# Input resistance of a transistor circuit

I have a question from the razavi's microelectronics book:

I'm puzzled because in the solution provided, the first line states that:

"We know that the input resistance Rin = R1||R2||$r_π$."

While I understand if I transform the circuit's base input into Thevenin equivalent, R1 and R2 are in parallel (R1||R2). But shouldn't the small signal resistance $r_π$ be in series rather than in parallel?

e.g.
Vth --- Rth --- $r_π$ --- ground of emitter

Am I missing something fundamental?

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Isn't r$\pi$ in parallel with $R_2$? –  Spehro Pefhany Mar 18 at 6:06
yes it does. But when I transform it into a small signal model, with a Thevenin equivalence as the inputs, rπ seems to be in series.That's the part I'm having trouble understanding. –  jshen Mar 18 at 6:11
To put it in another way, if R2 and ground doesn't exist, then the input impedance should be R1+rπ. The Thevenin equivlance of 5.113 looks just like that, so why is its input resistance not calculated the same way? namely in series. –  jshen Mar 18 at 6:14

You need to follow a two-step process to solve this. First you find the bias point, and calculate the value of r$\pi$. For that, you use the Thevenin equivalent and Vt.

Once you have r$\pi$, you move to small-signal thinking, and in small-signal terms, the non-input side of R1, R2 and r$\pi$ are all connected to ground, so they are all in parallel.

Here's a nice MIT EdX open courseware lecture you might find useful.

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Look at the circuit, but mentally disconnect the base of $Q_1$for a moment. The signal is fed into the node shared by $R_1$ and $R_2$. Both these resistors load the input signal. So you treat them in parallel. Now, add the base back in. That base must also load the signal, too. So that must be treated in parallel, as well. The input impedance must be reduced by adding the transistor, not increased.

Imagine what you are suggesting as an alternative. Suppose you could make $r_{\pi}=\infty$? Then according to your thoughts, the input impedance would also be $\infty$! But that's exactly the situation with the $Q_1$base disconnected, isn't it? If you were right, then all voltage dividers would have infinite impedance to a source connected to them. But you know that doesn't make any sense.

The base of $Q_1$loads down the divider. It must always cause the input impedance to decline, not increase. Your scenario would help you compute the base current, though, if you somehow were told the value of $r_{\pi}$without it.

For 5.13, just by way of example, with $g_m \ge \frac{1}{260\Omega}$, this means that $I_c \ge \left(\frac{26mV}{260\Omega} = 100\mu A\right)$. At $\beta=100$, then $I_b=1\mu A$. Now $r_{\pi}=\frac{26mV}{1\mu A}=26k\Omega$ and $\therefore R_{th}\ge 16.25k\Omega$ to meet the $10k\Omega$ input impedance minimum.

Also, you now know that $V_{be}=\frac{kT}{q} \cdot \ln\left(\dfrac{100\times 10^{-6}A}{20\times 10^{-18}A}\right)\approx 760mV$.

So $V_{th}=V_{be}+R_{th} \cdot I_b=760mV+1\mu A \cdot 16.25k\Omega=776.25mV$. And that means that $R_1\approx 52.34k\Omega$ and $R_2\approx 23.57k\Omega$.

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While I understand if I transform the circuit's base input into Thevenin equivalent, R1 and R2 are in parallel (R1||R2). But shouldn't the small signal resistance rπ be in series rather than in parallel?

I see what you've done and why you're puzzled.

When you look out of the base and replace the source and base divider resistors with their Thevenin equivalent, you've made the resistors part of the source circuit; the Thevenized source impedance is then $R_1||R_2$.

But, this isn't what you want. What you want to do is to find the Thevenin equivalent looking into the node where the base and base divider resistors are connected.

Then, it is easy to see that $R_1, R_2, r_{\pi}$ each have one terminal connected to that node and that each have the other terminal connected to (AC) ground. Thus, they are all in parallel. This is the input impedance connected between that node and ground.

Am I missing something fundamental?

Yes, your approach isn't valid. The impedance seen by $V_{th}$ is, in general, not the same as the impedance seen by the actual source as you've discovered.

For example, consider this circuit:

simulate this circuit – Schematic created using CircuitLab

The resistance seen by $V_1$ is, by inspection

$$R_1 + R_2||R_L$$

Now, replace $V_1, R_1, R_2$ with their Thevenin equivalent circuit:

simulate this circuit

The resistance seen by $V_{th}$ is

$$R_1||R_2 + R_L \ne R_1 + R_2||R_L$$

Thus, it is a mistake to assume that the impedance seen by a Thevenin source is the same as the impedance seen by the actual source.

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