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### Network # Op amp circuit

J

#### Jim Donaldson

Jan 1, 1970
0
Hi All,

I've just encountered an op-amp circuit that has confused me slightly.
The input signal comes into the non-inverting input, and there is
feedback via a potential divider into the inverting input (so far, so
good - a standard non inverting amplifier.)

However, there is in addition a DC voltage into the inverting input
which is set by a two resistor potential divider between the power
supply rails (0VDC and 5VDC.)

I presume that this provides a DC bias for the circuit (is this
correct?), however, I am a little confused as to how to analyse it. My
problem being that if we use the standard ideal op amp analysis
technique, then V- = V+ (the voltages on the op amp inputs become
equal because of -ve feedbak.) However, we have this DC voltage input
connected directly to V-, which means that V- doesn't equal V+.

Thanks for all help in advance,

Jim

J

#### John Popelish

Jan 1, 1970
0
Jim said:
Hi All,

I've just encountered an op-amp circuit that has confused me slightly.
The input signal comes into the non-inverting input, and there is
feedback via a potential divider into the inverting input (so far, so
good - a standard non inverting amplifier.)

However, there is in addition a DC voltage into the inverting input
which is set by a two resistor potential divider between the power
supply rails (0VDC and 5VDC.)

I presume that this provides a DC bias for the circuit (is this
correct?), however, I am a little confused as to how to analyse it. My
problem being that if we use the standard ideal op amp analysis
technique, then V- = V+ (the voltages on the op amp inputs become
equal because of -ve feedbak.) However, we have this DC voltage input
connected directly to V-, which means that V- doesn't equal V+.

Thanks for all help in advance,

Jim

You are correct in your understanding, so far. The only other
equation that you need to finish is based on the assumption that the
V- input has a very high impedance, the current into this pin can be
considered to be approximately zero, so any current arriving at that
node via one branch (the resistor to the pot, for example) must exit
via some other branch (the feedback resistor, for example).

I

#### Ian Bell

Jan 1, 1970
0
Jim said:
Hi All,

I've just encountered an op-amp circuit that has confused me slightly.
The input signal comes into the non-inverting input, and there is
feedback via a potential divider into the inverting input (so far, so
good - a standard non inverting amplifier.)

However, there is in addition a DC voltage into the inverting input
which is set by a two resistor potential divider between the power
supply rails (0VDC and 5VDC.)

I presume that this provides a DC bias for the circuit (is this
correct?), however, I am a little confused as to how to analyse it. My
problem being that if we use the standard ideal op amp analysis
technique, then V- = V+ (the voltages on the op amp inputs become
equal because of -ve feedbak.) However, we have this DC voltage input
connected directly to V-, which means that V- doesn't equal V+.

Thanks for all help in advance,

Jim

The DC bias to the non inverting input is a standard way of biasing an op
amp for use in single supply situations. Normaly though, the resistor you
have from the inverting input to ground would be decoupled by a capacitor.
In the way the DC level at the output would be the same as the pot divider
connected to the non inverting input. The pot divider creates a false
ground for the op amp, usually at about half the supply volts. If the
resistor fron the non inverting input really is connected directly to 0V
then this effectively is a negative input voltage relative to the false
ground. SO just subtract the false ground potential from all inputs and
analyse as normal.

Ian

G

#### Glenn Gundlach

Jan 1, 1970
0
Ian Bell said:
The DC bias to the non inverting input is a standard way of biasing an op
amp for use in single supply situations. Normaly though, the resistor you
have from the inverting input to ground would be decoupled by a capacitor.
In the way the DC level at the output would be the same as the pot divider
connected to the non inverting input. The pot divider creates a false
ground for the op amp, usually at about half the supply volts. If the
resistor fron the non inverting input really is connected directly to 0V
then this effectively is a negative input voltage relative to the false
ground. SO just subtract the false ground potential from all inputs and
analyse as normal.

Ian

I sometimes use an offset bias like this even when powered from dual
supplies. Its fairly common in video amps and error amplifiers in
feedback loops.
GG

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