How to remove the negative component of a sine wave?

The following is the schematics of a simple circuit I'm working on, It filters and amplifies an electret condenser microphone's signal to get an output ranging from 0 to 9V (to drive an equalizer like leds not a speaker):

simulate this circuit – Schematic created using CircuitLab

The mic signal passes first through a low pass filter with a cutoff frequency of around 20Khz, then gets coupled with a 47nF capacitor, then gets offset by 1/2 VCC by a voltage divider, then amplified 1000 times using 3 noninverting amplifiers with a gain of 10 each.

Now the next stage I want to clamp the negative part of the signal, to do that I put a reverse biased diode after a coupling capacitor. But after testing the signal appears to pass through unchanged. Why is it not getting clamped?

Update:

If I disconnect the non-inverting input of the last opamp from the diode and measure the voltage at the junction between C6 and D1 I get a nice clampped singal between -0.6V and 3V (I had to increase C6 to 100uF though).

• You want to clamp your circuit or clip your circuit? Shift signal up or simply remove values below 0? Commented Mar 22, 2017 at 12:22
• how big is the signal? What is the forward conduction voltage of that diode?
– user16324
Commented Mar 22, 2017 at 12:22
• R11 and C7 form a low pass filter, not a high pass.
– JRE
Commented Mar 22, 2017 at 12:22
• @BrianDrummond the signal is -1mv to 1mv and the diode is 1N4148 with a 0.6 forward drop voltage. Commented Mar 22, 2017 at 12:28
• So now you need to deal with the opamp's Input Bias Current. You need a DC path for this current otherwise it will just charge up your coupling cap C6. Put a resistor in parallel with D1 - probably a 1M will work. Commented Mar 22, 2017 at 14:34

The LM358 OpAmp you're using has an Input Bias Current spec (from table 6.5 on pg.5 of the DS) of typically 45nA.
If you refer to the internal schematic diagram in Fig.16 on pg.13:
you can see where this current comes from.
The current from the 6uA source is split between the 2 input legs and even though the majority of it will flow to ground through the input transistors, some of it still needs to flow out of the input pins in order to bias those transistors.

When you use the opamp in a circuit, you must provide a path for this input bias current.

In your circuit, all of the opamps inverting (-) inputs have a feedback resistor which does this job, and the first 3 have their non-inverting (+) inputs are either dc-coupled to each other or fed by a resistor voltage-divider.
But your 4th opamp is AC-coupled by C6 and has no DC path for the bias current being sourced by the non-inverting input.

The easiest way to fix this in your circuit would be to put a resistor in parallel with D1.
To calculate the value of this resistor ou need to take the value of the bias current and decide for yourself what the maximum permissible offset voltage you can tolerate at that point is. Ohm's law will then produce a value for you.
You'll also need to be aware that this will form a high-pass filter with C6.

Take D1 off and try to create an ideal diode with the last of your op amp. https://en.wikipedia.org/wiki/Precision_rectifier improved circuit of course !

I have tried to modify your schematics but I do not have circuit lab account.

• You need a split supply for that right? Commented Mar 22, 2017 at 16:02
• As is yes, as your own design ! It's just an inspiration. You can for example connect positive input to half value of Vbat.
– doom
Commented Mar 22, 2017 at 17:06