltage on the cap after triggering as 1.45V.

Leakage contributes to longer than expected charge times, not less, so it's not that. Maybe you're misreading a resistor or capacitor value? Or someth ing else in the circuit is providing charge to the capacitor, something you 've forgotten about.

The exponential solution gives 1.45 volts in 150 seconds. My linear approximation makes 1.45 volts in 138 seconds, which I figured was close enough.

If it actually takes 409 seconds, there is probably some serious capacitor non-ideal behavior. Leakage or dielectric absorption maybe.

A better math hack would assume that the average supply voltage is

1.45/8.3 * R * C = 150.4 seconds.

Not ESR, but parallel resistance.

Charge the cap with a flashlight battery (or a power supply!) for a few seconds, to 1.5 volts, and measure the cap voltage with a DVM. Then disconnect the DVM and set the cap aside for an hour. Then measure it again. A good cap won't self-discharge much.

Or apply 9 volts to the cap through 790K and plot the cap voltage vs time, with a DVM. A 10M DVM won't load things much.

Any leakage reduces the current that's available to go into the capacitance, so it charges slower.

If the thevenin charge voltage were 1.46, it could take a really long time to reach 1.45.

Maybe the cap can be reformed. Apply rated voltage overnight.

?? It WAS a ?longer than expected? time.

Electrolytics... -20% +50% tolerance ??

Thanks for the follow-up advice.. Back on the case to pursue those suggestions Saturday morning.

Terry, East Grinstead, UK

t = .175677 * R * C with no hacks involved.

Sure, but that's too easy.

We like to do this sort of stuff standing up, in our heads.

ests I?e done so far.

after 409 s. But the formula predicts 138s. I measured the voltage on the c ap after triggering as 1.45V.

ng

's not that.

We don't know which of the 2 times has been shortened more by leakage, look s like it's the first cap not the 2nd.

NT

Good to try different capacitors. Do you have a volt-ohm meter that can measure current?

If so, attach one probe to your power source and one to the capacitor (when out of circuit) and the other end of the capacitor to the opposite end of the source. Observe polarity. Start with your current meter on a high range and work down.

It takes about 165k of equivalent parallel resistance to get your measured time (assuming your capacitor of 1090 uF is correct).

I also takes only 6.5uA of leakage to do the same.

Measure things.

Please excuse me for being a bit pedantic.