thank you,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.
Maybe input voltage is so small that we need such a "large" voltage gain.thank you,
but why the power gain Av is so big?
Av= -(1M/12k)
= -83.33
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.and can i know one more thing, C1 is used to filter the input signal?
Appreciate.
but my input voltage is 9v, voltage after pass through R2 still have 2.63V.Maybe input voltage is so small that we need such a "large" voltage gain.
thanks~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.
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.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~
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.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.
by Jake Hertz
by Jake Hertz
by Duane Benson
by Duane Benson