# Basic Transistor Questions

1. Dec 18, 2005

### infamous_Q

Ok so i recently actually learned what transistors due, when used as amplifiers (i always knew they acted as sort of switches). now my question is...how does this happen? Is it true that if i hooked up a 9v battery to the input side (lets say it puts out .1A due to resistance in the transistor and circuit), thats a total of .4W, and hook up a 12V battery to the output side of the transistor, (which in its circuit would normally have a current of .05A, which is a power output of .6W) then the input will force the output circuit's current to be .1A yielding a power output of 1.2W instead of .6W on the output circuit? (hopefully you can keep track of my explanation above). Not to mention that you can do it with voltages where output V >> input V and input A >> output A, true? It just seems like i'm not gettin something here since you're technically getting power out of nothing, any help on what im missing would be great. Thanks guys!

2. Dec 18, 2005

### Ouabache

Interpretation of your description is confusing. Do you mean a bipolar transistor or a field effect transistor? There are three connections to a bipolar transistor (base, emitter, collector) and three for an FET transistor (gate, drain, source). Your reference to connecting a voltage to the input side and output side is ambiguous.

Let me give you an example using an NPN bipolar transistor as an amplifier. When a small signal is applied to the base terminal and the collecter is properly biased, the small signal current is amplified in the collector circuit, resulting in current amplification. This gain is often represented as Hfe or $\beta$ = Ic/Ib (collector current divided by base current). http://www.mtmi.vu.lt/pfk/funkc_dariniai/transistor/bipolar_transistor.htm [Broken]

Last edited by a moderator: May 2, 2017
3. Dec 18, 2005

### Averagesupernova

Yeah I'd say you aren't getting something there. There is never an overall power gain. And I don't know why you are hooking a battery to the output. I also can't see where you get .4 watts. 9 volts at 100 mA is .9 watts. A transistor can be configured several different ways. A voltage amplifier will take an input voltage and multiply it to get a larger output voltage. But this is not without a seperate power source. You do not have an overall power gain. Any extra power coming out of the transistor amplifier has to be supplied by an external power source.

The same thing goes with a current amplifier.

4. Dec 18, 2005

### infamous_Q

oops my bad. i meant .9W then (sorry i went back and changed something and forgot to change everything else it related to).

as for what type of transistor it is lets say its a Bipolar Junction transistor (for sake of simplicity, and for more simplicity lets say we're not taking into account any current leakage).

also to clarify more, or attempt to, let ssay the 9v battery is in the base side and the 12v is attached in series with the load on the collector side.
I'm aware that for amplication to proceed there must be an extra power source on the output/collector side, which is why the 12v battery is there.

does that help to understand my question?

5. Dec 18, 2005

### Averagesupernova

Well by definition a bipolar junction transistor is a current device. As long as the base to emitter voltage gets above about .6 volts the base will start to conduct. Then, the collector circuit will start to conduct. That is VERY simplified of what is happening. Why do you think you are getting power out of nothing? I now realize that your 12 volt source is part of the output circuit. So again, why do you think you are getting something for nothing?

6. Dec 18, 2005

### Ouabache

The way you describe this example, you are putting the transistor into saturation where it behaves like a switch. You wanted it to behave as an amplifier, so it should be biased to stay in the active mode. Take a look at Example 1 on page 7 of this http://ece-classweb.ucsd.edu:16080/spring05/ece53b/Chapter10.pdf [Broken]. It shows both active and saturation situations.
What we haven't indicated here is that you normally amplify a time varying signal The amplification is $\beta$ times the base current. If you input a tiny sinusoid applied to the base, you get a large sinusoid output. That is a very useful result. The 12Vdc you applied to the collector through your load impedance, supplies the potential to generate this large output.

Last edited by a moderator: May 2, 2017
7. Dec 19, 2005

### infamous_Q

ok, so time varying...what if its the voltage is basically constant? (ie, the input is a constant voltage, and/or the output is a nearly constant voltage)

as for the power out of nothing...lets try the example with a bigger gap. 9V on the base (with the resistance in that circuit lets say its 1A), which is 9W. now for the power supply on the emitter side lets say its running at 24V (and with that resistance lets say its running at .5A) which is 12W. Now add those together, you get 21W total power in the circuit. BUT if the current from the base forces the current in the emitter to be 1A (at 24V) then you have 24W of power on that side alone, plus the regular 9W from the other side, giving a total of 33W. This gives a total difference of 12W, so it seems to me i'm missing something...
thanks for all your help guys.

8. Dec 19, 2005

### Averagesupernova

I'm still not quite grasping what you are saying. But I think you have some misconceptions about transistors. First, the base current is never going to be the same as the emitter current let alone more than the emitter current unless the collector is left open. That serves no purpose. The transistor beta is what determines the ratio of emitter to base current. The collector current is ALWAYS the emitter current minus the base current.

Lets modify your example slightly. Suppose we have a transistor set up with a base current supply of 9 volts in series with a 1000 ohm resistor. The transistors emitter is connected directly to the negative side of the base power supply. Now suppose we have a collector circuit that also has the negative side of a 9 volt power supply directly connected to the emitter. The positive side goes through an 8 ohm resistor to the collector. We have a base current of about 8.3 mA. With a beta of about 75 for instance (power transistors typically have lower betas than small signal transistors so this is really not typical) we will have an emitter current of about 622.5 mA. The collector current is usually assumed to be the same as the emitter current but to be technical we will call the collector current 614.2 mA. The 8 ohm resistor in series with the collector will drop a current of about 4.91 volts. But the total power that the collector circuit power supply has to deliver is about 5.6 watts. 3 watts end up in the collector resistor and the rest is dissipated inside the transistor. The base power supply delivers about 75 mW. 68.8 mW is dissipated in the base resistor and the rest is dissipated in the transistor. What I have described is the typical although simplified and somewhat useless practical-wise configuration of an NPN BJT.

This transistor will run warm and all BJTs will have their beta vary all over the place with temperature changes. So it is kind of a bogus example. Usually in order to stabilize the operating point of a transistor configured as a common emitter amplifier, an emitter resistor is added, and voltage divider bias is used on the base.

I read a long time ago in a Radio Shack book by Forrest Mimms that you can think of a transistor like an aerosol can. A small force (you pushing the button) controls a large force (stuff that comes out of the can). It's not too bad of an analogy actually. That is really all a transistor does. The word transistor really describes what it does. "Trans" meaning changing or variable and "istor" being the word "resistor" shortened.

If you are going to school for electronics spend as much time as possible on discreet transistors. People will tell you otherwise since op-amps make it so easy to get voltage or current gain in a convenient package. But knowing transistors well will benefit you in more ways than you can imagine.

Any questions now?

9. Dec 20, 2005

### Cliff_J

infamous - if you open the water facet above a sink and water comes running out of the tap, did you create power from nothing? No, you inputted a tiny amount of power to open the valve, and the stored power in the water was released. That stored power comes from the city's water pumps and your facet is only controlling it (and your hand is controlling the facet).

The transistor is similar, a small control current can control how much current flows through the transistor. In fact, in Britian the transistor has a slang term of valve. But the power that is transmitted through the transistor is coming from the power supply its connected to.

I think the confusion results because a transistor may be described as an amplifier, one that can make bigger signals from smaller ones. But this description leaves out the true nature of how the transistor works. Instead, a more accurate description might be it allows enough flow from a large power source in a way that copies the input signal, just bigger.

How much larger the output is from the input can be referred to as gain, but again the power is being released from the power supply in a controlled amount by the transistor. You could also think of it as a variable resistor like a potentiometer, except its output is controlled by electricity and not a mechanical knob.

Also, its easier to see this work in the voltage domain, not power. So if we had a voltage gain of 20 it might be 0.2V on the input = 4V on the output, and 0.3V input = 6V output type of thing. This would be true until the output voltage reaches the power supply voltage (or more correctly close to it).

Hope a simplified explanation helps the other posts make more sense.

Last edited: Dec 20, 2005
10. Dec 20, 2005

### infamous_Q

k. thanks guys, the amplifier use is what really got me. which i'm still confused about. say for a car amplifier (without adding an extra battery) you have the little 5V signal from the cd deck, or a preamp out, and the car battery (no idea on the voltage or current of these) hooked up the amplifier. How would you get like 700W out of that to power speakers, etc (yes i'm aware that at some point you HAVE to add a capacitor or extra battery to power these huge car speaker systems).

also, (just trying to get something straight) say on the emitter side you have a circuit which is 9V @ 6 ohms, running at 1.5 A (on its own, without a transistor hooked up). You then have on the base side a circuit running 6V @ 3 ohms running at 2A (on its own without a transistory hooked up) (assuming this is in an attempt to increase current). if you hooked up a transistor to the emitter side and base side, appropriately, to these two circuits and the load you were trying to drive was on the emitter side what would the current be in that circuit?

11. Dec 20, 2005

### Cliff_J

I'll answer the first question for now and maybe the second a little later on. I'll ignore losses so its just a basic example.

Inside a car amplifier is a switching power supply. It has a transformer that it runs +12VDC through in little pulses in both directions - making for pulsed AC, and this is later rectified back into DC. Since a transformer can alter voltage/current (the product of volts*current or VA stays the same) you might start with 12V and 90A to end up with 72V and 15A. (its actually going to be something closer to +36VDC and -36VDC in most amplifiers since they are class B designs but not important for this example).

So while its called just an amplifier, it also has its own power supply just like home amplifiers do. Only the tape/cd/radio unit in the dash or very small amplifiers would not have its own power supply (although very high-power aftermarket radios do have a power supply). And as your learning, this lack of a power supply means a lack of power because there isn't enough voltage to make big power.

So you feed in your 5V AC signal with music, and the transistors swing the output voltage from 0V to +72V and now it gets interesting. A musical AC signal is made of sine waves so it makes sense to represent this in RMS so we have 72V * .707 = 51V and if we apply this to a 4 ohm speaker we get 12.75A of current, and 51V * 12.75A = 650.25W or what would normally be called 650W RMS.

If you didn't convert the voltage to RMS, you get twice the wattage. This is common in inexpenisve gear, its called peak wattage or some other wording, and its generally accepted to be a false number since music will never allow you to extract that much power (only half) and you'd need to play square waves to hit the peak number. In fact, because there isn't regulation, peak can be exaggerated even more than that and limited to a short burst. Remeber how I referenced losses briefly eariler? Well for a very short burst (like 1/20th a second) the power supply can deliver more power than over time.

When the car is running, the altenator supplies all the current (until you draw too much and then its the battery). So extra batteries is for listening with the car off. And capacitors are a waste of money, they are far too small to do any work at 12V - capacitors store energy much better at high voltages and should be inside the amp power supply. In our example, using .5 * C * V^2 as our figure, you can see that it would be 36 times more effective to have capacitance inside the power supply than outside of it. Good quality amplifiers are built like this and show little benefit from an external capacitor, cheap ampfliers have little to gain so the external capacitor is marketing fluff to be avoided.

12. Dec 20, 2005

### infamous_Q

ok i guess that makes more sense. the reason i was so confused i believe is because of the way my professor explained a basic BJT as an amplifier (looking back at it again i think i may have screwed up what iw as saying quite badly...lol, lemme try this again):
(this is on a pnp bjt, and please ignore the ... between lines, there there just so that it stays spaced out properly)
|------(+voltage1-)-----|--------(+voltage2-)----|
|...............I................|.................I................|
|<---Emitter.................|<---Base.....................| <----Collector
|----------------------PNP-----------------------|
hope that you guys can follow that. Now the way it was stated was that voltage 1 was smaller than voltage 2. and the current I which is caused by voltage 1 in the emitter-base circuit is forced through the transistor causing a current I in the collector-base circuit. thus with the higher voltage (voltage2) and the current I also, the power is amplified.
thats where i learnt it from, is that right or do I need some clarification?
Thanks again guys.

13. Dec 20, 2005

### Averagesupernova

That's kind of a crazy way to look at it. I don't think there is technically a current in the base/collector junction. Also, you say that there is a higher voltage and current in the collector circuit. Under most circumstances there is ALWAYS more current in the collector than the base. Forget about voltage though, because a BJT is not really a voltage device. We configure their connections so they become voltage amplifiers and things of this nature. Cliff hit it right on the head with the water faucet analogy. Have you drawn a schematic of what I posted earlier? Do you know the correct symbols for NPN and PNP BJTs?