# BJT Common Emitter Output Resistance

Hello, I am attempting to determine the small signal output resistance $$\R_o = v_x / i_x\$$ of the Darlington pair common emitter's small signal model.

I have written the equation KCL at node c,

$$\ i_x = g_{m_1}v_1 + (v_x - v_2)/r_{o_1} + g_{m_2}v_2 + v_x/r_{o2}\$$

and KCL at node a,

$$\ v_1r_{\pi_1} + (v_x - v_2)/r_{o_1} + g_{m_1}v_1 = v_2r_{\pi_2}\$$

How many more equations do I need to find a solution? How to I apply node voltage analysis to this circuit? Thank you much!

Do you really want to compute the oupur resistance - including ro1 and ro2 - by hand? You can expect a rather huge expression.

Here is the result from a symbolic calculator:

Note that the transconductance gm is here expressed by h21/h11=FHfe/Rhie and the output resistances as ro=1/Ghoe

( + Fhfe_Q1 Rhie_Q2 + Ghoe_Q1 Rhie_Q1 Rhie_Q2 + Rhie_Q2 + Rhie_Q1)

( + Ghoe_Q2 Fhfe_Q1 Rhie_Q2 + Ghoe_Q1 Fhfe_Q2 Rhie_Q1 + Ghoe_Q2 Ghoe_Q1 Rhie_Q1 Rhie_Q2 + Ghoe_Q1 Rhie_Q2 + Ghoe_Q1 Rhie_Q1 + Ghoe_Q2 Rhie_Q2 + Ghoe_Q2 Rhie_Q1)

Your first equation (KCL at node c) has two remaining variables to eliminate: $$\v_1\$$ and $$\v_2\$$. So you need two more equations total. Your second equation (KCL at node a) is one of these two (apart from the typos you need to correct).

KVL can give you the additional equation. This last equation is quite trivial, since Q1 base and Q2 emitter are both at small-signal ground. Therefore application of KVL between those points give you $$\v_1 = -v_2\$$.

We can try to split the circuit to simplify the equation and calculations.

I split your circuit and first what we will do is to find $$\R_{X1} = \frac{V_X}{I_X}\$$ for this part of a circuit:

simulate this circuit – Schematic created using CircuitLab

Additional we see that $$\r_{\pi 1}\$$ is in parallel with $$\r_{\pi 2}\$$ thus:

$$V_1 = - I_X \: (r_{\pi 1}||r_{\pi 2})$$

And

$$I_{ro1} = I_X - I_C = I_X - g_{m1}V_1$$

So we can write the KVL equation:

$$V_X = I_X \: r_{\pi 1}||r_{\pi 2} + (I_X - g_{m1}*V_1)r_{o1}$$

$$V_X = I_X \: r_{\pi 1}||r_{\pi 2} + (I_X - g_{m1}*(- I_X \: r_{\pi 1}||r_{\pi 2})r_{o1}$$

$$V_X = I_X \: (r_{\pi 1}||r_{\pi 2}) + (I_X + g_{m1} I_X \: (r_{\pi 1}||r_{\pi 2}))r_{o1}$$

$$V_X = I_X \left((r_{\pi 1}||r_{\pi 2}) + r_{o1} + g_{m1}(r_{\pi 1}||r_{\pi 2})r_{o1} \right)$$

And finally, we have:

$$R_{X1} = \frac{V_X}{I_X} =(r_{\pi 1}||r_{\pi 2}) + r_{o1} + g_{m1}(r_{\pi 1}||r_{\pi 2})r_{o1} = r_{o1}+(1 + g_{m1}r_{o1})(r_{\pi 1}||r_{\pi 2})$$

Now we need to find the equivalent resistance for this circuit:

simulate this circuit

$$I_X = \frac{V_X}{r_{o2}} + \frac{V_X}{R_{X1}} + g_{m2}V_2$$

We notice that $$\ V_2 = I_X*(r_{\pi 1}||r_{\pi 2}) = \frac{V_X}{R_{X1}}(r_{\pi 1}||r_{\pi 2}) \$$

Therefore we have:

$$I_X = \frac{V_X}{r_{o2}} + \frac{V_X}{R_{X1}} + g_{m2}\frac{V_X}{R_{X1}}(r_{\pi 1}||r_{\pi 2})$$

Thus the Darlington pair output resistance is equal to:

$$R_X = r_{o2}||R_{X1}||\frac{R_{X1}}{g_{m2}(r_{\pi 1}||r_{\pi 2})}\approx r_{o2}|| \frac{R_{X1}}{\beta_2} \approx r_{o2}$$

Where:

$$\R_{X1} = r_{o1}+(1 + g_{m1}r_{o1})(r_{\pi 1}||r_{\pi 2})\$$