#### xinyi9393

Joined Jun 19, 2013
3 may i know in this circuit, the amplifier is inverting or non-inverting? i'm very confusing about the input. pin 2 &3 both does have + input voltage.

and what is the uses for R4 1M ohms? why it's so big?

#### Jony130

Joined Feb 17, 2009
5,089
LM741 in this circuit work just like a inverting amplifier with voltage AC gain
Av = R4/R1. And R2 and R3 provide proper bias point needed in this circuit becaoues of single supply use in this circuit.

• xinyi9393

#### xinyi9393

Joined Jun 19, 2013
3
LM741 in this circuit work just like a inverting amplifier with voltage AC gain
Av = R4/R1. And R2 and R3 provide proper bias point needed in this circuit becaoues of single supply use in this circuit.
thank you,
but why the power gain Av is so big?
Av= -(1M/12k)
= -83.33
and can i know one more thing, C1 is used to filter the input signal?
Appreciate.

#### Jony130

Joined Feb 17, 2009
5,089
thank you,
but why the power gain Av is so big?
Av= -(1M/12k)
= -83.33
Maybe input voltage is so small that we need such a "large" voltage gain.
and can i know one more thing, C1 is used to filter the input signal?
Appreciate.
Yes C1 and R1 form a hight pass filter. But the main C1 function is to "block" any DC voltage present at the input. We don't want to gain any DC voltage.

#### xinyi9393

Joined Jun 19, 2013
3
Maybe input voltage is so small that we need such a "large" voltage gain.
but my input voltage is 9v, voltage after pass through R2 still have 2.63V.
so the Av = -83.33x 2.63V= -220V am i correct?

Yes C1 and R1 form a hight pass filter. But the main C1 function is to "block" any DC voltage present at the input. We don't want to gain any DC voltage.
thanks~

#### #12

Joined Nov 30, 2010
18,167
Yes, your math is correct and your reasoning is correct. If C1 was a wire instead of a capacitor, the amplifier would attempt to multiply 2.63 volts times 83.33. The resulting answer would be useless.

#### Veracohr

Joined Jan 3, 2011
694
but my input voltage is 9v, voltage after pass through R2 still have 2.63V.
so the Av = -83.33x 2.63V= -220V am i correct?

thanks~
It's not going to amplify 2.63V. It will amplify (+Input) - (-Input). If you have 2.63V on the +Input, it will amplify 2.63-signal. I'm guessing the AC signal voltage of the piezo will be much less than 2.63.

#### Ron H

Joined Apr 14, 2005
7,014
It's not going to amplify 2.63V. It will amplify (+Input) - (-Input). If you have 2.63V on the +Input, it will amplify 2.63-signal. I'm guessing the AC signal voltage of the piezo will be much less than 2.63.
Due to feedback, the DC voltage on pin 2 will be identical to the voltage on pin 3 (2.53V). Due to the fact that C1 is infinite impedance at DC, the DC output voltage will also be 2.53V, plus the input bias current multiplied by the value of R4. The typical bias current is 80nA, so the DC output voltage will be typically be ≈2.61v. It could be a little lower, or as high as 3V.
The AC gain to the piezo input signal will be -83.3, as mentioned. However, piezo elements tend to be high impedance, so the gain may be less.