0
\$\begingroup\$

I need to build the cheapest amplifier possible knowing Rin, Rch, Vin <= 1 mV, Vout = 12 V and the simplified datasheet of four differents op-amps:

Question

datasheet

Here is my book's answer:

The answer

It is claimed that:

A non-inverting first stage is chosen so that the input impedance of the circuit is equal to that of the op-amp. to that of the op-amp

I don't understand that need. Maybe it has a link with impedance matching ? Yet, the impedance matching rule for tension is R2 >> R1, not R2 = R1 (which is for power matching).

Could I have drawn the circuit below ?

my proposal

\$\endgroup\$
1
  • 2
    \$\begingroup\$ You can do any circuit you want as there is no requirements made for the system. It is not forbidden to have input impedance of zero in the task, but choosing non-inverting buffer the input impedance is infinite. Or in real world whatever the op-amp has. So unless the task defines what kind of requirements there are, it is difficult to know why the book chose the answer it did. \$\endgroup\$
    – Justme
    Dec 20, 2022 at 16:35

2 Answers 2

2
\$\begingroup\$

The basic problem with the inverting configuration is that your source impedance Rin will also play a role in your gain. Therefore, the gain of the 1st stage of the amplifier you propose will be: $$Gain_{AOP1} = -\frac{R_1}{\left(R_2 + R_{in}\right)}$$.

If your Rin isn't well-defined (i.e. it can vary with respect to frequency or some other parameter), then this is undesirable. Of course, if you know the max. value your Rin can take, then you can take R2 to be around 2 orders magnitude higher than Rin so that the gain error due to your source impedance Rin is minimized.

In order to not deal with this, your book's answer uses the fact that, ideally, the input impedance of the op-amp is very high. Therefore, it doesn't matter whether Rin varies or not. You can, easily, see this if you imagine that there's a large R from the (-) input of the op-amp to ground. If you have a resistive voltage divider with a large R to ground, then you can imagine that the ratio towards the (-) input is very close to 1.

There other more complicated considerations to chose between both op-amp amplifier configurations, but your book is limiting to consider the input impedance of the amplifier to be as high as possible. Even though they don't say it, the closed-loop gain of the amplifier has to do with it, as I mentioned in my previous paragraphs.

EDIT:

As the other answerer posted, impedance matching isn't useful here since you're looking for a optimum voltage transfer and not power transfer. Power in this situation is just incidental. You just need to take care that your amplifier is able to provide the required output power, which is 0.72W (or 60mA to give a 12V output) given your load.

\$\endgroup\$
3
\$\begingroup\$

No, not impedance matching. Impedance matching is useful to maximize power transfer (as well as minimize reflections in RF applications). In this case you don't want to maximize power, you want to maximize voltage. What works best for that is an "infinite" input resistance, which results in zero attenuation of the input voltage, even if the source has some series resistance.

The input resistance of the non-inverting topology for an op-amp circuit is very high. In contrast, the input resistance of the inverting topology is much lower, equal to the value of the resistor between the input and the "-" input of the op-amp.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.