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let's consider this logarithmic multi - stage amplifier:

enter image description here

Each stage is an amplifier with this mechanism: if the input signal is small (under a certain threshold), it is amplified with high gain A; if it is large (over the previous threshold), it is amplified with unitary gain.

The purpose of this multi - stage amplifier is that of getting a Voutput - Vinput curve like that in the previous picture. But I do not understand the presence of the Adder circuit. I think that simply the cascade of the previous amplifier will generate a curve like that, in which small signals are amplified a lot, and high signal are amplified less.

Can you help me on understanding the presence of the adder?

This circuit is analyzed here.

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  • \$\begingroup\$ are you asking about a Relative Signal Strength Indicator RSSI? \$\endgroup\$ – analogsystemsrf Sep 9 at 16:56
  • \$\begingroup\$ Yes, I wanted to use it for power measurement \$\endgroup\$ – Kinka-Byo Sep 9 at 16:58
  • \$\begingroup\$ the NXP NE504 has one, if I recall correctly. \$\endgroup\$ – analogsystemsrf Sep 10 at 2:00
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If you take only the output from the final amplifier, then the transfer function is:

Output at 4th stage only At some value of input, the output stage will saturate and not increase any further.

If we then add the other outputs, we will get (not to scale):

Additive gain

As each amplifier saturates, then the slope reduces until the next amplifier in the chain saturates until all 4 amplifiers are in saturation. As each amplifier output is lower then the following devices the output slope from the addition reduces.

By adding these outputs together we get a log amp shown (not complete and also not to scale) by the purple line.

This is simplistic, but it is the fundamental principle of operation.

The output slope is for a setup where all the amplifiers have a gain of A:

All amplifier still active: 10A (4+3+2+1)

4th amplifier saturated: 6A (3+2+1)

3rd and 4th amplifier saturated: 3A (2+1)

2nd, 3rd and 4th amplifier saturated: A

So the output slope reduces as the level of input is raised.

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  • \$\begingroup\$ Perfect, thank you very much \$\endgroup\$ – Kinka-Byo Sep 10 at 4:39

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