Mode of operation for transistor

[magnifik]

I am trying to determine the mode of operation for the following circuit and find the voltages and currents:

I am given that β = 50

I know that V_EC > V_EC,sat for a transistor in active mode
I applied KVL on the right loop and got:
V_CC = 300I_c + 1000I_c + V_EC
V_EC = V_CC - 1300I_c

i'm not sure if the above is right or where to go from here. what is throwing me off is the 300 Ω resistor connected to the 1kΩ resistor. is the node between the two B or C? or are B and C equal in this case?

thanks in advance.

 

[rude man]

Assuming the transistor is on, what must be the base voltage without writing down any equations at all?

 

[magnifik]

Assuming the transistor is on, what must be the base voltage without writing down any equations at all?

V_BE = V_BE,on = 0.7 V (as defined in the text)

 

[rude man]

That's the voltage between the base and the emitter. So what's the voltage at the base?

 

[magnifik]

That's the voltage between the base and the emitter. So what's the voltage at the base?

i'm unsure. is it the voltage at that node between the two resistors?

 

[rude man]

Why yes, the base is the node between the two resistors.

 

[magnifik]

what confuses me about this is the position of Vc. normally, it is the voltage across the bottom resistor, but since there are two resistors with a node in between, is Vc the voltage across the 300-ohm resistor? or does Ic occur in both the 300-ohm and 1k-ohm resistor

 

[rude man]

Vc is the collector voltage to ground. "c" stands for "collector".

The base voltage is Vb = Vcc - Vbe = Vcc - 0.7V.

Ic is the current flowing out of the collector.

 

[magnifik]

V_cc = V_be,on + V_b
--> V_b = 2.5 - 0.7 = 1.8

V_b/1000 + (V_b-V_c)/300 = 0
1.8/1000 + 1.8-V_c/300 = 0
--> V_c = 2.34

I_c = 1.8/1000 = 1.8 mA

2.5 = V_ec + 300I_c + 1000I_c
--> Vec = .16

since Vec = 0.2 in this problem and Vec is less than this, the transistor is not in active mode so i re-solve for saturation mode?

 

[rude man]

V_cc = V_be,on + V_b
--> V_b = 2.5 - 0.7 = 1.8

V_b/1000 + (V_b-V_c)/300 = 0
1.8/1000 + 1.8-V_c/300 = 0
--> V_c = 2.34

I_c = 1.8/1000 = 1.8 mA

2.5 = V_ec + 300I_c + 1000I_c
--> Vec = .16

since Vec = 0.2 in this problem and Vec is less than this, the transistor is not in active mode so i re-solve for saturation mode?

Yes that's exactly right, sorry for the delay in answering.

So now what do you do? (Hint - the base current is no longer assumable to be ic/50).

 

[magnifik]

i solve for I_e then do the KVL for the CE loop. then solve for I_c from that equation. then get I_b from the relationship I_b = I_e - I_e. if I_b and I_c are both positive then the saturation mode assumption is correct.

thanks for pointing me in the right direction!

 

[rude man]

Yer' welcome! Post your results if you care to.