Voltage divider bias:what if Rc is short circuited?

Hello, for a 4 resistor voltage divider bias circuit I have a question
that I need some help with.

Vcc = 18V
Vbe = 0.7V
Rc = 5k6
Re = 100
R1 = 27k
R2 = 4k7
Gain is not stated

1.State what will happen if resistor Rc has become short circuited?

My calculations are:
Rb = (R1 * R2)/(R1 + R2)
Rb = (27000*4700)/(27000+4700)
Rb = 4.003k

Vb = Vcc*(R2/(R1+R2))
Vb = 18 * (4700/(4700+27000))
Vb = 2.67V (2dp)

Ve = Vb - Vbe
Ve = 2.67 - 0.7
Ve = 1.97V (2dp)

Ie = Ve / Re
Ie = 1.97/100
Ie = 19.69mA (2dp)

Vc = Vcc = 18V

Vce = Vc - Ve
Vce = 18 - 1.97
Vce = 16.03V (2dp)

Ib = (Vcc - Vbe)/Rb
Ib = (18 - 0.7)/4003
Ib = 4.32mA (2dp)

Ic = Ie - Ib
Ic = 19.69mA - 4.32mA
Ic = 15.37mA

Vce cutoff = Vcc = 18V
I am not sure of my results, it seems as if the transistor is not shut off but as this question is only worth 2 marks I am probably missing something.
Even if my calculations are correct I don't know what answer to give

attached is an image of my schematic with Rc replaced by a short circuit
Thanks for any help any one can offer.

Since you are not told the value of Beta I would have just stopped at Ie=19.69mA [also assume Ic=Ie] and Vce=16.03V.

Not sure why you then went on with a calculation for Ib, which was incorrect.

Remember in this case the implicit initial assumption is that Ib is so small that it can be ignored and Vb is entirely determined by the voltage divider. If Beta is sufficiently small than the assumption is incorrect - but in this case the question seems to be framed such that the assumption is OK.

Your transistor is biased wrong and therefore will not work.
You have 19.7mA in its emitter and in its collector so the 5.6k collector resistor causes the collector voltage to be almost the same as the emitter voltage because the transistor is saturated.

Reduce the base voltage or increase the value of the emitter resistor.

Thanks for your response t_n_k
I tried to recalculate Vc with Rc put back in the circuit
and the result is:
Vc=Vcc-(Ic*Rc)
Vc= 18-(0.01969 *5600)
Vc = -92.25

So I guess that the answer to the question is that the replacing of Rc with a short circuit means that Vc = Vcc instead of -92.25V
and Vce = 16.03V instead of -94.22V

Thanks also Audioguru , the question is what will happen if Rc is replaced with a short circuit so I guess that I am expected to explain the outcome of the change to the circuit. I don't know if the resistor selection is intended to catch me out.

If you short RC then the current in the transistor will be 19.7mA then it will increase a little as the transistor heats up.

Actually, the circuit is so badly designed that the 5.6k Rc will limit the current in the transistor to only 3.1mA. Then the emitter voltage will be only 0.31V and the collector voltage will be only 0.41V.

The second part of this question is:
State what would happen if: The 4k7 resistor R2 has become a 'short circuit'.

My calculations:
Vb = 18V

Ve = Vb - Vbe
Ve = 18 - 0.7
Ve = 17.3V

Ie = Ve / Re
Ie = 17.3 / 100
Ie = 173mA

Ic ≈ Ie = 173mA

Vc = Vcc - Ic*Rc
Vc = 18 - 0.173 * 5600
Vc = -950.8V

Vce = Vc - Ve
Vce = -950.8 - 17.3
Vce = -968.1V

What conclusion can I draw from this result? does this mean if R2 is short circuited that the transistor will burn out.

Edited circuit image attached

hehe, no if you short R2 then BJT will be cut-off (off).
So nothing will burn.

I thought that Vce shutoff voltage is equal to Vcc so in this case 18V
but the value I calculated for Vce is -968.1V

What causes the transistor to shut off. does it matter if Vce is negative?

First of all that 2.7V. at the base is not going to stay at that same value, because if you have 2 volts for VE across 100 ohms, that's around 20mA. If the transistor were a short the most current you can have is around 3mA.

So forget having 2V. at the emitter, it won't happen once the base circuit is connected to the divider, your going to have a diode and a 100 ohm resistor in parrallel with R2, which is going to drop the base voltage considerably, if the collector resistor was small enough to keep the collector MORE POSITIVE, than the base term. than the transistor would conduct into it's linear region, and the divider would not be loaded to the point of dropping it's voltage so much.

So yeh, if the collector was shorted than the transistor would conduct, the 20mA. to keep the VE at 2V. while the VB sits at around 2.7V. according to the voltage divider, biasing it at that.

ShowerOf said:

I thought that Vce shutoff voltage is equal to Vcc so in this case 18V
but the value I calculated for Vce is -968.1V

What causes the transistor to shut off. does it matter if Vce is negative?

Click to expand...

Apparently you haven't realized that you can't get a negative voltage (relative to ground) in this circuit, because the power supply is positive.

hobbyist said:

First of all that 2.7V. at the base is not going to stay at that same value, because if you have 2 volts for VE across 100 ohms, that's around 20mA. If the transistor were a short the most current you can have is around 3mA.

So forget having 2V. at the emitter, it won't happen once the base circuit is connected to the divider, your going to have a diode and a 100 ohm resistor in parrallel with R2, which is going to drop the base voltage considerably, if the collector resistor was small enough to keep the collector MORE POSITIVE, than the base term. than the transistor would conduct into it's linear region, and the divider would not be loaded to the point of dropping it's voltage so much.

So yeh, if the collector was shorted than the transistor would conduct, the 20mA. to keep the VE at 2V. while the VB sits at around 2.7V. according to the voltage divider, biasing it at that.

Click to expand...

Hi hobbyist,

I'm a bit confused as to the conclusion you have come to.

Do you mean it's possible for the circuit as originally shown by the OP to have Ie around 20mA or not? You final statement seems to say that it is possible but the preceding paragraphs seem to say that it's not.

If the transistor Beta is high enough then it will be in the linear region - with Rc =0.

Putting in some numbers with Vcc=18V, R1=27k, R2=4.7k & Re1=100

If Beta = 300 & Vbe=0.669V [for simplicity of calculation] then .....

Ib=58.65uA, Ie= 17.65mA, Ve=1.765V, Vb=2.435V, Vce=16.23V, Ic=17.6mA

If Beta=50 & Vbe=0.669V [again] then ....

Ib=219.8uA, Ie= 11.21mA, Ve=1.121V, Vb=1.791V, Vce=16.88V, Ic=10.99mA

Hi hobbyist,

On re-reading your post I see you are talking in the first instance about the case where Rc is not zero.

Guess there is now no confusion on my part.

rgds,

t_n_k

With Beta=300, Vbe=0.67, etc., I calculate saturation would occur with Rc somewhere around 980 ohms and above.

t_n_k said:

With Beta=300, Vbe=0.67, etc., I calculate saturation would occur with Rc somewhere around 980 ohms and above.

Click to expand...

Actually on re-calculation, it was probably closer to about 900 ohms. For the assumed transistor parameters Vc is about the same as Vb at that point - all fairly insignificant really. I wonder what the value (from the student's perspective) in this exercise is all about.

I have corrected my schematic and added one with its collector resistor shorted.

t_n_k said:

Hi hobbyist,

On re-reading your post I see you are talking in the first instance about the case where Rc is not zero.

Guess there is now no confusion on my part.

rgds,

t_n_k

Click to expand...

Hi ,T_N_K

When I first looked at his original schematic, I was calculating 2V. across 100 ohms, 20mA. IE, but when I got that unreal voltage drop across RC, I thought, wait a minute, theres no way the emitter will be at 2V. with RC being that high,
then the conclusion I drew from it was , like a light came on, sure the base voltage, will drop considerably due to the high resistance of the 27K resistor, in the voltage divider with respect to the base emitter diode and 100 ohm resitor, parralleling the bottom resistor.

Ron H said:

Apparently you haven't realized that you can't get a negative voltage (relative to ground) in this circuit, because the power supply is positive.

Click to expand...

Thanks for your response Ron H, I understand your statement, but how do I
deal with the negative values for voltage resulting from my calculations?

If in calculations you get negative voltage, this mean that:
1. BJT is in a saturation and then Ic=Hfe*Ib don't hold any more.
So to find the solution you need to apply KVL, KCL and Ie=Ib+Ic
And solve the circuit.
2. Mistake in the calculation
http://forum.allaboutcircuits.com/showpost.php?p=148440&postcount=6
http://forum.allaboutcircuits.com/showthread.php?p=165135#post165135 (post 24)

Thanks Jony130 your answer is giving me something to think about
and I am reading the threads you gave links for.

I have much to learn