## Base Emitter Voltage in BJT?

Hi guys,

I'm really confused as to what "Base-Emitter Voltage" is. I thought it was the voltage difference between the base and the emitter, but apparently it's no such thing? If you could explain it, that would be fantastic.

The reason why I ask is I'm not sure if I'm going to blow up my transistors!

I have a 12V car battery through a pushbutton switch on my collector, and a 5v output from Vdd on a Basic Stamp 2 on my base (max ~90mA). What's the base-emitter voltage? I figured it would be abs(5-.2-12) = ~6.8V, which would be higher than the 5V max rating, but Yenka sim software says it will work no problem. Now I'm just confused :/
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 Recognitions: Science Advisor I'm really confused as to what "Base-Emitter Voltage" is. I thought it was the voltage difference between the base and the emitter, but apparently it's no such thing? If you could explain it, that would be fantastic. Yes, that is right. You would not drive a transistor's base directly from a voltage source or the transistor would blow up and never work again. You put a resistor in series with the base and this controls the current that can flow into the base. Suppose you had a 5 volt source and you wanted 0.1 mA of base current into the base of a silicon NPN transistor. The base emitter voltage will be very close to 0.6 volts for quite a big range of base currents. This is convenient because it lets us calculate a suitable resistor. Voltage across the resistor = 5 Volts - 0.6 Volts Current in resistor and base = 0.1 mA or 0.0001 amps Resistor = 4.4 volts / 0.0001 amps or 44000 ohms (Ohm's Law) So, you could use a 47 K resistor or a 39 K resistor.
 Oooooooh. That makes more sense now xD I've always been calculating it with 5v = IR not (line voltage - Vbe) = IR How does a voltage divider affect the Vbe? for example, 12V with a 180ohm on the base, then a 180 ohm attached between the base and the emitter?

Recognitions:

## Base Emitter Voltage in BJT?

I did a simulation using those figures, using a 2N2222 transistor with no collector load.

Base-emitter voltage = 1.718 V.

Base current = 47.575 mA

Current in top 180 ohms = 57.121 mA

Current in bottom 180 ohms = 9.546 mA

Collector current in a 2N2222 = 1.752 Amps.

As you can see, the resistors are too small, but you can get the idea. The current in the top resistor divides between the base and the bottom resistor. (47.575 + 9.546 = 57.121).

Also note that the current in the bottom resistor still obeys Ohm's Law. Current = 1.718 volts / 180 = 9.54 mA

I tried it again but with more reasonable resistors. 100 K at the top and 10 K at the bottom.

Base-emitter voltage = 0.712 V.

Base current = 41.6 ľA

Current in top 100K = 112.8 ľA

Current in bottom 10K = 71.2 ľA

Collector current in a 2N2222 = 9.07 mA.

You can see that the transistor now has a current gain of 9.070 / .0416 or 218.
The base emitter voltage is higher than I would expect at 0.712 volts. 0.6 V is pretty standard.
Again, 41.6 ľA + 71.2 ľA = 112.8 ľA.
 Well I needed 800mA collector current (TIP41C - Assumed gain of 15), so I used 180ohms. EDIT: Oh, and it was a 12V circuit. But I don't understand, why does Vbe change?
 Recognitions: Science Advisor I changed the simulation to use a TIP 31C which is hopefully similar to the TIP41. (I don't have a model for the TIP41.) Using your 180 ohm resistors, the base emitter voltage was 2.31 volts with a base current of about 41 mA and a collector current of about 2.66 amps. This may be a good simulation or it may not. The way to find out would be to build it and measure it. You have no control over what happens inside transistors, but the makers provide data sheets where you can look up the base-emitter voltage for a given base current. So, with large base currents, you need to get the data sheets and use them.

 You have no control over what happens inside transistors, but the makers provide data sheets where you can look up the base-emitter voltage for a given base current. So, with large base currents, you need to get the data sheets and use them.
Let me rephrase: Why does adding the low resistor make the Vbe higher?

For a transistor with one resistor on the base, from what you said initially, and with my needing to pass 800mA, at 15 hFe

12V - .6V = (.8/15) * R
R = 214 ohm

Or are you saying that at 800mA collector current, Vbe is higher simply because of the transistor characteristics?

12V - .8V = (.8/15) * R
R = 210 ohm
 Recognitions: Science Advisor The base emitter voltage depends on the base current and on temperature. (Higher temperatues drop Vbe and this is affected by the external resistance). If you look in the data sheet for your transistor, you will sometimes find a graph of Vbe vs base current. I didn't see this graph in a data sheet I downloaded for the TIP41. The extra resistor has an effect on the base current (it reduces the base current) but after that the internal characteristics of the transistor are all that matters. The main reason the second resistor is added is if you have an emitter resistor, then you can calculate the collector current without knowing the exact current gain of the individual transistor. Otherwise, the lower resistor is just wasting drive current. The only time I would use one, without an emitter resistor, is if the drive voltage did not drop to zero when it was supposed to be off. Early 555 chips would do this and they could leave a transistor turned on when it was supposed to be off. So, you could add a voltage divider to help it turn off.