In attached circuit, I have RC circuit which turns-on transistor as soon as Cap Voltage is >=0.6. Cap voltage keeps going up until it reaches 2.8V. So far so good.

I want to extend this circuit that capacitor discharges itself as soon as it has reached 2.8V. In essence I want to add capacitor discharge sub-circuit to attached circuit.

Would anybody be able to help me out.

 

I think that would be called an "oscillator". My first instinct is to use an SCR with a minimum hold current more than the current through the resistor on the upper left. If it's difficult to find that, you can increase the resistance until the current is low enough to starve the SCR after it has discharged the capacitor.

How you trigger the SCR is dependent on how accurate you need the circuit to be. It might be as simple as a 2 resistor voltage divider or as complicated as a comparator circuit.

 

Google "555" and see how they do exactly what you want to do.

 

ErnieM: I have checked "555", and getting understanding of it. However, internally it uses Voltage comparator which is far too complex to understand as per my current level of understanding in electronics.

What I am trying to do it to have basic understanding, and build simple circuits.

 

I think I should apologize for telling you a slick trick. You probably aren't thinking at that level yet.

 

#12: I think that would be called an "oscillator"
You are correct. Can it be achieved without an SCR? Sorry I haven't got any SCR understanding.

 

As usual, there are a dozen ways to do it. They vary in accuracy, expense, and complexity. Look up, "2 transistor model of an SCR". (You can build and SCR out of 2 transistors and some resistors.) Doing that will educate you about SCR's.

Why do I keep beating the SCR horse? Because it's the easiest way to do this.

Edit: I just remembered the Programable Unijunction Transistor. You can make a capacitive discharge oscillator out of those. Then there is the neon relaxation oscillator. These circuits do not resemble the circuit you posted, but they oscillate.

 

Just went through SCR's. Hmm!!! How it can be used in provided schematic? The answer could be simple; still looking for an answer.

 

Maybe like this. I have guessed at all the resistor values. You can play with it in a simulator?

 

Thanks..I will give it a try.

 

I can't get it to work. Would you be able to try to check if works?

 

What results are you getting? Stays stuck "on"?

 

yeap!!! That's correct

 

As soon as Capacitor reaches approx 0.6V (transistor Vbe), SCR turns-on and remains turned-on.

 

I am attaching a list of SCR's that require 30ma to 80ma to stay on. If you set the supply resistor value to 9 ma, the capacitor will charge up to the gate turn on voltage of the SCR and then the SCR will dump the charge in the capacitor. It will then shut off because the current supplied by the resistor is way less than its advertised hold current.

What we have here is 9 ma (decreasing) going into 470uf. Two resistors set the leakage to ground in the SCR trigger circuit to 1 ma. When the capacitor voltage rises to about 6.2 volts, the SCR will trigger "on". That will dump the charge in the capacitor. The 15 ohm resistor limits the dump rate so the current will not melt the SCR. When the dump current drops below the hold current, the SCR will turn off and the capacitor will resume charging.

 

This is one of the problem that I find very difficult to comprehend because I am from computing background. In computing, we usually apply logic like IF something Then do something.

I am trying to do same here, that is, if SCR turned-on then drop Capacitor voltage. Please note, I don't want to use AND/NAND/OR chips as yet. If I do, then that would defy my purpose of learning.

Thanks in advance for helping me out, and putting in your inputs.

 

I'd fix the circuit I gave you first but that would require thinking.

 

Thanks a lot #12
Trying to understand this, and your comments. Will get back with more questions.

 

What we have here is 9 ma (decreasing) going into 470uf.

Understood

Two resistors set the leakage to ground in the SCR trigger circuit to 1 ma.

Understood

When the capacitor voltage rises to about 6.2 volts, the SCR will trigger "on".

Didn't get this. Shouldn't it be around approx 2.7volts, given SCR trigger voltage is 0.7 volts? Please ignore my ignorance.

That will dump the charge in the capacitor.

Haven't got this either. Shouldn't current will flow from 1K resistor into SCR?

The 15 ohm resistor limits the dump rate so the current will not melt the SCR. When the dump current drops below the hold current, the SCR will turn off and the capacitor will resume charging.

Sorry what is dump current and hold current?

 

The trigger voltage on an SCR is 1.5 volts.
The current from the 1k resstor does flow into the SCR, and it does that while the charge in the capacitor is flowing through the SCR.
Dump current is what the SCR does with the charge that has accumulated in the capacitor. Dump current is also a noun. The dump current flows through the SCR.
Hold current is the minimum current required to keep the SCR turned on.