# Transistor acting as a switch in closed circuit (with a thermistor)

1. Nov 3, 2012

### DJ-Smiles

1. The problem statement, all variables and given/known data

Ok so I have been given a diagram which if it worked should be attached. I am having a fair bit of trouble understanding it. Now I am aware that there is both electrical and conventional currents but I am confused as to whether this is electrical or conventional. I have been told that the circuit is acting as a switch which when reaches a low temperature (resistance is high) will turn the heater on (in schematic it is a lamp). I have looked at the schematic and if it was conventional current the thermistor would not play a very big role would it? It appears to make more sense if the circuit was electrical current but then again the transistor seems to be collecting at the emitter and emitting at the collector as the transistor is an NPN. If I am correct in assuming it is an electrical current, what happens along the way. I know the capacitor is stored but as you move along and get to the junction where it goes up to the transistor and keeps going straight to the thermistor, what is the voltage that the transistor is exposed to at the emitter?

However if it is a conventional current, what purpose does the thermistor have? To me, it looks like to should be electrical but I am having trouble understanding if it is or if it's conventional. To follow up, where would I move the transistor if I wanted the heater (well when I move on it will be an air conditioner) to turn on when the temperature is high (resistance is low). If it is is electrical, I am thinking you swap the normal resistor with the thermistor.

Thanks to anyone who can help me!.

P.S. Sorry if the format isn't perfect, I am stressing and it may seem a bit jumbled.

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2. Nov 3, 2012

### DJ-Smiles

Also don't take note of the number of each resistor (R1 or R2) I just drew that up and put the normal resistor first.

3. Nov 3, 2012

### DJ-Smiles

No worries for those who see this in the future, crisis averted!

4. Nov 3, 2012

### aralbrec

I had to look it up to make sure I understood what you meant by conventional current.
http://www.mi.mun.ca/users/cchaulk/eltk1100/ivse/ivse.htm

First it should be said all circuit analysis is done with 'conventional current' or at least I am not aware of anyone who does it the other way around. Second, charge carriers in a circuit do not have to be electrons. It can be holes in semiconductors (which are positively charged and flow in the direction of conventional current), it can be ions in a solution like a battery, etc.

In summary, I find that webpage above very misleading.

Conventional current or not, it doesn't matter at all. A voltage source establishes a field orientation in the wires and with conventional current we assume the carriers are positive so that current flows away from the +ve terminal. Later on, if it matters what the actual charge carriers were doing, we reconcile conventional current with the actual carriers.

For example, if a part of the circuit is made of metal then the charge carriers are electrons. In this case, if conventional current points to the right, then electrons in the wire will actually move left. This is equivalent.

If the current is passing through a doped semiconductor like a transistor then we have holes and electrons moving. Holes are positively charged. If our circuit analysis finds conventional current flowing to the right, then electrons in the semiconductor will flow left and holes will flow right. We have both directions going. This is equivalent to conventional current to the right.

So having said that, get rid of your hangups with conventional current and from now on we can just speak of current.

To your circuit: first assume Q1 acts as a switch. In reality it will act as open switch (no current flowing), then it will be partly on (in active mode where it acts like an amplifier) and then it will be all the way on like a closed switch (the voltage across Q1's collector and emitter will be about 0.3V and not change much from there). How "on" the transistor is is determined by voltage across Q1's base and emitter. Because the current through the transistor's collector is exponentially related to the voltage across its base and emitter, the base/emitter voltage of Q1, once on, will not change much from about 0.7 to 0.8 volts.

So now you will see that Q1 will turn on (in active mode or switch closed) when the voltage across R2 is at least 0.7 volts. Calculate the corresponding value of R2 and at what temperature that is achieved. Will the transistor turn on for higher or lower temperatures than this point?

The lamp is probably neon or some such thing and has specific characteristics which maybe is provided? For example, it may not conduct at all unless Q1 is able to pull a certain minimum current through it. So Q1 being a little on (active mode) may not be enough to start the lamp. It may have to be a lot on (closed switch, called saturated). This happens when the 0.7 volt drop rises a little to maybe 0.8 volts. Notice I am saying a lot of about and maybes here. As you can see, the difference is tiny and probably won't make much difference to the value of resistance R2 you find. But you may want to calculate the sensitivity of the turn on point as a function of the base emitter voltage to see if it will matter at all.

Another is neon lamps show some hysteresis. Once it fires, the current to keep it going is smaller. This is very useful. Suppose the circuit turns on at 18C. What if the temp drops to 17.9C in 30 seconds and then rises to 18.1C a minute later? Maybe this goes on for a while. The circuit will turn on and off fluctuating around the set point. The bit of hysteresis the neon lamp provides means the current must be high to start it but there is a gap to a lower current that must be reached to stop it. So this fluctuation will not occur. If the set point to start heating is 18C, maybe the setpoint to stop heating is 21C.

You will need the characteristics of the lamp to determine that.