Engineering How to Determine the Base Current in a Saturated BJ Transistor Circuit?

[tsuwal]

Homework Statement

Find the base current of the following circuit

Homework Equations

The Attempt at a Solution

The solution to the problem is 4 micro amperes. Why? I get base current = 0 A because the transistor is satured and colector current=emissor current=1 mA

 

[darkxponent]

The Attempt at a Solution

The solution to the problem is 4 micro amperes. Why? I get base current = 0 A because the transistor is satured and colector current=emissor current=1 mA

Base current zero in saturation mode. you mean beta is infinite!. Sure?

 

[tsuwal]

base voltage = 7-1,5=5,5
emissor voltage=base voltage-0,6=4,9
emissor current=1 mA
If it is saturated then
colector voltage = 5,1 and current also 1 mA

?

 

[gneill]

There are a lot of assumptions being made, including the assumption that the transistor is in saturation.

Suppose that you assume that it is not saturated, and that I_C = β I_B. What values do you calculate for I_B and I_C? How about V_CE? Are the results self-consistent, or is saturation confirmed?

 

[tsuwal]

I have done the math since the beggining, I just want to know if the solution is wrong or not. If is not satured then ib= 2 micro Amperes and ic=998 micro Amperes. Colector voltage becomes 5,11 and Vce=0,21. It is in the active zone. solution is wrong right?

 

[gneill]

I have done the math since the beggining, I just want to know if the solution is wrong or not. If is not satured then ib= 2 micro Amperes and ic=998 micro Amperes. Colector voltage becomes 5,11 and Vce=0,21. It is in the active zone. solution is wrong right?

It does seem to me that a base current of 4 μA is too high for this circuit, and that your 2 μA is a more reasonable value.

 

[CWatters]

I agree with 2uA calculated as follows...

The base voltage is fixed at 5.5V so the emitter voltage will be 4.9V (cannot be lower than 4.9V).

Therefore

Ie = 4.9/4900 = 1mA

Ie = Ib + Ic

Ic = Ib * hfe

Ie = Ib + hfe*Ib

Ib = Ie/(1+hfe)

Substitute values

Ib = 1mA/501 = 2uA

Ic = Ie - Ib = 0.998mA

That gives a collector voltage of 5.11V. Not quite saturated.

 

[CWatters]

On reflection I think it unlikely that the LED would drop 1.5V at 2uA. So the base voltage would be pulled higher and transistor probably would be saturated.

Will try and have another look at it.

 

[darkxponent]

On reflection I think it unlikely that the LED would drop 1.5V at 2uA. So the base voltage would be pulled higher and transistor probably would be saturated.

Will try and have another look at it.

That is a good point but i don't that the question is asking for the non-linear operation of the diode. But go have a look, i am also interested in knowing the behaviour of the ciruit.

 

[CWatters]

Ok try this. Let's not assume the transistor is saturated for the moment. Let's replace the LED with an unknown resistor value Rb and assume Vce is unknown. Then I think you can write the following three KVL equations...

(Ib+Ic)*Re + Vce + (Ic*Rc) - 10V = 0

(Ib+Ic)*Re + Vbe + (Ib*Rb) -7V = 0

7V - (Ib*Rb) + Vbc + (Ic*Rc) = 0

Also

Ic = Ib*hfe

Vbe + Vbc + Vce = 0

I believe there are 5 unknowns, Ib, Ic, Rb, Vbc, Vce.
Assume Vbe = 0.7V.

Then solve for all unknowns.

I've not tried it !

 

[darkxponent]

I don't think one can take take the resistance of a diode unless it is given there in the question itself. We just replace it by a voltage source for forward biased mode and
open circuit for reversed biased mode. Even if we have to take the resistance of diode we have to take the voltage source of 1.5V with it in series.

So the equation 7V - (Ib*Rb) + Vbc + (Ic*Rc) = 0 would modify to 7V - (Ib*Rb) - 1.5 + Vbc + (Ic*Rc) = 0.

But still i don't agree with the notion of taking Rb, without being given in the question.

 

[CWatters]

I wasn't really suggesting you calculate a fixed value for the diode resistance. There is no need to. It should be possible to solve for Ib without actually calculating an equivalent value for Rd. I dissagree on Vd. You would not need to take the Vd=1.5V into account. My point was that the base current is so low that it's not valid to use a model that has a constant voltage drop.

If you substitite Ib*Rb with 1.5 V that means you have 5 equations and 4 unknowns. Is it possible that would give you multiple answers?

 

[rude man]

The transistor is most definitely saturated.

So to get the base current, use ie = ib + ic. If saturated, what is Vc? And therefore ic? And ie? And then the grand finale : ib?

Hint: ib is way, way bigger than 2 uA! And beta has nothing to do with the problem.

 

[darkxponent]

The transistor is most definitely saturated.

If you could tell us how you got the saturation mode. Because i don't find any problem with the solution given by CWatters.

I agree with 2uA calculated as follows...

The base voltage is fixed at 5.5V so the emitter voltage will be 4.9V (cannot be lower than 4.9V).

Therefore

Ie = 4.9/4900 = 1mA

Ie = Ib + Ic

Ic = Ib * hfe

Ie = Ib + hfe*Ib

Ib = Ie/(1+hfe)

Substitute values

Ib = 1mA/501 = 2uA

Ic = Ie - Ib = 0.998mA

That gives a collector voltage of 5.11V. Not quite saturated.

Now Vcb = 5.11-5.5 = -0.39, which is greater than -0.4.
If you could tell us where CWatters is wrong becouse i agree with CWatters.

 

[rude man]

If you could tell us how you got the saturation mode. Because i don't find any problem with the solution given by CWatters.




Now Vcb = 5.11-5.5 = -0.39, which is greater than -0.4.
If you could tell us where CWatters is wrong becouse i agree with CWatters.

ie = (V - 2Vd)/Re = 5.5V/4.9K = 1.12 mA
ic = (10-5.5)/Rc = 4.5V/4.9K = 0.92 mA
ib = ie - ic = 1.12 - 0.92 = 0.20 mA

EDIT: I did not notice that the drop across the LED is 1.5V. That makes C Watter's analysis essentially correct. (I would assume Vbe = 0.7V rather than 0.5V).

There is about 0.4V headroom for Vce so the transistor is not in saturation and the use of beta = 500 is correct.

Sorry folks.


Q is saturated because the computation ic*Rc ~ ie*Rc = 5.5V which is > (10 - Ve) and so impossible.

 

[CWatters]

I used Vbe = 0.6 as that's stated in the problem.

Personally I'm still undecided how best to answer the question..

If you take the question at face value and use 1.5V for the LED drop then the transistor is not quite in saturation and you get Ib=2uA

However i think the LED drop is unlikely to be 1.5V at 2uA in which case Ib is likely to be higher. See my post #10 for a way to calculate Ib in that case.

Will be interesting to see what the official answer is.

 

[rude man]

I used Vbe = 0.6 as that's stated in the problem.

Personally I'm still undecided how best to answer the question..

If you take the question at face value and use 1.5V for the LED drop then the transistor is not quite in saturation and you get Ib=2uA

However i think the LED drop is unlikely to be 1.5V at 2uA in which case Ib is likely to be higher. See my post #10 for a way to calculate Ib in that case.

Will be interesting to see what the official answer is.

I agree, 1.5V for an LED runing only 2 uA is unlikely. For another thing, no way is it putting out visible light.

Per the OP's post the 'official' answer is 4 uA.

I have shown how to compute ib if the transistor is saturated in my previous post, wherein I assumed a drop of 0.8V for the LED (1.5V total LED plus Vbe).