# Simple Capacitor + Transistor Circuit

Discussion in 'Homework Help' started by rlm42, Feb 16, 2013.

1. ### rlm42 Thread Starter New Member

Feb 16, 2013
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0
Hi I need help explaining a simple switch circuit to activate a solenoid with a capacitor and FET transistor. L1 is the solenoid and is activated when the X1-1 and X1-2 short is opened, sending current through the transistor causing the cap to instantaneously discharge and activate the solenoid.

I have not yet learnt transistors, and have only done 1 subject on capacitors so far. I am not sure how the transistor causes the capacitor to discharge when the short is opened? Also it was explained to to me that R1 & R2 form a voltage divider dissipating 6V across each. Why are these not included when we find the time constant for charge or discharge (I thought you would have found Req (R1 + R2 || R4), seeing as they are dissipating voltage.

Also I am trying to calculate the time it takes to discharge the capacitor when the circuit is switched off (Should be over 60 secs).

I have the time it takes to charge T = CR = 2200 * 10^-6 * 2400 = 5.28 secs (Time constant). Then, 5.28 * 5 = 26.4 secs to fully charge.

I have been unable to calculate the time it takes to discharge though, have been trying to use the formula, V(t) = V(0) = e^(-t/RC)

I have uploaded the circuit for you guys to see.

Many thanks.

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2. ### MrChips Moderator

Oct 2, 2009
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5,450
The transistor is a FET which means that there is high resistance between the gate of the transistor and the drain and source (the other two pins of the FET). Hence R1 and R2 are isolated from the rest of the circuit (except at turn off).

When switch S1 is closed, C1 charges up through R4 and L1. Hence the charging time is determined by C1 and R4. One would need to know the holding current of L1 in order to determine the ON time of L1. R x C will give an approximate time.

Why did you multiply R x C by 5?

Last edited: Feb 16, 2013
3. ### rlm42 Thread Starter New Member

Feb 16, 2013
10
0

Ok so it has the charge has nothing to do with R1 and R2.. But what about the discharge? Because I was told they drop 6V across each.. I have no idea how to calculate the time it takes the capacitor to discharge.

I multiplied RC by 5 because there are 5 time constants.

BTW L1 has a resistance of 12 Ohms.

4. ### MrChips Moderator

Oct 2, 2009
17,554
5,450
5 time constants for what to happen?
You really need to know the holding current, not the time it takes for the current to decay to zero.

When the X1 jumper is removed, the FET conducts and C1 will discharge through the FET and L1. The discharge time will be very short.

5. ### #12 Expert

Nov 30, 2010
18,093
9,683
Just a note..that circuit won't work. You can't hold the gate 6 volts above the source and have a jfet that works. It's forward biased. Relatively massive currents are going through the gate-source junction. It's good that you neglect the influence of the jfet in your RLC analysis because the circuit is bogus.

Jumper? Sorry. Didn't notice that part.

6. ### rlm42 Thread Starter New Member

Feb 16, 2013
10
0
5 time constants for the circuit to reach full charge, really not sure what you mean by holding current? I know that the discharge time is instantaneous when the X1 jumper is removed. What I am trying to find is the actually time it takes to discharge when S1 is switched off. I measured it and it takes over 60 seconds but I am wanting to do the calculation.

The max current was 12/2400 = 5mA. I'm not sure what holding current means.

7. ### MrChips Moderator

Oct 2, 2009
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We will assume that the FET is an enhancement mode FET.

8. ### rlm42 Thread Starter New Member

Feb 16, 2013
10
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The circuit works, I didn't design it..

9. ### MrChips Moderator

Oct 2, 2009
17,554
5,450
As C1 charges up the current will diminish inverse exponentially to zero, over a period of 5 time constants.
The relay will de-energize long before that depending on the minimum current it takes to "hold" the relay in the energized state.

10. ### #12 Expert

Nov 30, 2010
18,093
9,683
Don't worry. MrChips covered it by declaring an enhancement mode transistor.

11. ### rlm42 Thread Starter New Member

Feb 16, 2013
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How would I go about finding the holding current?

12. ### MrChips Moderator

Oct 2, 2009
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1) you can go to the spec sheet
2) measure it
3) guess

13. ### rlm42 Thread Starter New Member

Feb 16, 2013
10
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Hmm I probably won't be able to do any of those.. Am running out of time. I would have thought this could be solved using v(t) = v(0)e^(-t/RC), is there any possible calculation that can be done to get a good estimation of the time?

14. ### MrChips Moderator

Oct 2, 2009
17,554
5,450
I would go with option 3), guess by assuming one time constant, i.e. when the current drops to 37% of the initial current.

When S1 is open, we will assume that the X1 jumper is in place.
C1 will discharge through two paths, via R1 and R4, and LED, R3 and R4.

I doubt that L1 would be on for 60 seconds.

15. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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728
Rule of thumb for small relays: 12V Rated coil will switch at about 9V (75%), and release around 6V(50%) (it takes more to move the contacts than to hold)

However, if you weren't told that explicitly, I'm not sure how you are supposed to decide what voltage the relay will release at.

Feb 16, 2013
10
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17. ### takao21203 AAC Fanatic!

Apr 28, 2012
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The circuit is nonsense.

A) To use a capacitor in the load branch.
B) To use a JFET that way.

I recommend to learn how to use components in a regular way.

You said you have not yet learned transistors.
But you deal with a JFET here.

Digital MOSFETs are not so much difficult. You do not really need formula. They work very similar to a relay.

Examine these. You can also use a capacitor on the gate.

I also suggest to examine one new component at one time, with components that are already understood.

Play with circuits. You will build many nonsense circuits maybe.

But the discussion here is not of much use really. No one would use a capacitor, relay and JFET that way.

18. ### MrChips Moderator

Oct 2, 2009
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I am assuming this is homework exercise.

19. ### takao21203 AAC Fanatic!

Apr 28, 2012
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Yes. But it is nonsense to put a big capacitor in series with a relay coil. It is finicky because you need to deal with the latch and release voltages.

Relays are not so much interesting these days. Normally a MOSFET is used, or a small signal relay if a zero Ohms contact is wanted.

Only power electronics sometimes still deal with (large) relays.

For an Ampere or two, you use a IRF MOSFET + a BJT - done.

20. ### rlm42 Thread Starter New Member

Feb 16, 2013
10
0
It's not nonsense it works for what it was made for, that is to trigger the solenoid when the circuit is cut at X1, X2. The solenoid then releases a rod which it is holding up..

Yeah it is kind of homework, It is way over my head, someone else designed it, I was asked to examine it and talk about it in a presentation. MrChips, it is safe to use that discharge plot on my previous post, and to say that at 37% of the total voltage, the current will be at 37% aswell, therefore that being 1 time constant? So 12*0.37 = 4.44V (Reaches this at approx 13secs).. Therefore 13 seconds being 1 time constant. 13 * 5 = 65 seconds approx. to discharge?