# current limiter after a voltage divider for op amp input

I am doing a two-stages amplifier for a very small signal (-+300uV), coming from a radar sensor.

The following two images are taken from application notes.

I know R2 and R3 are voltage divider. 2.5V is required to shift the input signal.

What is the purpose of R5 in both circuits?

Why is it not used in stage 2 at the amplifier A and used in both stages at the amplifier B?

Is it a current limiter? If so, who can I design its value?

Also note that I have applied the amplifier B with LM358 instead of OPA2365 and I sensed a very weak (micro Volt) periodic (10 Hz) oscillated signal and I did not see that at amplifier A. and the difference is clearly the first stage (inverting and non-inverting).

However, the main question is: what are R5 doing?

R5 provides DC bias of Vss/2=2.5V to Vin+ used by Amp A.

Amp B does not use Vin+ and rather uses 2 stages of inverting gain thru Vin-. Thus R5 does not affect the signal impedance on Amp B. Mainly it has more sensitivity with the extra filtered gain stage with the dual Op Amp then finally a comparator to produce a pulse.

Basically Amp A is missing the comparator but Amp B looks better.

The filter bandwidths are also different but cannot say why but Amp B at least has a spec. --- although an odd looking 999 Hz even with 1% tolerances.

• The voltage divider R2, R3 provide the DC bias 2.5 V. What if I did not add R5 ? Commented Oct 11, 2019 at 14:10
• In Amp A that would short out the signal with 100uF. What is your real issue? Commented Oct 11, 2019 at 14:16
• So is it a current limiter? how can I design its value? Commented Oct 11, 2019 at 14:50
• No , R5 purpose in A is different than B . Are trying to make A perform like B? Then use cct B. Commented Oct 11, 2019 at 16:19
• R5 is needed to prevent all of your signal energy going thru the 100uF across R3. Commented Oct 12, 2019 at 4:02

Technically, R5 has the same role in both circuits:

In amplifier A, R5 plays as an RC low pass filter for the IF source signal. The fc is 1/2*3.14*330k*100u = 0.0048 Hz. The IF signal is normally more than that frequency (3-1000 Hz), so it cannot pass that filter and disturb the 2.5 V DC bias.

In amplifier B, we have V+ = V-, so the IF source signal will appear on IN+, R5 also plays an RC low pass filter with C4, fc = 1/2*3.14*4.7k*10u = 3.3 Hz. Thus, it cannot pass that filter and disturb the 2.5 V constant DC bias.