# Determine the input impedances of a circuit

I'm studying a circuit with BJT's and I'm asked to determine the input resistances of the two amplification steps of the circuit. The circuit I'm analyzing is the following one:

To determine the input resistances I performed an AC analysis and obtained the following results:

I know that the values are not perceptible but they are Rin1=23.272231 kohm and Rin2=220.77106 kohm. I decided to choose a somewhat random value in the zone of the graphic where the curves are constant (I'm not sure if this is correct...).

Well this value are very different from what I obtained when I analyze the theoretical values of the input resistances.

Using:

$$R_{i2}=(β+1) R_{E3}//R_L+r_{π2}$$

where r_{π2} is 9.375 kohm and beta is 300. I obtained R_i2=240.375 kΩ

And for Ri1

$$R_{i1}=R_1//R_2//(r_π1+R_{E1} (1+β))$$ I obtained R_i1=20.183kohm.

I used the small-signal analysis models to determine this...

Now this is odd. I'm getting an error of about 15% on Ri1 and 7% on Ri2. Am I doing something wrong or are this deviations perfectly normal and dependent on the methods used on the theoretical analysis and on LTSpice. Can someone help me clarify this?

• Your uploaded picture needs re-saving and re-uploading. Have you used the transistors, separately, to determine their values? Models usually differ from real life. Also, you could try to use a current source at the input, and plot 1/V(vs), but I suspect it will be the same. – a concerned citizen Apr 22 '18 at 6:42
• How do you determine the r_pi value? Also, LTspice includes ro resistance and VT at 27 degrees, – G36 Apr 22 '18 at 8:53

How does the SPICE model differ from hybrid-pi? Are those differences important?

• Beta is slightly different (294.3 versus 300). Also the values of r_pi might be a little bit different, because the currents are slightly different than 0.8 mA. Will that be enough reason? – Granger Obliviate Apr 22 '18 at 2:44

At $1 \textrm{kHz}$ the $C_E$ capacitor reactance is around :

$X_C \approx \frac{0.16}{F \cdot C}\approx 1.6\textrm{k}\Omega$

So the $Z_E$ impedance will be around:

$Z_E = R_{E1} + R_{E2}||X_{C_E} \approx 1.2 \textrm{k}\Omega$

And input impedance is:

$R_{in1} = R_1||R_2||(Z_E + r_e)(\beta+1) \approx 23.232 \textrm{k}\Omega$

Rin1 = 1/( 1/68 + 1/39+1/((1.2+0.0325)*301))

I assume $I_C = 0.8 \textrm{mA}$ and $r_e = \frac {26\textrm{mV}}{I_C} = 32.5\Omega$

As for $R_{in2}$ is equal around:

$R_{in2} = (R_{E3}||R_L||r_o)(\beta+1)+r_{\pi2}$

Because the LTspice includes $r_o$ (Early effect).

So to get almost exactly the same result you need to know $ro$ or Early voltage $\textrm{VAF}$ in LTspice.