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Correct me if I am wrong, but from my understanding a transistor is current controlled in that the current flowing from Collector to Emitter is varied by the amount of current going from Base to Emitter. I initial though that a transistor was controlling the current flow based upon B->E and not changing voltage.

I recently was looking into transistors further and read the following:

The power dissipated in the transistor is the voltage drop across the collector emitter junction times the collector current

http://www.physics.unlv.edu/~bill/PHYS483/transbas.pdf

I thought that there wouldn't be a voltage drop when using a transistor since B->E is controlling current flow, not voltage.

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  • \$\begingroup\$ In order for the transistor to control the current, \$V_{ce}\$ has to adjust to whatever is needed to make the current flow in the external circuit. \$\endgroup\$ – The Photon Mar 31 '16 at 18:22
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Assume this schematic:

schematic

simulate this circuit – Schematic created using CircuitLab

What I1 does and why isn't important, we just know it sources exactly 10μA.

The transistor Q1 is a type that, in this set-up happens to have an exact current gain factor of 100. Since this is a theoretical question, we can just assume that.

Now, that means that 1mA will flow into the collector right?

We also know that Ohm's Law teaches us that 1mA through 100 Ohm will only cause:

V = I * R = 0.001A * 100 Ohm = 0.1V

Add to that that the Battery BAT1 is 12V, we'll assume it ideal, so that we can say that across the resistor and transistor there is a total of 12V. If Q1 and R1 have 12V across them and R1 has only 0.1V across it, where does the rest go?

The only place it can: It falls across the Transistor.

Now, of course in the real world this voltage will slightly influence the current gain through some more complex formulas, but this is just to sketch the idea.


In the real world, there are quite a few other similar situations where such things happen.


And even if you use it as a hard-switch, the transistor will have a saturation voltage, the voltage that will always fall across the Emitter-Collector region, even if the path is fully turned on.

There are plenty transistors where this is only 0.1V in many use-cases, but if you'd put 100A through it, that's still 10W, which can be enough to kill any transistor. (Though types that can't handle that usually also don't handle 100A too well to begin with).

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  • \$\begingroup\$ In your example, would the power dissipation be 11.9v * .001A = 0.0119 watts? \$\endgroup\$ – Justin Mar 31 '16 at 18:41
  • \$\begingroup\$ @Justin That is very nearly correct. The difference is ignorable, but since you didn't round away the microwatts: The base current times the base-emitter voltage will also cause a few microwatts of heating. \$\endgroup\$ – Asmyldof Mar 31 '16 at 18:43
  • \$\begingroup\$ And now for a practical question, If I have a 12v 5 watt (therefore 0.5 amps) LED strip that I would like to control the brightness of, can I do this safely with a transistor? \$\endgroup\$ – Justin Mar 31 '16 at 18:47
  • \$\begingroup\$ @Justin I think for that it'd be smart if you look up "PWM" for "Pulse Width Modulation" on teh interwebz. Or just this site. You can do it with a properly heat-sinked transistor and current-control, but PWM will be much more energy efficient. (ps: 12V, 5W is not exactly 0.5A, but 5W/12V =~ 0.42A) \$\endgroup\$ – Asmyldof Mar 31 '16 at 18:58
  • \$\begingroup\$ Thanks for the reply, yes I was thinking about doing PWM, but the arduino outputs at 5v and doing PWM on a relay would a bit noisy. \$\endgroup\$ – Justin Mar 31 '16 at 19:00
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Ahem.

There is no contradiction in the collector current being controlled by Ibe and the existence of a voltage drop over CE. In fact it is a basic effect of KVL that there will be a voltage drop.

If an ideal transistor had 0V over CE all the time it was no controlled current source. Look up a V-I diagram of a (controlled) current source, which is the CE part of a BJT in not saturated mode, and you will see, that the voltage can be any positive or negative voltage for any given current the source is designed (or controlled) to deliver. For a NPN-BJT, of course, voltage has to be positive to prevent the transistor from smoking.

More precisely spoken: That the current over CE is controlled means that the voltage isn't. By no means the voltage is zero. If it was 0 regardless the behaviour and design of the surronding circuitry it was somehow a controlled voltage.

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That quote is specifically related to transistors operating in the linear region. In the linear region of operation, the transistor behaves a bit like a resistor. Since the collector current is controlled by the base current, then there must be some voltage component from collector to emitter. Ohm's law tells us that if we have a voltage and a current flowing, there must be a resistance (or effective resistance in this case).

A very good explanation of BJT operation can be seen here: https://electronics.stackexchange.com/a/13082/82922

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  • \$\begingroup\$ What is a linear region? \$\endgroup\$ – Justin Mar 31 '16 at 18:34
  • \$\begingroup\$ Take a look at this. In this case, the linear region is before it reaches saturation. In the graph to the right, you can see how prior to the device reaching saturation (when collector current will not increase despite an increase in base current), the device behaves almost linearly. \$\endgroup\$ – Brendan Simpson Mar 31 '16 at 18:35
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Up here at 300 degrees above absolute zero (normal room temperature) EVERYTHING has resistance (=voltage drop), even semiconductors operating as digital (on/off) switches. Remember to consider the difference between theoretical explanations and what we experience here in the Real World.

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  • \$\begingroup\$ The non-zero ohmic resistance of a semiconductor ist not the primary cause of the voltage drop on CE on a NPN-BJT. The primary cause is the current imprinted into the whole circuitry by the controlled current source. Given a circuitry of resistors, voltage sources and similar building a loop around CE the voltage drop can be derived using KVL. The resulting resistance between CE is purely virtual then. \$\endgroup\$ – Ariser Mar 31 '16 at 18:40

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