# Circuit analysis of mono audio preamp (K1803 Velleman)

For a school project I have a question about the first part of the circuit below (amplifies at 20*log(220k/2.2k)=40dB). I know that U2A is the amplifying part of the circuit. The problem is U1A and what is the role of this? I think in many ways it resembles an active low pass filter but it lacks a capacitor on the feedback. Isn't the RC on the input a low pass filter and how can I tell what the role of the first op-amp is?

EDIT: I will not use a lm324 but a lm358n (ST), the battery is just a symbol for 12V DC input.  • I hope you get the battery connection right if you're going to build it. Also using an LM324 at such high gain in an audio circuit makes it a poor choice of op-amp - it barely has enough open-loop gain at 5kHz let alone 20kHz. May 7 '13 at 10:27
• Maybe I should have specified this in the question: I will not use a lm324 but a lm358n (ST), the battery is just a symbol for 12V DC input. May 7 '13 at 10:42
• The LM358 is a poor choice too. The battery you have drawn is connected the wrong way round May 7 '13 at 11:04
• If you can spring a dollar or two, get an NE5532. Also, it's quite silly to put all the gain on one op-amp. If you spread the gain between the two it is better.
– Kaz
May 7 '13 at 21:10
• Another thing: the C4 capacitor is quite pointless. Any output offset at the output of U1A will likely be negligible. You can worry about offset in audio circuits if you have a larger number of stages. You can let the offset accumulate over a bunch of stages, rather than decoupling between all stages. And C4 could be a lot smaller if you didn't have to use that small R7, which is 10X smaller relative to R10 in order to achieve a gain of 10 (20 dB). You could have a gain of 10 dB on each stage instead: a resistor ratio of only 3.16.
– Kaz
May 7 '13 at 21:13

Pins 2 of both U1A and U2A are both so called 'virtual ground'. Due to the amplifier architecture, the opamp tries to keep the non-inverting input and the inverting input at the same potential: the voltage at the voltage divider R3/R4. C2 makes this potential coupled to ground for AC.

We are talking AC and pin 2 is 'virtual ground'. This means that the input impedance of the left stage is fully defined by R2 and the input impedance of the right stage is defined by R7.

Due to the high amplification of the right stage with the given ratio R10/R7 and the requirement for practical resistor values R7 will be relatively low. Thus the right stage has a relatively low input impedance equal to R7 (2k2).

The left stage however has a low amplification R5/R2 = 1 and again for practical resistor values are chosen. The advantage here however is that both resistors can be relatively high (100k).

So the role of the left amplifier stage is a high input impedance for the source voltage. The source voltage is probably a microphone, low voltage and only able to deliver very low current. Only with a high input impedance that source will deliver a signal with a low distortion.

The actual input impedance is R1//R2 = 50k.

• What exactly are you disagreeing with? The function of C2? May 7 '13 at 21:16
• I disagree that the only function of the left amplifier stage is adding DC bias to the signal. May 7 '13 at 21:17
• As I didn't say any such thing, there is nothing to disagree with. May 7 '13 at 22:09
• Changed my answer. May 8 '13 at 8:53

U1A is a unity gain buffer, given by R5/R2. C2 is there to put the non-inverting inputs of U1A and U2A at AC ground, because the power supply isn't symmetrical: it's not a low-pass. U2A has gain of 100, given by R10/R7, i.e. 220k/2k2. If you consider the OP-amps one at a time it becomes clearer.

• It's a strange circuit. C4 is redundant. I don't know what you need 100x gain for in audio, but if you do, it should be in the first stage, not the second, for noise reasons. I don't even know why there are two stages really. May 7 '13 at 22:22