Capacitor discharge

Hello im new here and to electronics. Im having a problem with my transistor capacitor circuit. I have two leds in the circuit one before the capacitor and the other after. So I see the caps have charged when the lights go out. But when I "discharge" it with my transistor used as a switch. It does not seem to discharge all the way. When using a push button after it's discharged the lights start out bright and then dim. With the transistor after discharge it starts out dim till caps full and goes dark. If I take the cap out and manually discharge it the lights will be bright then dim. I have tried a few things but I can't seem to fix the problem. One thing im thinking is I need a higher Volt Cap. So maybe it will discharge quicker. My highest is 50 volts lol I tried hooking them in series but still does not seem like enough. thanks for any answers I just hope I don't get this answer "Your grammar suks so does your life" or something like that since Im new I don"t know what to expect :sweat_smile:

My grammar skillz are very very li lim limited ... none 8)

How about a schematic of your test setup?
Keep in mind that current flow rate is a function of voltage, I = V/R,
as the cap discharges, the voltage drops, so the current drops, and that last bit of voltage drags out exponentially.

I think you are running up against the Uncertainty Principle. Your LED's are drawing way too much current and influencing the measurement by discharging the capacitor. If you really want the LED's you might have to put a high impedance buffer in front of them to isolate them from your circuit.

Wow I did not expect the fast reply's. :astonished: Thank you. Im not sure how to do a schematic on here. I seen others that had like a cool picture schematic with the breadboard that was very detailed but. I looked and could not find a free program to do it with. If you could show me one I would be more then happy make the diagram. If I had a camera I would draw one and upload it. I can try to describe it kinda. The anode of led one is hooked to +/power the cathode is hooked via wires to the anode of the cap and the first lead of the NPN transistor the middle pin of the transistor is hooked to output pin of the board and the last lead of the transistor is hooked to the anode of led2 along with the cathode of the cap then led2 cathode goes to - / ground Not sure if im using the terms right im not using a resistor right now to try to get a faster discharge circuit down the road I prob will or try to code a lower power output from my board :slight_smile: i have lot of leds so i don't mind burning a few out in testings thanks again for any answers XD oh and the cap im using is 50V 4.7uf

KeithRB:
I think you are running up against the Uncertainty Principle. Your LED's are drawing way too much current and influencing the measurement by discharging the capacitor. If you really want the LED's you might have to put a high impedance buffer in front of them to isolate them from your circuit.

imma do some research on it since im still noobish on electronics Thanks for your reply :slight_smile:

Hi, just a picture of a hand drawn circuit diagram will help SB, and we will see if we can help you understand.

Tom...... :slight_smile:

From your description,
If you have this: Try this suggested modification:

Resistors are recommended to limit current and protect components.

dlloyd:
From your description,
If you have this: Try this suggested modification:

Resistors are recommended to limit current and protect components.

The diagram you did on the left is the one I have. I have tried doing a similar one like on the right but it slows down my discharge and charging. The second led on the other side of the cap is just for lighting up to show it discharged.

TomGeorge:
Hi, just a picture of a hand drawn circuit diagram will help SB, and we will see if we can help you understand.

Tom...... :slight_smile:

I did not have a camera to upload a drawing :cry: luckily

dlloyd drew one for me its the diagram on the left in his post

Note that in your circuit (and in my suggestion), there is about 4V dropped across the LEDs and 0.7V dropped across the transistor when discharging. So the capacitor could only charge to 1V then discharge to 0.7V.

Try changing both resistors to 330 ohm and removing LED2. You'll see a big difference in how LED1 responds.

To calculate the time constant and energy in the capacitor, look here.

Also, the energy level is quite low for a 4.7µF capacitor to have much effect on the illumination / de-illumination of an LED. Larger values 47µF - 1000µF may surprisingly give the results you need.

Note that the resistor in the LED circuit limits its maximum current (brightness) and also sets the time constant for charging.

I did not have a camera to upload a drawing

Your cell phone doesn't have a camera ?

Hi, Sb, is just an omission or have you got the arduino gnd connected to the charging circuit ground.
Also where are you getting the 5V charging current from?

Tom.... :slight_smile:

I have the end of the circuit the second led hooked to negative and that is hooked to the Gnd pin the power is from the 5v pin on my board

raschemmel:

I did not have a camera to upload a drawing

Your cell phone doesn't have a camera ?

Nope Im not much of a talker so I don't really use a phones so I have just the basic XD

dlloyd:
Note that in your circuit (and in my suggestion), there is about 4V dropped across the LEDs and 0.7V dropped across the transistor when discharging. So the capacitor could only charge to 1V then discharge to 0.7V.

Try changing both resistors to 330 ohm and removing LED2. You'll see a big difference in how LED1 responds.

To calculate the time constant and energy in the capacitor, look here.

Also, the energy level is quite low for a 4.7µF capacitor to have much effect on the illumination / de-illumination of an LED. Larger values 47µF - 1000µF may surprisingly give the results you need.

Note that the resistor in the LED circuit limits its maximum current (brightness) and also sets the time constant for charging.

I will try that when I get my board back im letting a friend well my only friend use it cause he does not have one yet XD

Thanks dlloyd your suggestion worked. :slight_smile: Thank you everyone who posted on the thread. Im happy that their are people out there willing to help.