# Why am I experiencing instantaneous capacitor discharge?

#### koya-47

Joined Oct 15, 2021
22
Hi, I have a circuit with a capacitor and resistor in series, and a bulb in parallel (it was more complex but I simplified it to try identify what is happening with the capacitor). When I attempt to charge then discharge the capacitor it discharges instantaneously. I've tried increasing the time constant by increasing the resistance but it still discharges instantly. The values are R=10k ohms/10M ohms (tried the 10 M ohm resistor to increase time constant), capacitance = 0.33 uF and up to 370 V charge capacity, voltage source = 24V. For the time constant I used T=RC and got about 3 seconds.

When I disconnect the voltage source, the voltage across any component instantly falls to 0 V and the bulb turns off instantly. I've tried multiple capacitors in parallel, ensured the resistor is providing the rated resistance, allowed the capacitor to charge for over 20 minutes (even though it should take about 15 seconds based on my calculations). I can't figure out what else is wrong.

I'm no expert with circuits, I only started recently so I might be missing something obvious here. I would hugely appreciate if anyone here could point out what I'm doing incorrectly or how I could stop my capacitor discharging instantly.

#### MrSalts

Joined Apr 2, 2020
2,767
draw your circuit. A bulb has very low resistance and, when in parallel with your resistor, the total resistance becomes even lower resistance than the bulb alone (even if your resistor is 1M ohm). If I understand your description, the parallel resistance of bulb and resistor explains your short time constant but, schematics are the language of this hobby, please clarify your question with a schematic.

#### ElectricSpidey

Joined Dec 2, 2017
2,264
Is the bulb in parallel with the cap and resistor or just the cap?

#### ElectricSpidey

Joined Dec 2, 2017
2,264
When you disconnect the power the cap, resistor and bulb form a series circuit, and the bulb will not light with a 10k resistor in series with it.

#### Papabravo

Joined Feb 24, 2006
19,827
A: You have a low impedance path to ground.
I can assure you that it is not instantaneous. It may perhaps be a rapid exponential decay (discharge), but that is quite far from instantaneous.

#### BobTPH

Joined Jun 5, 2013
6,280
Need a schematic to say anything useful, Your word description is open to interpretation.

Also, what kind of bulb? You are not going to light even an LED for long with a 0.33 uF capacitor.

Bob

#### koya-47

Joined Oct 15, 2021
22
draw your circuit. A bulb has very low resistance and, when in parallel with your resistor, the total resistance becomes even lower resistance than the bulb alone (even if your resistor is 1M ohm). If I understand your description, the parallel resistance of bulb and resistor explains your short time constant but, schematics are the language of this hobby, please clarify your question with a schematic.
Here is the circuit diagram of the circuit I am attempting to use.

And here is a circuit diagram of another circuit I attempted.

The problem I have with the second diagram is the circuit is supposed to maintain power to a device during momentary power losses (device will be in place of the bulb). The device has a resistance of about 460 ohms, requires a voltage between 8 V and 24 V and draws a current of up to 0.1A from the power supply. When the resistor is in series I don't seem to get that amperage.

#### koya-47

Joined Oct 15, 2021
22
A: You have a low impedance path to ground.
I can assure you that it is not instantaneous. It may perhaps be a rapid exponential decay (discharge), but that is quite far from instantaneous.
Are there any ways I can lengthen the rate of decay whilst maintaining a decent current supply to the bulb? I posted another reply above with the details, ratings and circuit diagrams.

#### MrSalts

Joined Apr 2, 2020
2,767
I don't know what k/M Ω is. Is that kilo or mega or ...?

if kilo, then your time constant is 0.33 uF x 10k = 0.00000033F x 10000 = .0033 seconds for your 24v to drop to 8v and will have essentially dropped to zero in 5x that much time (0.015 seconds) which is also known as "immediately" I measured by the human eye.

if it is 10M ohm, then the time constant is 10,000,000 ohms x 0.0000033 =3.3 seconds to drop to 8 volts.

unfortunately, in both circuits, you have a rather large resistor (10k or 10M ) in front of the bulb so you will not really see any light from it because the resistor will limit the current to a very low value.

#### Papabravo

Joined Feb 24, 2006
19,827
The 0.33 uf capacitor and the 10 K resistor have an RC time constant of 3.3 milliseconds. It takes about 5 time constants to charge or discharge the capacitor. That is about 16.5 milliseconds. That would sure look instantaneous if you had now way of actually measuring how long the discharge takes. Do know why the product of R and C is called a time constant, and has units of seconds?

Joined Jan 15, 2015
7,135
Don't you want your lamp in parallel with your capacitor? When S1 is closed the capacitor will charge through R1. When power is removed the capacitor will discharge through the lamp. The larger the capacitance the longer the lamp will remain lit and your cap is way to low of a capacitance, as was just pointed out, to hold the lamp lit very long. Then too, I may not understand what you are actually expecting?

Ron

#### koya-47

Joined Oct 15, 2021
22
I don't know what k/M Ω is. Is that kilo or mega or ...?

if kilo, then your time constant is 0.33 uF x 10k = 0.00000033F x 10000 = .0033 seconds for your 24v to drop to 8v and will have essentially dropped to zero in 5x that much time (0.015 seconds) which is also known as "immediately" I measured by the human eye.

if it is 10M ohm, then the time constant is 10,000,000 ohms x 0.0000033 =3.3 seconds to drop to 8 volts.

unfortunately, in both circuits, you have a rather large resistor (10k or 10M ) in front of the bulb so you will not really see any light from it because the resistor will limit the current to a very low value.
It was intended to be kilo or mega since I stated I tried both in the question. I had the same calculations as you so glad to know I was on the right track.

Is there any way I can go about designing this circuit so that the capacitor will discharge with a time constant over 1 second but also provide a current of at least 0.1A?

#### LowQCab

Joined Nov 6, 2012
2,880
There is no such thing as "a device",
the Load has various Specifications, not just "Resistance".

An Incandescent-Light-Bulb is not a good "Dummy-Load" because its
Resistance changes all over the place,
depending upon the temperature of the Filament.

Why are You using only a 330nf Capacitor ?
For your purpose, something in the range of 1000uf + plus would be called for.

This Circuit will not work with any Resistors.
Resistors will only degrade its performance.
.
.
.

#### koya-47

Joined Oct 15, 2021
22
Don't you want your lamp in parallel with your capacitor? When S1 is closed the capacitor will charge through R1. When power is removed the capacitor will discharge through the lamp. The larger the capacitance the longer the lamp will remain lit and your cap is way to low of a capacitance, as was just pointed out, to hold the lamp lit very long. Then too, I may not understand what you are actually expecting?

View attachment 250397

Ron
The lamp is just a visual indicator that the circuit is working as intended. The actual circuit will have a device connected where the lamp is and the capacitor is intended to power the device during very short power losses (no longer than a second or two). The device resistance is about 460 ohms so wouldn't the capacitor discharge swiftly if there is no resistor next to the lamp?

I was hoping I could find a solution without needing to replace the capacitors but I am learning now that, as you suggested, perhaps my capacitance is too low.

#### koya-47

Joined Oct 15, 2021
22
There is no such thing as "a device",
the Load has various Specifications, not just "Resistance".

An Incandescent-Light-Bulb is not a good "Dummy-Load" because its
Resistance changes all over the place,
depending upon the temperature of the Filament.

Why are You using only a 330nf Capacitor ?
For your purpose, something in the range of 1000uf + plus would be called for.

This Circuit will not work with any Resistors.
Resistors will only degrade its performance.
.
.
.
Thank you for the information. I was using some extra components I had lying around to try test the theory of the circuit before ordering any parts. I will try a higher capacitance without resistors.

#### koya-47

Joined Oct 15, 2021
22
Thank you everyone for all the comments, I feel like I've learned a lot. I will try using a larger capacitor and post the results once I've tested it.

#### MisterBill2

Joined Jan 23, 2018
14,199
First, the cappacitance is much too small to provide any longer hold up time for the voltage. The supply that I had created for one project had several large capacitors, each 10,000 microfarad, charged to 24 volts for one output and to 12 volts for the other output. That amount of storage would power a small transistor radio for several minutes. Those capacitors were about six inches long and about three inches in diameter, rated at 30 volts working voltage.
Today there are "supercaps" that have far greater capacitance but at a much lower voltage rating. But physically they are a lot smaller.
And no, your discharge is not instant, it is just fast because of a very short time constant, which is several milliseconds.

#### Tonyr1084

Joined Sep 24, 2015
7,247
the circuit is supposed to maintain power to a device during momentary power losses (device will be in place of the bulb)
The bulb is a poor substitute. Without knowing what the device is and its characteristics you can't get an answer to your question.
Don't you want your lamp in parallel with your capacitor? When S1 is closed the capacitor will charge through R1. When power is removed the capacitor will discharge through the lamp. The larger the capacitance the longer the lamp will remain lit and your cap is way to low of a capacitance, as was just pointed out, to hold the lamp lit very long. Then too, I may not understand what you are actually expecting?

View attachment 250397

Ron
Ron: R1 will limit the current. The bulb may not light. The cap will hardly charge because the bulb will drain it faster than it can charge.

WITHOUT R1, the cap will charge and the lamp will remain illuminated for a period of time as yet undetermined. Given that we've been told it's an incandescent light bulb, we can't determine anything without accurate information.

NOW, if you want to use a light emitting diode (LED) as an indicator, there must be some modifications to your circuit. Assuming we're talking about a standard 5mm LED with 20mA current on a 24V source, you'd need a 1.2KΩ resistor. The cap in parallel with the LED should charge to the same voltage as the LED and when power is interrupted the LED should remain illuminated for some period of time, depending on the value of the cap. And this is a quick and dirty explanation. There are yet more things to be considered when building this sort of circuit.

If you want to power a device from a cap for a short period of time then please be forthcoming with the information that will yield a better understanding of what you want to do and result in a good answer. For now - we're flying in the dark amidst some pretty tall mountains. Success is not expected.

#### crutschow

Joined Mar 14, 2008
31,493
The device resistance is about 460 ohms
The answer, per you RC equation, is staring you in the face.
You just didn't use the correct R.
For a one second time constant the capacitor would need to be 1s / 460Ω = 2.18mF or 2,180µF.

#### MisterBill2

Joined Jan 23, 2018
14,199
Actually, to avoid having an excessive voltage drop during the momentary power loss, the capacitor value should be quite a bit more. I am guessing that the acceptable drop is less than ten percent. And one more consideration is that the load must be able to operate on DC, because that is what the cap will be supplying.
And I just now noticed that I don't see a reference to what the 460 ohm load voltage is and wants to be maintained at. I had guessed that it was the mains voltage but now I am seeing that is probably not the case. So just what voltage does the TS want to hold for a second or a few seconds, when the mains power fails?? That matters quite a lot.