# Finding relevant voltages and current in a BJT

I'm supposed to find Vb, Ve, Ic, Vc and Vce for this circuit but I have never seen anything like this, so I'm lost on how to even approach it.

Do I start from KVL and write two loops? But then there's way too many unknowns. Here's a very poor attempt at writing one:

Top loop: -VR1 - Vbc - VRc + 20 = 0 Bottom loop: -VR2 - VRe - Veb = 0

Do I assume the NPN transistor is in active mode and set Vbe = 0.7v? We don't know that until we find the actual relevant voltages around the transistor.

This somewhat resembles a voltage divider, but it doesn't seem like one.

Rough guideline is fine, I'd like to know a starting point on how to tackle this problem.

• @Vbe=0.7V? The transistor doesn't know either if it will be in its active mode until the circuit is actually powered up. There is only one way to find out and that is by calculating voltages and currents and see if they make sense. From top of mind calculations I think the transistor will be in its active region. – jippie Jan 23 '16 at 18:57
• Dou you have the current gain of Q1? If not, this can only be solved approximately. – Martin Zabel Jan 23 '16 at 18:57
• @MartinZabel No current gain given, but the answer has actual voltages/currents on them. They are Vb = 2.564V, Ve=1.864V, Ic=1.695mA, Vc=12.35V, Vce=10.486V. – Xiagua Jan 23 '16 at 19:00

R1 and R2 make up a voltage-divider which is loaded by Q1. That, is some current may flow into the base of Q1. So, at first, you should check, if Q1 could be in active mode. Without any load, the voltage divider gives a voltage of $$V_{mid} = 20V*\frac{R2}{R1+R2}=2.6V$$ at its mid-point. A current of $$I_{12}=\frac{20V}{R1+R2}= 2.6 mA$$ will flow through the divider. So, as long as the base current of Q1 is much less than this, it would not sensibly change the voltage at the mid-point.

Q1 must now be conducting. Because if not, the voltage across Re would be 0V and the base-emitter voltage 2.5V which is much greater than the often necessary 0.6V.

Let's assume, that the base-emitter voltage $V_{BE}$ is nearly 0.7V, because the collector current changes exponentially with $V_{BE}$. Then, you can calculate the current through Re to

$$I_{Re} = \frac{V_{Re}}{Re} = \frac{V_{mid} - V_{BE}}{1.1kOhm} = 1.7 mA$$

This current flows through the emitter of Q1 and also roughly to the collector of Q1 if the current gain is high enough (e.g. greater than 100). Now, you can calculate the collector-emitter voltage to:

$$V_{CE} = 20V - V_{Rc} - V_{Re} = 20V - 1.7mA*(4.7kOhm+1.1kOhm) = 10V$$

Thus, Q1 is in the active region.

• I didn't recognize the voltage divider because that capacitor C1 threw me off. What exactly is the purpose of putting that capacitor there? – Xiagua Jan 23 '16 at 19:32
• @Xiagua The capacitor lets only AC pass through from the input. Thus, the DC offset of the input does not affect the operating point of the circuit. – Martin Zabel Jan 23 '16 at 20:17

There would be at least three loops since Vcc is really an ideal voltage source connected to ground. But mesh analysis is not a good way to solve this problem.

You need to understand the relationship between Ib, Ic, and Ie, which is based on the transistor's beta (hFE). This circuit is a standard common emitter amplifier topology. If this is homework, you probably should have covered CE amplifiers in class by now. If you haven't, that's unfortunate. :-)

The first step is to know that the equivalent input resistance looking into the base is equal to beta * Re. This lets you calculate the base voltage. From there, assume the transistor is in active mode and calculate the other voltages and currents one at a time.