Loading effects of a signal source

Question

I'm reading "The Art of Electronics" (3rd edition) by Paul Horowitz and Winfield Hill. In the 2nd chapter on BJT (page No. 79, section Emitter follower). The author says:

".. In general, the loading effect of the following stage causes a reduction of signal. For this reason, it is usually best to keep Zout << Zin (a factor of 10 is a comfortable rule of thumb)."

Here Zin and Zout are the input and output impedance os an emitter follower circuit.

My question is, isn't it desirable to keep output impedance more than the input impedance to prevent loading effects? For example, consider the following:

If RL << R2, the equivalent resistance of the lower part of the divider would go down and hence there would be a reduction in Vout with the load. Can anyone please explain what am I missing here? Thanks for your help.

You are certainly missing something but, your equations tell you that with a lower load impedance you will get less signal to that load. However, your words are contradictory. — Andy aka, Nov 28 '18 at 16:11
@Andyaka, sorry for the confusion. As you said, the lower load impedance will deliver less signal to the load. My question was, why does the author say that it is best to keep Zout< — Vinit Shandilya, Nov 28 '18 at 16:16
First of all just consider a Thevanin equivalent, forget the divider. Then repeat your calculations. Reduce it to the trivial case where the output impedance is zero. Now your output voltage is independent of load resistance. Zout << Zload (Zin of the next stage) is approximating this condition. — John D, Nov 28 '18 at 16:17
draw the actual circuit, and then draw the equivalent models — analogsystemsrf, Nov 28 '18 at 16:39

score 3 · Accepted Answer · answered Nov 28 '18 at 16:26

Consider two cases:

a) \$R_L\$ = 0 (zero)

and

b) \$R_L\$ = infinite

It is obvious that a) will result in \$V_{out}\$ = 0 as the output is shorted, so this is a useless situation.

It should also be obvious that b) will result in the desired \$V_{out}\$ as there is no load on the output.

So the conclusion is that \$R_L\$ should be as large as possible.

Now we can get to the \$Z_{in}\$, \$Z_{out}\$ discussion. There you have to be vary careful what you call \$Z_{in}\$ and what you call \$Z_{out}\$!

The book uses:

\$Z_{out}\$ = the output impedance of the voltage divider.

\$Z_{in}\$ = the input impedance of the load = the load impedance.

You confuse yourself by mentioning the emitter follower.

For any amplifier there is a \$Z_{in}\$ and a \$Z_{out}\$

Now we connect the amplifier's input to the voltage divider's output.

Note that then the voltage divider's \$Z_{out}\$ needs to be smaller than the amplifier's \$Z_{in}\$.

Remember that \$Z_{in}\$, \$Z_{out}\$ are usually referred from the viewpoint of the circuit you're discussing. So one circuit's \$Z_{out}\$ connects to the next circuit's \$Z_{in}\$.

score 0 · Answer 2 · answered Nov 28 '18 at 18:28

This is quite easy to visualise if we take the Thevenin equivelant circuit. Thevenin said:

Any combination of batteries and resistances with two terminals can be replaced by a single voltage source e and a single series resistor r. The value of e is the open circuit voltage at the terminals, and the value of r is e divided by the current with the terminals short circuited.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. The equivalent circuits.

The open circuit voltage, V2, is given by \$ V_2 = \frac {R_2}{R_1+R_2}V_1 \$.

The value of 'r' is \$ R_3 = \frac {V_2} {\frac {V_1}{R_1} } = \frac {V_2}{V_1}R_1 \$.

With that out of the way it should now be clear that the lower the load resistance the more the voltage delivered to the load will drop.

If RL << R2, the equivalent resistance of the lower part of the divider would go down and hence there would be a reduction in Vout with the load. Can anyone please explain what am I missing here?

But H&H said "... it is usually best to keep Zout << Zin". That is, keep R₂ << R_L. You got it mixed up.

score 0 · Answer 3 · answered Nov 28 '18 at 19:32

I think you are considering Zout to be solely determined R2 - this faulty analysis then dictates that to lower Zout, one should lower R2 BUT that, in turn, lowers the output voltage irrespective of the load being applied.

But the output impedance (Zout) is dictated by both R1 and R2 AND, if they have values that set the output voltage at 50% (say for instance) then lowering both together keeps the ratio as it was and, importantly, lowers the output impedance.

Loading effects of a signal source

3 Answers3