BJT bias - VDB open goes to emitter-feedback

Thread Starter

sbixby

Joined May 8, 2010
57
I'm working on a problem in Chapter 8 of Malvino/Bates "Electronics Principles", and one in particular is driving me nuts.

In this problem, we start with a VDB (voltage-divider bias) and open the R2 resistor; the result is essentially an emitter-feedback, with the BJT saturated.

There is an example that specifies expected voltages at the transistor's connections for this situation, as well as a problem which does the same thing with another example, and I can't seem to calculate voltages for either example that makes sense. I've tried it a number of ways using the equations in the chapter.

Attached are two screen-caps from Paul Falstad's applet that I was using to visualize things. For the "normal" one, I have no problem getting matching numbers, but for the R2-open one, I'm stuck - I'm not clear on how the saturated transistor gives the numbers shown.

Could someone give me an example of getting the voltages at CBE in the "r2open" attachment?

Note that the applet's model is using a Vce of about 0.6v rather than the "standard analysis assumed value" of 0.7v, but that still doesn't give me reasonably close numbers using the "emitter-feedback bias" equations from the book.
 

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Audioguru

Joined Dec 20, 2007
11,249
When the transistor is saturated then it is simply a diode with no current gain. Therefore the base current adds to the emitter current which raises the emitter voltage. When the emitter voltage rises and since the transistor is saturated then the collector voltage also rises.

All the current numbers add correctly.
 

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t_n_k

Joined Mar 6, 2009
5,455
So putting some numbers to your example.

One has to assume a VCEsat value.

To make it simple let's say VCEsat≈0, VBE=0.6 (as you say)

Then VE=(Ib+Ic)*1k

But

Ic=[10-(VCEsat+VE)]/3.6k

&

Ib=[10-(VE+VBE)]/10k

Or

VE=[10-(VCEsat+VE)]/3.6+[10-(VE+VBE)]/10

VE=[10-VE]/3.6 + [10-0.6-VE]/10

Which solves to give

VE≈(10/3.6+9.4/10)/1.38=2.69V

From there you can find the other values.

If you assume VCEsat is ~50mV you would get a slightly different answer.

Also keep in mind that the value of the stated current gain HFE = 100 doesn't have much meaning in saturation.
 

Thread Starter

sbixby

Joined May 8, 2010
57
From simulations, I have no doubt the numbers add up, but I'm not sure how we get to those numbers in the first place.

IOW: How did you establish the currents? I know that Ie=Ib+Ic, and those numbers jive.

How did you get Ie (or Ib or Ic) to start with?
 

Thread Starter

sbixby

Joined May 8, 2010
57
t_n_k replied just before I posted my follow-up.


Thanks, guys! Putting numbers into equations pointed out my misunderstanding - that we use 9.4v across R1 along with 10v across Rc to get the total current. After that, as you say, the rest of the numbers fall out.


Gracias!!
 
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