Understanding Transistor Biasing in a Circuit

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

This is a solved example in my Physics textbook . I don't understand how this problem can be done without specifying the battery V_BB in the Base Emitter loop .

The author assumes that the 20 V battery is connected between both Base-Emitter as well as Collector Emitter ?

Instead it should be V_BE = V_BB - I_BR_B ?

V_BB is the biasing battery in Base Emitter loop .

Please help me understand this .

 

[cnh1995]

The author assumes that the 20 V battery is connected between both Base-Emitter as well as Collector Emitter ?

Here only one battery is used.
What exactly is your confusion?

 

[Jahnavi]

Here only one battery is used.
What exactly is your confusion?

How is 20V battery connected between Base and Emitter ?

How does 20 V battery come in picture while writing KVL in the Base Emitter loop ?

In fact where is the loop to write KVL ?

 

[cnh1995]

How is 20V battery connected between Base and Emitter ?

Like this:

 

[Jahnavi]

Like this:
View attachment 217486

I am talking about Base-Emitter not Collector-Emitter .

 

[cnh1995]

I am talking about Base-Emitter not Collector-Emitter .

How many paths from +ve terminal to -ve terminal do you see here? The battery can send current through BE junction. In fact, many of the practical transistor circuits use single power source.

 

[Jahnavi]

How many paths from +ve terminal to -ve terminal do you see here? The battery can send current through BE junction. In fact, many of the practical transistor circuits use single power source.

But not in the way the book seems to apply KVL

Could you draw the loop which the book assumes so as to apply KVL in Base Emitter loop ?

Please show me how is V_BE = 20 - I_BR_B ?

 

[cnh1995]

Please show me how is VBE = 20 - IBRB ?

Let's call the +ve and -ve terminals of the battery P and G respectively.
Starting from P, you can see there's a path
P→R_B→Base→Emitter→G.

Apply KVL to this path and see.

 

[Jahnavi]

Let's call the +ve and -ve terminals of the battery P and G respectively.
Starting from P, you can see there's a path
P→R_B→Base→Emitter→G.

Apply KVL to this path and see.

You are right

 

[gneill]

Very strange analysis given in yourbook for this example problem. Giving a circuit with all the part values including the transistor beta, then imposing a particular collector current is not right. Consider that a typical npn silicon transistor is unlikely to survive an 18.5 V base-emitter voltage! It's essentially a forward biased diode, and the resulting current would be huge.

Instead, I would have started with an approximation for the base current, assuming forward bias:

$I_B = \frac{(20 - 0.7)\; V}{120\; kΩ} \approx 160 \; μA$

Then found that if the transistor was operating in its linear region that this base current, via the transistor's β, would want to produce an $I_C$ that is closer to 32 mA rather than 2.5 mA . Clearly that can't happen because 32 mA flowing through the collector resistor would produce a potential drop of about 160 V, and there's only 20 V to work with. From this you should be able to draw your own conclusions about what state the transistor will be in, and calculate an approximate value for the actual collector current.

 

[Jahnavi]

Thanks gneill and cnh1995 .

 

[CWatters]

I agree with gneill, the circuit and the question are inconsistent with each other.

If you just look at the circuit and ignore the problem statement you get a base current of about (20-0.7)/120*10^3=0.16mA. The transistor will try to achieve collector current of 200 times that or 32mA. However the collector resistor limits it to about (20-Vcesat)/5000 ~ 4mA. So according to the circuit the transistor is fully on/saturated.