# BJT help

Discussion in 'General Electronics Discussion' started by go2255, Mar 1, 2014.

1. ### go2255

9
0
Oct 28, 2013

This is the result from a lab,but I am not sure the results are correct.
Also,when I try to find I(c) , I(b) and I(e),by measured V(RC) and V(RE),I find the results strange.
I guess the combination of R1=10k and R2=10k is under linear region,but the dc gain is so small.
For combination of R1=10k and R2=1k,Idon't know why the I(b) is negative.Is it the cut off mode?
And for combination of R1=1k and R2=10k,it should be in saturation?
I guess R2 is like a battery so current I(b) flow ,but how about R1??

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2. ### Harald KappModeratorModerator

10,565
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Nov 17, 2011
I think not quite, see below. If you have a chancec, go back to the lab and repeat the measurements.

This is not possible, at least not such a high value (35µA) as you show in your table. Since Ie=Ic+Ib it is very likely that you either noted a wrong sign or connected the meter in reverse because the numbers without sign add upo to the correct sum.

Probably. Saturation is defined as Vce<Vbe. But your measurement for this case is implausible. Your table shows Vb=Ve=8.3V which in turn gives Vbe=0V. The transistor would be off under this condition and Ic=0A.

No. A resistor is a resistor is a resistor is not a battery. A battery can source current to a load. A resistor can only sink current from a current (or voltage) source).

Last not least the arrow for Vce is unfortunate because it points from emitter to collector, giving Vec instead so Vce as indicated by the arrow would be negative. But I know that in the anglosaxon world this arrow thing is handled a bit sloppy which often leads to confusion over the right sign of a voltage or current.

3. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,418
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Jan 21, 2010
Firstly, unless you measured the voltages incorrectly, your results are correct. It is your interpretation of these results which needs to be looked at.

Having said that, the measurements of Vb and Ve on the third line and the subsequent value for Vre both seem problematic. The calculation of Vre on the first line also looks wrong.

I note no measurement of Vcc. From your figures it was 9.93V, 9.94V, and 10.89V -- clearly something is wrong on the last line.

If your voltage was actually 10V then the shunt error caused by the impedance of the voltmeter may also have introduced errors.

It seems you measured voltages then calculated currents.

How did you determine Ib? I see two possible ways:

1) by looking at the difference between Ir1 and Ir2
2) by looking at the difference between Irc and Ire

It looks like you used method 2.

Why not try method 1?

Remember that your resistors have a certain tolerance. Even if you're using 1% resistors, in the second case Re could be between 9.9k and 1.1k, so if you measured 0.4V across it, the current Ie could be anything between 3.96e-5 and 4.04e-5. The same is true of all the other resistors. Actually measuring the resistors to get a more accurate value (say one read 9.98k and the other 10.03k) would have been wise.

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,418
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Jan 21, 2010
Something else which points to your errors in methodology is the ratio between Ib and Ic.