0
\$\begingroup\$

I'm having some doubts about this circuit and about capacitors charge/discharge in general

SCR CIRCUIT; Igt= 100MicroAmps;Vgt = 0.7V

Ok I have like 5 questions

1) Is flowing current through R2 and R3 before C1 gets fully charged?(I know in t=0 c1 will be like a wire but as long as it gets more charged it will have more oposition to current so will current split or it will it go completely through capacitor until its charged?), if the current does split will it charge as fast as if they werent there? using the famous exponential equation for RC,(R1=R)? 2) Ignoring the scr part, Will c1 charge until Vcc or will it charge until the voltage that R2+R3 will have(around 35V)? or first till 90V and then when it gets like open it will discharge until 35V?

3)Doing the math I get 700MicroAmps trhough R3 and 800Trhough R2( since I have fixed 0.7V on R3 and adding it up with 100MicroAmps through the gate I get 800Micros through R2)so that is 3.34Volts for Vc to activate the SCR, my doubt is: since I'm asked for the graph of voltage in C1( or AK) will C1 discharge by the switching since SCR will get activated?(to 0Volts or VFon I don't know) which will get on 0V along time. or can the charge(maybe it will try to charge more and all current will go through it to charge it) or discharge of c1 (or any other component in this circuit) get IH lower , causing a loop of charge/discharge in the graph?

4)Imagining SCR is not 100% ideal and it doesn't have 0V, how can I know if current will flow through R2 and R3 as well?will it be lower or higher than AK current, maybe 50-50?

5)Imagining SCR is not there and instead of R3 there is another Capacitor identically to C1, how can I know the time (from the time I connect the source) of charge of this C2, do I have to assume that c1 is charged first and then c2 so I will have to add up the times by separate(tc1 + tc2)? or is there any kind of relationship between them? since C2 would have more resistance and hence more time of charge... what is the way of knowing it with certainty?

I apologize if I didn't express my questions correctly or clearly, and also if I asked to many questions in 1 threat, many thanks to everyone that can help me to watch this clear!!

\$\endgroup\$
1
\$\begingroup\$

Leaving the SCR out of the circuit for the moment, there is nothing in the circuit that is non-linear, so there is no point at which for example a cap will stop charging and then current will flow through a different part of the circuit. Everything will change steadily from t=0 to steady state.

In the first question, initially the voltage across R2 and R3 is 0V, but as C1 charges the voltage increases, and so there will be current flowing through R2 and R3. The current there is simple to find at any given point in time using Ohms law (C1 voltage / (R2 + R3), however finding that voltage is less trivial. Initially all the charging current (through R1) flows into C1. As it charges, the extra impedance of R2 and R3 mean extra current through R1, which means additional voltage drop and it slows the charge rate of the cap. It also limits the final charge voltage.

An easier way to think about it is to remove C1 initially, and calculate the voltage between R1 and R2, i.e. 90 (4.3 / (4.3 + 6.8)) = 34.86V as it seems you already did. Now putting C1 back in, you know that the final voltage on C1 will be 34.86V (assuming ideal components). It will charge exponentially to that level as if there was no R2 and R3, and the supply voltage was 34.86V, except that the timing would be different. C1 would charge faster (voltage / time) if R2 and R3 is there or not, but the time constant would be the same whether R2 and R3 is there or not. So yes, C1 would charge with time constant RC, where R = R1.

Skipping to Q5, if now you add a cap across R3, you can determine steady state values as done before, by leaving the caps out and calculating voltages. You already determined the voltage across R2 and R3, so you can find the voltage across R3. That will be the final voltage for the new cap. But C1 will not charge 'first', followed by C2. As I mentioned before everything will change steadily. C2 will start charging at the same time that C1 starts charging, however C2 may finish charging later than C1 (never earlier), depending on the value of C2.

Jumping back to Q3 and Q4, when the SCR is added, the circuit is no longer linear. Initially the gate voltage will start to increase as C1 charges and the SCR will be triggered. Yes, when the SCR turns on, it will reduce the voltage across C1 (rapidly), which will cause the gate voltage to drop again. It won't go to zero, but low enough that the gate would be below the triggering voltage. Yes, because it's not 0V, there will be some current also through R2 and R3. The ratio depends on the specs of the SCR.

Depending on the holding current, the SCR may remain latched at that point simply passing current through R1 and the SCR. If the holding current is high enough, the SCR may turn off, allowing the cycle to repeat again.

\$\endgroup\$
  • \$\begingroup\$ Hey thanks! Skipping to Q5 what would be the equation to know the time of charge of C2, the same exponential formula but adding up R1+R2 as R for RC equation? And having an ideal or hypothetical VFAK = 0V, then there is no way that the scr turns off , right?(since no current will go through r2 or r3) thanks very much :) \$\endgroup\$ – N. C. Mar 14 '17 at 17:47
  • \$\begingroup\$ Once the second cap is added, it becomes a second-order circuit, and will result in a second-order equation, so not as straightforward. I suggest looking through this and/or other references for second-order RC circuits online. For the SCR, no, even if the voltage is reduced to 0, it doesn't necessarily mean it will never turn off. It will turn off if the AK current goes below the holding current. \$\endgroup\$ – AngeloQ Mar 15 '17 at 5:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.