# How to calculate the input impedance of a transistor in saturation

I have a simple circuit where a single transistor is used as an inverter. The transistor is in saturation when the Base of the transistor is high enough and the output, through the collector is driven low. But my question is how do I calculate the input impedance, when the transistor is saturated?

Here is my circuit

• Just calculate the ratio between the Vce and Ic. – Eugene Sh. Feb 7 '17 at 17:57
• @EugeneSh. But that will give me the output impedance. The output is taken from the collector. – Arjob Mukherjee Feb 7 '17 at 18:00
• Ow.. you want the input.. sorry. – Eugene Sh. Feb 7 '17 at 18:01
• Zin = 330 ohm plus the dynamic resistance of a diode (approximately). – Andy aka Feb 7 '17 at 18:02
• @Andyaka Yes I thought that too, but if I drive the base by a source of $Z_{out} = 330E$ (same as the input impedance), the voltage at the input stays at 3 volts not 2.5 volts. Ignoring the dynamic resistance of 2.6 Ohms ( $r_e = {26mv}/{10mA} = 2.6E$ ) – Arjob Mukherjee Feb 7 '17 at 18:12

It sounds like you are talking about a source that is generating a 5 volt square wave and you are expecting, due to a potential divider effect, the see 2.5 volt peak, yes?

Yes, you are correct.

Take a 1N4148 diode for example: -

When your signal generator is putting out a 5 volt peak, the current into the diode could be somewhere between 5/660 amps and (5-0.7)/660 amps. Thats a range of 7.6 mA to 6.5 mA.

As you can see, with this sort of current flowing, the diode produces a DC voltage of about 0.7 volts so this immediately adds to the 2.5 volts you expected giving you 3.2 volts.

This is a first level approximation. In reality, there will be about 0.7 volts on the diode and what remains (4.3 volts) is split equally in half by the two resistors so you would get 0.7 volts + 4.3/2 volts = 2.85 volts.

With a transistor, the base - emitter voltage my be a little higher so, as you can see, about 3 volts sounds reasonable.

• Ok I understand. So the input impedance is around 330E? But why is the dynamic resistance model not giving a clear idea in this case? – Arjob Mukherjee Feb 7 '17 at 18:32
• I have no idea what you mean. – Andy aka Feb 7 '17 at 18:37
• Sorry, my mistake. I think I have got what you meant. – Arjob Mukherjee Feb 7 '17 at 18:43