# The concept of the input and the output resistance of a transistor amplifier circuit

I am a student of physics. I do not have a deep foundation in electronics. Is there a simple way to understand what is meant by the input resistance and output resistance of a circuit (in particular, that of a transistor amplifier)?

I have read the two-port hybrid parameter analysis of the BJT amplifier in common emitter (CE) mode. I have derived the expression for the input resistance $$R_i=h_{ie}-\frac{h_{re}h_{fe}R_L}{1+h_{oe}R_L}$$ and an even more complex expression for the output resistance $$R_o=\frac{h_{fe}R_L}{h_{ie}+(h_{ie}h_{oe}-h_{fe}h_{re}) R_L}.$$

The derivation is fine! But I don't have a good intuition for these expressions. Physically, what do these expressions really mean? Do they represent some equivalent resistances? I so, the equivalent of what? Also with these expressions, I don't have any feel for why the input resistance must be very high and the output resistance must be very small, for a good amplifier.

• I think you have two questions. One is hard and the other is easy. The intuition part is hard and what you need is sit-down time so that someone who has really good (and multiple) visualizations can pass along to you what they "see" in their mind. The why about input resistance being high and output resistance being small is easy. You can design sections in isolation (ignorance) of what surrounds the section if and only if the input resistance of your section is high enough and the output resistance of your section is low enough. For signals as voltage, anyway.
– jonk
Oct 21, 2022 at 4:57
• Do you know about the voltage divider rule? Oct 21, 2022 at 6:02
• Note that those are small signal quantities. They're not 'real' resistances, that is for a transistor you can't really stick a resistance (whether i/o it's irrelevant). Strictly speaking. I think your best bet is to first read up on small signal analysis(SSA), then what it means when applied to bipolar transistors (which you're referring to, and I assume you know the physics involved), and then finally at amplifiers if you like pain and analog electronics. Sedra-Smith is a good undergrad book, or Grey Meyer for grad I'd say.
– edmz
Oct 21, 2022 at 12:24

In short (without formulas) - just for your understanding:

• The input resistance r_in of a circuit is determined and defined by the input current i_in , which flows into the circuit when the signal voltage v_in is applied (r_in=v_in/i_in). For a voltage amplifier, this resistance should be as large as possible.

• The output resistance of an amplifier stage determines how much the amplified signal output voltage drops when a load (e.g. a second amplifier stage with finite input resistance) is connected. The law of parallel connection applies, so this output resistance should be as small as possible to minimize the influence of the load.

• This requirement is usually not fulfilled with a transistor stage in common emitter configuration - but much better with the common-collector (emitter follower).

• The output resistance is measured, calculated or simulated by applying a test voltage to the output to be measured (with signal input source grounded) and determining the current flowing into the circuit.

• Remember the operational amplifier which is a very good voltage amplfier because the input resistance is extremely large (assumed to be infinite during most of the calculations) and the output resistance is very small (assumed to be zero in most cases). Both values are significantly improved by the influence of the negative feedback.

I don't have any feel for why the input resistance must be very high and the output resistance must be very small, for a good amplifier.

Transistor amplifiers amplify the signal, and different amplifiers can be cascade connected to further increase the amplification factor. For example, if you need an amplification factor of 1000 (e.g. amplify a 1mV signal to 1V before feeding to power amplifier) then you can cascade connect, for example, 2 amplifiers having amplification factors (or gain) of 40 and 25, respectively (40 x 25 = 1000).

When it comes to cascade connect two or more amplifiers the input and output impedances become important. I'll try to illustrate this in the following diagram:

simulate this circuit – Schematic created using CircuitLab

NOTE: Ideal amp has infinite input resistance, and zero output resistance. By placing Zin in parallel to infinite the actual input resistance becomes Zin. Likewise, by placing Zout in series with zero output resistance the actual output resistance becomes Zout.

As you can see from the diagram above, the signal source's output impedance, RS, forms a voltage divider with AMPLIFIER-1's input impedance, Zin1. So the voltage seen by AMPLIFIER-1 will be

$$V_{i1}=V_S\ \frac{Z_{in1}}{Z_{in1}+R_S}$$

Likewise, the AMPLIFIER-1's output impedance, Zout1, forms a voltage divider with AMPLIFIER-2's input impedance, Zin2. So the voltage seen by AMPLIFIER-2 will be

$$V_{i2}=V_{o1}\ \frac{Z_{in2}}{Z_{in1}+Z_{in1}}$$

As you can see, if the input impedances are not high enough to neglect output impedances, or in other words, if the output impedances are not low enough to be neglected by input impedances then the voltage seen by the amplifiers will be lower. This will, therefore, result in lower output voltage which is not a desired outcome.

There's another thing: Loading the previous section. For example, if the signal source doesn't like to be loaded (e.g. because of being current-limited) then feeding its output to a low-input-impedance amplifier may damage or even destroy it. That's why you can't connect a dynamic microphone's output directly to a speaker because the mic will be loaded and therefore won't work properly.