VDB circuit

While going through the text, the author says that when \$R_1\$ is shorted \$V_B=10, V_E=9.17, V_C=9.2.\$
I understand that \$V_B\$ is \$10\$ but I am getting \$V_E\$ as \$9.3\$ and I have no idea how to get \$V_C.\$
Edit: Since there is a slight confusion in my question I'm including one more image with the introduction by author.
" Let us discuss troubleshooting voltage-divider bias because this biasing method is the most widely used. Fig 8.16 shows the VDB circuit. Table 8.1 lists the voltages for the circuit when it is simulated with MultiSlim. The voltmeter used to make the measurements has an input impedance of \$10M\Omega. \$" Table 8.1

  • \$\begingroup\$ I think you should give us the transistor's characteristics \$\endgroup\$
    – DavideM
    Nov 2 '16 at 9:09
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    \$\begingroup\$ Now we see the whole context it makes much more sense. Note that the values are simulated, not calculated. The example you gave would be almost impossible to calculate as it needs iteration t find the stable point - which is just what the simulator does when needed. | The case you gave has the transistor acting as a be diode and effectively c open circuit. Work with that and you get about what he got. ie the example IS a good one because it is an abnormal condition simulation. \$\endgroup\$
    – Russell McMahon
    Nov 2 '16 at 22:39
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    \$\begingroup\$ O've posted a separate answer that answers the question that you should have asked originally :-). \$\endgroup\$
    – Russell McMahon
    Nov 3 '16 at 6:06
  • \$\begingroup\$ @RussellMcMahon Thank you very much for all the effort put in. \$\endgroup\$
    – Bijesh K.S
    Nov 3 '16 at 7:57

You have used 0.7 volts as the base-emitter volt drop - this is an over-simplification. For instance, if you look at the data sheet for the 2N3904 it tells you that the base-emitter voltage might be as high as 0.85 volts with 1 mA flowing into the base and 0.95 volts with 5 mA flowing: -

enter image description here

So, it's more complex/subtle than what you thought but who's really going to get in a pickle about 0.13 volts difference on a single transistor circuit?

On this occasion to work out what Vc is you have to realize the transistor is saturated and regard the collector-base region as potentially forward biased - in other words, the collector voltage cannot be lower than the base voltage by more than a volt and so 9.2 volts sounds reasonable.


This is a superb example of why it is almost always a very good idea to provide as much information as possible in a question. The original question was very narrow, asked how an unusual situation could have been calculated and gave no context.

Once the context is known it makes sense.
This is a troubleshhoting example form a 'textbook'.
The writer says:

  • Let us discuss troubleshooting voltage-divider bias because this biasing method is the most widely used. Fig 8.16 shows the VDB circuit. Table 8.1 lists the voltages for the circuit when it is simulated with MultiSlim. The voltmeter used to make the measurements has an input impedance of 10MΩ.

The voltages shown are arrived at using a simulator - some are close to impossoble to calculate by normal means - including some in the example originally given, as they require iteration - which a simulator uses as of right, but which is not usually needed for 'hand calculation.

Here is how the results are caused in the fault situatuons described.
He uses 'shorthand' terms to describe faults. eg

R1s = R1 short circuited - eg solder splash or bad PCB etching or just possibly a resistor failure (not usually).

Q1 = transistor used.

R2o = R2 open - R2 missing, dead, broken track, bad soldering.



R1s - Original example. Base short to V+ so Vbase = V+. Transistor is haaaaard on. Ibe is limited only by Vbe drop and Re. Vbe is almost 1 volt due to higher than usual current - about 0.85 Volt. The spec sheet allows only an approximate estimate of this value.
As Ib >> Ic (extremely unusual) Vce sat is very low - = 0.03V !!!.
Ic = V/R = (V+ - Vc)/Rc = (10-9.2)/3k6 = .8V/3k6 = 0.222 mA.
As Ie = Ve/Re = 9.17V/1k = 9.17 mA
then Ib = Ie - Ic = 9.17-0.222 = 8.95 mA.
SO Ic/Ib = 0.222/8.95 = 0.025 = forced beta of 1/40th.
Small signal transistors typically have Betas (Ic/Ib ) in the 50-600 range. "Forced Betas where Ib is purposefully set highertahn needed to achieve nomonal saturation or a specified Vce may be set to 100 or 50 or laybe even as low as 10 if a really high level of overdrive is wanted. In this case the forced Beta of 0.025 (40 mA of base drive for every mA of collector current) is over 1000 times higher than you'd usually see in extreme cases. The transistor is ON.

Which explains the most unusual sotuation originlly described.
Most of the others are much simpler.

R1o: R2 pulls base to ground. Q1 is hard off.
Vc = V+ as no I_rc
Ve = 0 as no I_Re

R2s: Same effect as R1o above - base is again at ground with same result.

R2o: Transistor is turned on by R1 and voltages staiblise at "expected" levels.
Ib = (V+-Vb)/R1 = 0.662 mA.
Ie = Ve/Re = 2/68/1k = 2.68 mA.
Beta = Ie/Ib = 2.68/0.662 = 4 = very low.
Vce = (2.73-2.68) = 0.050V = very hard on.

Res: Transistor hard on via R1/R2 divider drive.

Reo: Emitter is left "floating and emitter floats somewhat below Vb set by R1/R2 divider. e is high Z and meter measurement of voltage changes Ve.

Rcs: A1 is an emitter follower with c at V+. Vb set by divier and Re sets Ie.

RCO: Mainly be diode action as in 1st case above. Here Vb is set by divider so is < 10V so Ie is liower than in 1st example.

CES: Example is wromg. The say CEs means all terminals short but it should mean CE short only. Resut similar but not idential. Either way the resistor dividers are the main factor.

CEo: Transistor absent. C at V+. e at gnd. B at divider potential,

enter image description here

  • \$\begingroup\$ But when \$R_2\$ is open, isn't it the usual emitter-feedback circuit.And I have a set of equations to analyse it but the answers are not matching. When you found \$I_b\$, how did you know \$V_b\$? \$\endgroup\$
    – Bijesh K.S
    Nov 3 '16 at 7:35
  • \$\begingroup\$ @BijeshK.S Yes - R2 open situation is as you say. What figures do you get and how do you get them. Your method mUST make assumptions. What are they.? It would be useful to see what your methods and formulae are. \$\endgroup\$
    – Russell McMahon
    Nov 3 '16 at 10:26
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    \$\begingroup\$ I probably got Vb from his simulation BUT the figure can be arrived at APPROXIMATELY without much trouble. This cct is also iterative as , as Ie rises Ve rises and this affects Ib and that affects drive to Q1 and that affects Ic and ... . You can guarantee that the transistor will be driven hard so Vce ~= 0 (say ~~~= 0.1 - 0.3V expected) (he says 0.05V) so Ve will be > 1k/(1k+3k6) x 10V as Ib also flows in Re so Vb will be ABOUT 0.6V - 0.7V higher and Ic ~= (10-0.1-Ve)Ib ~= (10-0.6-Ve)/10k so Ib will be around 3 to 5x less than Ic so ... . \$\endgroup\$
    – Russell McMahon
    Nov 3 '16 at 10:26

I think they are wrong.
I'll 'wander around the subject' and see what comes out.
Wanders .... Concludes: He seems to be dealing in pedantic stupidity and this seems to be an exceptionally poor and low value example unless a very strong point is being made.
Where did you get this example.

He's decided that Vbe is 0.83V. That's higher than the traditional value used. It's temperature dependent and depends om the basic transistor model and assumptions. 0.6V to 0.7V is more typically assumed.
Reason - see below:


Once you have Ve you should be able to use
Ie = Ve/R1 = 9.17 mA and
as Ic ~= Ie
then the voltage drop across Rc =
V_Rc = Ic x Rc
~= Ie x Rc = 9.17 mA x 3.6k = 33V - which we have not got".

The most current you can have in Rc is V+/(RC + Re) = 10/ (3.6k + 1k) = 2.17 mA.

Something is "wrong".

He allows Vce = 9.2V - 9.17V = 0.03V saturation voltage.

He MAY be allowing base current to be added to Ie So the Vbe junction is acting as a diode. Most of the 9.17 mA in Re comes from the be diode. In the 2n3904 datasheet on page 2 we see Vbe sat at Ib = 5 mA and Ic = 50 mA is 0.95 V max.
The line above that gives 0.85V max.
NEITHER line is a good match to these conditions but Vbe of around 0.95V looks close enough. His 0.83V looks a bit low but he may have used another datasheet - this sort of spec is not usually of vast importance or interest.

He may have dug into the data sheet to get the very low Vcesat value used - but it will be about as meaningless as most of the rest of what he has done.


  • Vbe is in the order specified.
  • Most current in Re comes via the be diode.
  • This is a really bad and stupid and useless example for most purposes. Alas.
  • \$\begingroup\$ I don't know if something is wrong, I almost know nothing about the subject.Andy's answer made sense to me. I'm referring to Electronic principles 7th edition by Malvino. \$\endgroup\$
    – Bijesh K.S
    Nov 2 '16 at 9:28
  • \$\begingroup\$ My answer now ~= complete. I'm about to comment to Andy whose answer would be OK in most cases but misses an important point here. \$\endgroup\$
    – Russell McMahon
    Nov 2 '16 at 9:29
  • \$\begingroup\$ Actually, the author is discussing about this in the section titled "troubleshooting". There is a table listing troubles and this was one of them. The author has said the voltmeter used to make the measurements has an input impedance of \$10\$Mega Ohm \$\endgroup\$
    – Bijesh K.S
    Nov 2 '16 at 9:34
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    \$\begingroup\$ @BijeshK.S I'd have to see the surrounding text. Here the transistor is essentially a be diode with minimal other effects. This very very very hard driven mode CAN be useful in rare circumstances but you'd seldom meet it. Here knowing that V_Re ~~~= Vsupply is probably good enough. \$\endgroup\$
    – Russell McMahon
    Nov 2 '16 at 9:42
  • \$\begingroup\$ I don't think I've missed an important point - the base collector is, as I stated, in forward conduction thus, the transistor is no-longer operating as one would expect a normal transistor circuit to operate. \$\endgroup\$
    – Andy aka
    Nov 2 '16 at 9:51

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