# Circuit not working

Discussion in 'General Electronics Discussion' started by kingofjong, Jun 1, 2018.

1. ### kingofjong

30
1
Aug 14, 2014
Hi

I have build a circuit but for some reason it is not working. The capacitor is suppose to discharge through the bjjt pnp transistor. Hence the transistor acts as a switch. But the capacitor isn't discharging when I apply current to the base of the transistor. Circuit diagram is bellow.

Thanks

Last edited: Jun 1, 2018
2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,480
2,828
Jan 21, 2010
Your circuit is wrong in many ways.

Are you working from an original circuit or did you come up with this yourself?

3. ### kingofjong

30
1
Aug 14, 2014
I came up with it my self. What is wrong with circuit?

4. ### kingofjong

30
1
Aug 14, 2014
Hi

I figured out my mistake. So I don't need anymore help.

Thanks

5. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,480
2,828
Jan 21, 2010
I'd love to see what you came up with.

Also, since you're staying the circuit yourself, there's a few conventions normally applied that make the circuit more readable.

1. Power supply to one side
2. Positive rail at the top, negative at the bottom
3. Components horizontal or vertical
4. Transistors generally drawn so the base connection is at the left or right side (not at the top). This should also result in the PNP emitter being at the top, and NPN emitter being at the bottom.

I'll try to give you an example a little later.

6. ### davennModerator

13,786
1,936
Sep 5, 2009
I would also like to see what you did to solve the problem

and take good note on Steve's comments on drawing circuits

7. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,480
2,828
Jan 21, 2010
OK, here's a more conventionally drawn version of your circuit:

From this, it is clear that your circuit can't work. Pressing S1 will charge the capacitor, but the emitter of Q1 can never be more positive than the base, so the transistor cannot be turned on.

Probably, the switch needs to be in the base circuit, pulling the base low.

Let's see if I can draw something that might work...

In this circuit the capacitor charges up via R3 and R2, When the switch is pushed the capacitor discharges via Q1 and R2.

The capacitor will be discharged down to about 0.6V. Using an NPN transistor and pulling the base to +ve would discharge the capacitor to a few tenths of a volt. You could also rearrange the circuit so the capacitor wasn't tied to the -ve rail to do this.

And here's an option along those lines:

The solution you'd employ depends on the actual problem this circuit is intending to solve.

Last edited: Jun 2, 2018
davenn likes this.
8. ### WHONOES

1,115
310
May 20, 2017
In your solution, Q1 could really do with a pull up resistor otherwise when the switch is open, the base of Q1 is left floating and the device could be partially turned on.