How to divide signal level from function generator

Question

I looked for this answer but was unsuccessful so I want to ask the experts here. I have configured an op amp in non-inverting mode (the first in a chain of 3 stages). It is for amplifying extremely low level noisy signals so I'm using lower resistor values. The input resistor is a 100ohm and the feedback resistor is 3K ohms. I'm feeding the circuit on the bench with a sine wave from my signal generator which has an output impedance of 50ohms and a minimum level of 4mV. I need to go lower to about 5uV minimum. The generator increases in 1mV increments.

My question is how should I configure a voltage divider in order to divide this signal down to 5uV (minimum) and maintaining a 50ohm impedance match. I need to measure the output of the entire amplifier chain while knowing what level is going in versus what level is coming out. Hope this makes sense and hope I've included all needed info. Thanks for the help. FJ

Answer 1

Let's call the top resistor of a voltage divider R1 and the bottom R2. The output impedance is R1 // R2, and the gain is R2/(R1+R2). Set these two equations equal to what you want. Now you have two equations and two unknowns. Solve accordingly.

Note that R1 encompasses the 50 Ω of the source, so the actual top resistor (what you add directly in series with the signal generator) will be 50 Ω less than what you solve for R1. At your very low gain, this will make little difference though.

Answer 2

You probably don't need to maintain 50 ohm match, or if you want to, you can afford to have some small errors.

1) You need to attenuate by a large amount -say 1000x. If you want to maintain a 50 ohm load on the generator, start with a 50 ohm across the output. 2) to attenuate this signal, you want an R divider that is placed in parallel with this 50 ohm. You could use e.g. 500 ohm and .5 ohm, but 0.5 ohm is hard to wire up (because of wire resistance), and 500.5 ohm in parallel with 50 ohm would cause a small (not really problematic) error. 3) -- so use 50 kohm and 50 ohm in series. This does 3 things -- insignificant value in parallel with the 1st 50 ohm; 50k and 50 are practical values (could use 47k and 47 ohm in reality); 3) provides a 50 ohm output impedance also (actually 50 ohm // 50k -- close enough).