Transistor Timer Circuit, how does it work?

Discussion in 'Homework Help' started by catsfred7, Mar 19, 2016.

1. catsfred7 Thread Starter New Member

Mar 19, 2016
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Please take a look at this simple circuit.

http://startingelectronics.org/begi...ow/tut2-transistor-timer/transistor-timer.png

http://startingelectronics.org/beginners/start-electronics-now/tut2-transistor-timer/

I don't understand how the switch is turned on to begin with. If electrons flow from negative to positive and if the switch is not connected from the start how do the electrons find their way to give the base of the transistor a positive voltage?

Does conventional current have a say in this?
Some one please explain how current flows in this circuit?

What happens when I close the switch? Why does the light turn off?

I noticed the capacitor has a temporary reverse voltage when I close the switch? The current changes polarity in the capacitor when its shorted and discharged? Causing -V to turn off the transistor switch? Is this reverse voltage a bad thing for the circuit? If its that.

Thats my guess, Im really confused. Please some one explain exactly how this circuit works with electron current flow.

2. Papabravo Expert

Feb 24, 2006
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1,757
With the switch open, there is about 8.3 Volts across R2, which holds the base of the transistor at about 0.7V. The transistor is ON, and current flows through the LED.

After the switch closes the bsase drops to 0V and the transistor is off. As long as the switch remain closed, the base of the transistor will slowly return to 0.7V, while the + side remains at GND, and the transistor will turn ON again, lighting the LED. I think reversing the polarity on an electrolytic capacitor is a dangerous thing to do, but the current is limited by R2 @ 22K

When the switch is opened the + side of the capacitor will charge to +9V

Here is a simulation file.

• TTimer.asc
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3. catsfred7 Thread Starter New Member

Mar 19, 2016
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I still don't understand how the electrons find their way to R2 with the switch originally open? How would you draw electron flow on the circuit diagram? Does a small amount of electrons bypass the emitter to the base when the circuit first gets power from the battery even with the switch open?

Also what exactly happens when the switch is closed? Am i correct about the reverse voltage and the capacitor discharging?

4. catsfred7 Thread Starter New Member

Mar 19, 2016
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I was correct about this? This is what causes the switch to turn off? The amount of time it takes for the capacitor to discharge in the reverse direction?

5. Papabravo Expert

Feb 24, 2006
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One side of R2 is at +9V DC. the other side of R2 is restricted to the range of 0.0V to 0.7V DC. Why? Because the emitter of the transistor is connected to GND and once the base starts to rise above GND the current through the transistor increses exponentially as as a function of the base emitter voltage. Once it reaches 0.7V the base emitter voltage cannot increase further. So the maximum current through R2 is given by Ohm's law:

$\frac{8.3\; \text Volts}{22\;\text KOhms}\;= \;377 \mu\text Amperes$

Check out the following paper on the Ebers Moll equation. It shows how even small base emitter voltages let small currents flow.
http://www.mee.tcd.ie/~mburke/3c2/lectures/II Bipolar Junction Transistor/2 The Ebers-Moll BJT Model.pdf

The result of the simulation shows the current through the LED as the green trace I(D1), the voltage on the base is the blue trace V(n005), and the voltage on the + side of the cap is the Red trace. The off time for the LED is about 7.5 seconds. The voltage controlled switch has a period of 9 seconds.

• TTimer.png
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6. Jony130 AAC Fanatic!

Feb 17, 2009
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Because there is a path for a current flow from +9V DC voltage source--->through R2 resistor---> into Q1 base-emitter junction---> back to the negative terminal of a voltage source. So the base current is flowing, if so transistor is turn ON and this allows collector current to flow from +9V--->through R3 resistor---->D1 LED---->collector-emitter--->back to the negative terminal of a voltage source.

Feb 17, 2009
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8. WBahn Moderator

Mar 31, 2012
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With the switch open, the base current flows as shown:

After the switch as been off a long time, the capacitor looks like an open circuit. So there is no current flowing in R1. This means that there is no voltage drop across it and therefor that the positive terminal of the capacitor is at 9V. Because there IS a path for base current to flow, the base of the transistor is at about 0.7 V. So the capacitor has about 8.3 V across it.

The transistor has about 377 uA of current in it. If we first assume that this is sufficient to saturate the transistor, then the Vce will be about 0.25 V. If we assume a 2 V drop for the LED (without knowing the color we can only guess) then that would mean a collector current of about 6.75 mA, which would equate to a transistor beta of about 18, which is pretty consistent with the transistor being at, or very near, saturation. So the LED will be fully on one the switch has been open long enough.

When the switch is closed, the positive side of the capacitor is immediately brought to 0 V, but since the voltage across a capacitor can't change instantaneously, the voltage across it will remain 8.3 V which will take the base of the transistor to -8.3 V. Herein lies the first potential issue -- can the transistor handle a negative base-emitter voltage that large. The emitter-base breakdown voltage for a 2n2222 is generally rated as 6 V minimum, so you are exceeding this spec and could damage/destroy the transistor. At the very least, it is NOT guaranteed to behave as if it is in cutoff thus making this a poor design.

Let's assume that the transistor YOU happen to have can tolerate the 8.3 V reverse voltage (and it may well be able to). If so, then the transistor will be in cutoff and the LED will turn off. The capacitor will now charge on the right hand side through R2 starting at -8.3 V toward 9 V but will stop when it gets to about 0.7 V. This is about half way and it takes a first order circuit about 0.7 time constants to get to the halfway point. The time constant is the RC project of R2 and C2, which is 10.34 seconds, so the transistor should start to turn back on about seven seconds after the switch is closed. After this the capacitor will stop charging and all of the current in R2 will once again feed the transistor base just as before. But herein lies the second problem. The capacitor voltage is now at -0.7 V (positive plate relative to negative plate) and if it is a polarized capacitor as shown in the schematic it might not act like a capacitor. You MIGHT get away with this, for a while, but the cap might look like a short. It probably won't explode with this low of a reverse voltage, but it is a poor design.

Let's assume that the capacitor stays in one piece and continues to act like a capacitor. What happens when the switch is opened again? The left side of the cap will charge back to 9 V through R1. Since the right side of the cap (the base of the transistor) is still clamped at 0.7 V, the additional current through the cap will go through the base adding to the current from R2 and increase the LED current, but only to the degree that the transistor isn't already in saturation. The difference is likely not noticeable. The time constant for this process is the product of C1 and R1, which is about 0.22 seconds and it will need about five times this to "fully" charge, or about one second. At that point you are back where you started.

9. Papabravo Expert

Feb 24, 2006
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I agree, that it is an insidiously crappy design. How do they get away with promulgating this stuff? I'd be embarrassed to ha my name associated with it.

10. catsfred7 Thread Starter New Member

Mar 19, 2016
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thanks so much for the answers! it makes a lot more sence now.