Need help with Non-Inverting amplifier using single supply

AnalogKid

Joined Aug 1, 2013
8,538
A couple of things.
C3 and C1 form a 100-to-1 attenuator. Calculate the impedance of both at 1 Hz to see how this is.
C2 also is too small for a 1 Hz signal. Its impedance at 1 Hz should be less than 10% of R1, preferably less than 1%.
The connection from the 4.5V divider to the opamp + input pin should be a resistor. Start with 100K. This prevents an AC interaction between the input and the Vcc/2 decoupling capacitor C1.

ak
 

Bordodynov

Joined May 20, 2015
2,645
My electronic circuits are equivalent to the use of virtual land. It is simply masked. Voltage divider with two identical resistors R1 = R2 = R (e.g. 200 kohm) is an equivalent circuit consisting of a voltage source Vcc/2 and the series resistor Re = R1 || R2 = R/2. The positive pole of the equivalent source and acts as a virtual ground.
You can explain the operation of my circuits that produced by the resistors by the amount of shift voltage Vcc/2.
 

Thread Starter

Yuvraj Abhimanyu

Joined Jun 10, 2015
15
The connection from the 4.5V divider to the opamp + input pin should be a resistor. Start with 100K. This prevents an AC interaction between the input and the Vcc/2 decoupling capacitor C1.
ak
I Tried putting a 100K resistor between vcc/2 and gnd, Just the thing i needed :) yipeeee...!! :)
ver3.png
but what do u mean by "AC interaction"?

A couple of things.
C3 and C1 form a 100-to-1 attenuator. Calculate the impedance of both at 1 Hz to see how this is.
Firstly why 1Hz?
Impedence of c1 for 1Hz is 15.9K =z1(say)
Impedence of c3 for 1Hz is 1.59Meg = z2(say)
so Vo=(Vi*z2)/(z1+z2) will nearly give me 100, which means that Vo is 100 times Vi, how is this attenuating?

.
C2 also is too small for a 1 Hz signal. Its impedance at 1 Hz should be less than 10% of R1, preferably less than 1%.
why should the impedence be less than 10%? any specific reason or rule of thumb?
 

AnalogKid

Joined Aug 1, 2013
8,538
you: but what do u mean by "AC interaction"?

me: Your 4.5 V source has a 10 uF filter cap; that's good. But the idea is to create a virtual ground, which means anything connected to it is "shorted" to ground, within the limits of the capacitor's frequency response. As seen below, this is a 100:1 attenuator when connected directly to the 0.1uF input cap. The 100K resistor establishes a DC value for the + input and isolates the input by 100 K ohms from "ground".

you: Firstly why 1Hz?

me: Because I misread your V2 signal. oops.

you: Impedence of c1 for 1Hz is 15.9K =z1(say)
Impedence of c3 for 1Hz is 1.59Meg = z2(say)
so Vo=(Vi*z2)/(z1+z2) will nearly give me 100, which means that Vo is 100 times Vi, how is this attenuating?

me: Nope. It's Vi x z1 over z1+z2 because the network's output voltage is developed across C1. Think about it - a two component passive network can *never* have gain.

you: why should the impedance be less than 10%? any specific reason or rule of thumb?

me: Yes, the 10% rule. When guesstimating component values, the first-order approximation is that something that is 10% or less than something else is basically either not there or completely there. Simple resistor circuits: 1K in series with 10K is so much closer to the 10K value than the 1K value that it's like the 1K isn't there. 1K in parallel with 10K is so much closer to the 1K value than the 10K value that the 10K is almost shorted ("swamped") by the 1K. Remember, these are approximations and are inaccurate by 10%, just like the name of the rule.

ak
 
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