# Biasing of a circuit with two transistors BJT

I need to calculate the bias values. So far I have only written these equations, but something is missing. ($\beta=200$ for both transistors and $V_{BE\,on}=0,6\,V$) Capacitors=open circuits

$\frac{20-V_{C1}}{R_2}=\frac{V_{C1}-0,6}{R_1}+I_{B2}$

$I_{E2}=I_{C2}\times \frac{\beta+1}{\beta}$

What equations are missing me?

• The voltage across R1 equals the voltage across R3. – Rohat Kılıç Jun 5 '17 at 10:08
• You left out the collector current of Q1 in your first EQN, also saying Ib2 = 0 is a valid assumption. – sstobbe Jun 5 '17 at 14:11
• Your first equation should include the collector current for $Q_1$, as pointed out by sstrobe. – jonk Jun 5 '17 at 16:49
• @CarmenGonzález You are very close. Modify your first equation as sstrobe said. Also, you need to include an equation to construct $I_{B_2}$ from $V_{C_1}$. With those, you should find: $V_{C_1}=\frac{V_{B_1}\cdot\left[\left(\beta+1\right)^2 \cdot R_2\cdot R_3 + R_1\cdot R_2\right] + V_{CC}\cdot R_1\cdot R_3\cdot\left(\beta + 1\right)}{\beta\cdot R_1\cdot R_3 + R_1\cdot \left(R_2 + R_3\right) + R_2\cdot R_3\cdot\left(\beta + 1\right)^2}$. The result is highly $\beta$-dependent. – jonk Jun 5 '17 at 18:03
• @CarmenGonzález Just for your information, $\frac{\textrm{d}V_{C_1}}{\textrm{d}\beta}\approx -23.7\:\textrm{mV}$ at $\beta=200$. So that's how much $V_{C_1}$ shifts for a small change in $\beta$ near the calculated operating point. (Treating $V_{BE}$ as a constant for both BJTs -- in the above equation, $V_{B_1}=V_{B_2}=V_{BE}=600\:\textrm{mV}$.) That's a lot of change. – jonk Jun 5 '17 at 18:20

## 3 Answers

$V_{CE1} = I_{B1}\cdot R1 + V_{BE}$ (1)

Collector-emitter of Q1 is parallel to base-emitter of Q2 plus R3. So,

$V_{CE1} = I_{C2}\cdot R3 + V_{BE}$ (2)

From (1) = (2), $I_{B1} \cdot 1000k = I_{C2} \cdot 3.3k$.

I'm sure you can take it from there by yourself.

We are not here to do your homework for you.

There are various ways to approach this. Here is one.

As you say, the capacitors are open circuit for the purpose of finding the DC operating point. That leaves two transistors and three resistors.

For the exercise, unlike in a real circuit, the gain of the transistors is explicitly known. Once you know the voltage of the collector of Q1, everything else falls out trivially. Start by projecting R3 back to the collector of Q1. That becomes a load on R2. Don't forget the B-E voltage source in series with the projected resistance. R2 and this projected resistance together form a Thevenin source replacing R2 and the 20 V power supply.

Now think about what R1 and Q1 together look like as a load. These can also be simplified to a Thevenin source. Now you have two Thevenin sources connected together. Solve for the resulting voltage.

Once you have the Q1 collector voltage, the rest follows easily. By known that voltage, you know the current thru R1, which gives you the current thru Q1. Likewise, you know the voltage on the emitter of Q2, which is the voltage across R3, which gives you the emitter current of Q2.

While this is a good exercise to get used to simplifying circuits one step at a time, keep in mind that in real transistor circuits the transistor gains are never that well known. Good transistor circuits work from some minimum gain all the way to infinite gain without large changes in their operating points or characteristics. Part of designing good transistor circuits is making sure this is true.

If I needed to find the bias point of this circuit in reality, I could not make the assumption of some nice well known gain. I usually start with infinite gain, and see what that results in. That's actually easier to do than any finite gain. Then I start with the operating point found from the infinite gain approximation and change it incrementally by substituting some finite minimum gain the transistors must have. If the result doesn't change much, then the first approximation is good enough. If it does change a lot, then it's a crappy circuit.

You left out the collector current at the Vc node,

$$\frac{20-V_{C1}}{R_2}=\frac{V_{C1}-0,6}{R_1}+I_{B2} + \beta(\frac{V_{C1}-0,6}{R_1})$$

Start by assuming $I_{B2} = 0$, once you solve Vc1 you can calculate what Ib2 would actually be and you will see its negligible.

Try to recognize basic building blocks, Q2 is an emitter follower with an input resistance of $\beta R_E$ = 200*3k3 = 660k. 660k compared to R2 = 5k6, is negligible for hand analysis.

• I hadn't read your answer (it was at the bottom of the page) until after commenting. But this was an important catch of a flaw for the OP that others didn't bother clearly calling out. +1 – jonk Jun 5 '17 at 16:50
• @Jonk, yes, since the OP took the time to enter the EQN with mathjax it was easy enough to add the collector current term. KCL at the Vc node is how I would have approached this problem too. So, other than forgetting collector current, good effort from OP. – sstobbe Jun 5 '17 at 17:03
• That's almost the same thoughts I had. I try to first see if I can follow the OP and put myself into their mind and see how to guide from there. (Or just answer, if I'm in that mood.) Since the OP had bothered to use KCL on a node, the first thing I do is see if I can replicate it. So in this case, it just fell out right away. I agree that the OP is putting in good effort, too. So in a case like this, I might flip either way: writing a full answer, or not. I'm not so stuck on the idea of holding short, all the time. Especially if I imagine they can pick up the pieces after. – jonk Jun 5 '17 at 17:11