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Circuit diagram

Hello, here I attached a diagram of a series circuit containing a switch, 9 V bulb, resistor, and a capacitor.

When the switch is closed, the capacitor gradually starts charging from 0V, and hence, the bulb also glows. After some time, the capacitor charges to a voltage equal to the supply voltage 12 V and no current flows throughout the circuit anymore. Hence, the bulb turns OFF completely.

However, when the switch is opened, the capacitor still maintains its full charge of 12 V.

Now, my question is ... I want the capacitor to be DISCHARGED immediately whenever the switch is opened, and drops to 0 V. So that when the switch is closed again, the capacitor starts gradually charging to full voltage. And, when the switch is opened again, it will immediately discharge to 0 V .

Please help me with this discharging circuit which I can not think off my head. I really need a discharge circuit which will accomplish this discharging without using any ICs or chips. But, if possible, my preference is using relays, transistors or optocouplers, etc., etc.

Waiting eagerly for the discharge circuit diagram. Thanks.

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    \$\begingroup\$ With the switch located where it is and no other switches, the task seems to be impossible. After the switch has been closed for a while, both the current and the voltage across it will be zero, meaning you'll have no way to detect whether it's open or closed. Is it acceptable to move or reconfigure the switch somehow? Is it acceptable to add something which maintains some current through the switch so you'll be able to detect when it opens? \$\endgroup\$ Commented Sep 7, 2024 at 15:11

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You can swap out the switch with a change-over type and reposition it like this: -

enter image description here

The switch in the above image is shown in the discharge position. The low-value resistor is there to prevent unholy amounts of current flowing when discharging the capacitor. Maybe 1 to 10 Ω will do.

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  • \$\begingroup\$ Changeover switch is a mechanical manual switch. But I want to immediately discharge without the involvement of any manual switch. Hence, asking for the discharge circuit diagram. \$\endgroup\$ Commented Sep 7, 2024 at 11:17
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    \$\begingroup\$ There is no such thing as an immediate discharge of a capacitor and, whatever circuit you use, it should (a) respect that fact and (b) detect when the switch is open. My circuit does both by having an extra contact on the switch and, by having a low-value discharge resistor. \$\endgroup\$ Commented Sep 7, 2024 at 11:28
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    \$\begingroup\$ @ZOGAM But you already have a mechanical switch you need to close to charge the cap. It can be the same switch. \$\endgroup\$ Commented Sep 7, 2024 at 11:29
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    \$\begingroup\$ @ZOGAM I have rolled back your question to how it originally stood. If you want to add something to your question you may do so providing it doesn't invalidate answers already given. Your recent amendment changed the whole perspective of the question hence I rolled it back. BTW, that proposed circuit will not work. \$\endgroup\$ Commented Sep 7, 2024 at 12:07
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    \$\begingroup\$ @ZOGAM I suggest you raise a new post for this new circuit and, I'm sure, people will be ready-to-hand to explain why it won't work as effectively as you believe. Maybe you should also use a simulator and try the idea out for yourself. If we are done here, please take note of this: What should I do when someone answers my question. If you are still confused about something then leave a comment to request further clarification. \$\endgroup\$ Commented Sep 7, 2024 at 12:12
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You can use a couple of MOSFETs to do the current switching for you, under control of a simple push-button switch:

schematic

simulate this circuit – Schematic created using CircuitLab

R2 holds node G high at +12V, which switches M1, off and M2 on. When switch SW1 is closed, node G is pulled low to 0V, reversing the state of affairs, so that M1 is now on, and M2 off.

When M1 is on, the charging of C1 commences, via LAMP1 and R1, as per your circuit. When M2 is on, charging stops, and the capacitor is rapidly discharged via R3. The process looks like this:

enter image description here

The blue trace represents switch state, high (5V) being closed (pressed), and low (0V) being open (released). The orange trace is capacitor voltage, which can be seen charging slowly to 12V with SW1 pressed, and then much more rapidly discharging back to zero when the switch is released.

Depending on the size of C1, a lot of energy could be dumped into M1, R1 and M2, R3 during each charge and discharge cycle, causing them to heat up. I've chosen R3 to limit discharge current to \$\frac{12V}{4.7\Omega}=2.6A\$, but it's up to you to make sure that these components aren't over-stressed.

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  • \$\begingroup\$ The schematic shows R3 is 4.7Ω, so isn't the peak discharge current 12 V / 4.7Ω = 2.6 A rather than 1.8 A? \$\endgroup\$ Commented Sep 8, 2024 at 7:13
  • \$\begingroup\$ @ChesterGillon Yeah, you're right. I could have sworn... Anyway thanks for pointing that out. \$\endgroup\$ Commented Sep 8, 2024 at 8:22