TF EIM Chapt9

Operational Amplifiers (Op Amps)

The operational amplifier is monolithic (single chip) integrated circuit composed of transistors, resistors, diodes, and other components.

The small package produces a device with small capacitance and inductance which is superior to assembling the equivalent circuit with individual components.

Op-Amp

the "-" sign indicates the inverting input terminal and the "+" indicates the non-inverting input.

A positive going input voltage at the non-inverting ("+") input produces a positive going output.

A positive going input voltage at the inverting ("+") input produces a negative going output.

In other words the output is in phase when the input is non-inverting and 180 degrees out of phase when the input is inverting.

Equivalent circuit

Gain = [math]A_o \equiv \frac{V_{out}}{V_2-V_1} \equiv[/math] Open loop gain = gain when there isn't anything hooked up to the op amp to draw current from it.

Common Mode

Common mode input is defined as the situation where the two inputs are equal ([math]V_1 = V_2)[/math]. In this case[math] V_1[/math] is inverted and will cancel the input [math]V_2[/math] so the output becomes Zero in the ideal Op Amp.

Op Amps usually has two bias supplies in order to produce a high gain as determined by the difference in the input voltages. The output can't be larger than the supply bias.

Differential Op Amp

A single stage differential op Amp can be described as being composed of two bipolar transistors in a common emitter configuration as shown below.

Op Amp current and voltage rules

The simple rules used when analzing Op Amp circuits are

1.) The current into each input Op Amp terminals is Zero

2.) The voltage difference between the two input terminal is Zero. [math]V_1 = V_2[/math]

Note the impact of Rule 2 is that the open loop gain [math]A_0 = \frac{V_{out}}{V_2-V_1} = \infty[/math]

Inverting Amplifier

The above circuit is called an inverting amplifier. The input signal is applied to the inverting terminal ("-", [math]V_1[/math], or[math] V_n[/math]) of the Op Amp through a resister R_1. The non-inverting terminal ("+",[math] V_2[/math], or[math] V_p[/math] ) is grounded.

Applying the current junction rule at the inverting terminal one would have.

[math]I_1[/math]= current through resistor [math]R_1[/math] =[math] \frac{ V_1-V_{in}}{R_1}[/math]

[math]I_2[/math] = current through [math]R_2[/math] = [math]\frac{V_{out}-V_1}{R_2}[/math]

[math]I_{B1}[/math] = current entering the inverting terminal [math]V_1[/math] (the base of the bipolar transistor)

Current conservation[math] \Rightarrow[/math]

[math]I_2 = I_1 + I_{B1}[/math]

or

[math]\frac{V_{out}-V_1}{R_2} = \frac{ V_1-V_{in}}{R_1} +I_{B1}[/math]

A "perfect" Op Amp has [math]I_{B1} = 0[/math]

[math]\frac{V_{out}-V_1}{R_2} = \frac{ V_1-V_{in}}{R_1}[/math]

Using the equivalent circuit from above

[math]V_{out} = A_0(V_2 - V_1)[/math]

Since [math]V_2 = 0[/math] for the inverting amplifier circuit above

[math]\frac{V_{out}+\frac{V_{out}}{A_0}}{R_2}=\frac{ -\frac{V_{out}}{A_0}-V_{in}}{R_1} [/math]

[math]\Rightarrow \frac{1+\frac{1}{A_0}}{R_2} = \frac{ -\frac{1}{A_0}-\frac{V_{in}}{V_{out}}}{R_1} [/math]

[math]\frac{V_{in}}{V_{out}} =-\frac{1}{A_0}-R_1 \frac{1+\frac{1}{A_0}}{R_2}[/math]

[math]= -\frac{R_2}{A_0R_2}-R_1 \frac{A_0+1}{A_0R_2}[/math]

[math]= -\frac{R_2 + R_1 (A_0+1)}{A_0R_2}[/math]

Gain[math]\equiv \frac{V_{out}}{V_{in}} = - \frac{A_0 R_2}{R_1(A_0+1) + R_2} \approx - \frac{R_2}{R_1}[/math] : if[math] R_1 \ll R_2[/math] and [math]A_0 \gt \gt 1[/math]

The "-" sign means that the output is 180 degrees out of phase with the input.

This inverting amplifier has a gain which depends only on the ratio of resistors as long as the open-loop gain [math]A_0[/math] is a lot larger then unity and[math] R_2[/math] is a lot larger than [math]R_1[/math] (negative feedback).

NonZero input bias current I_B

Consider what happens if the terminal current is not zero (ie the Op Amp isn't perfect)

Returning back the the junction rule

[math]I_2 = I_1 + I_{B1}[/math]

[math] \frac{V_{out}-V_1}{R_2} =\frac{ V_1-V_{in}}{R_1}+I_{B1}[/math]

assuming for a minute that V_1=V_2 = 0 (first Op AMp rule is broken but second rule holds)

then

[math]V_{out}=-\frac{R_2}{R_1} V_{in} + I_{B1}R_2[/math]

Iff V_{in} >> I_{B1}R_2 then your Op Amp will operate close to a perfect one.

input bias offset current

The bias current [math]I_{B1}[/math] offsets the output voltage [math]V_{out}.[/math]

If we can find a way to cancel out this offset then we will be essentially "nulling" the bias current [math]I_{B1}[/math] so the Op Amp is more "perfect".

You can do this by using the other half of the amplifier, the non-inverting input current I_{B2}.

If you use a pull up resister ([math]R_3)[/math] to raise the non-inverting input from ground then there will be a potential difference created from the current [math]I_{B2}[/math]

If [math]V_{in} = 0[/math] then

[math]V_{out} = I_{B1}R_2[/math]

using the equivalent circuit description

[math]Gain = \frac{V_{out}}{V_2-V_1} \Rightarrow V_2 Gain = V_{out}[/math]

So to cancel the I[math]_{B1}R_2[/math] offset in[math] V_{out}[/math] you supply the opposite current with a voltage offset.

[math]Gain V_2 = - I_{B1}R_2[/math]

[math]V_2 = -I_{B2}R_3[/math]

Gain =[math] \left ( 1 + \frac{R_2}{R_1} \right)[/math] = gain for the non-inverting input (V_2)

\Rightarrow

[math]-I_{B2}R_3 \left ( 1 + \frac{R_2}{R_1} \right) = - I_{B1}R_2[/math]

or if [math]I_{B2} = I_{B1}[/math]

[math]R_3 = \frac{R_1 R_2}{R_1+R_2}[/math]

[math]I_{io} \equiv I_{B2} - I_{B1} \approx 0 [/math] = input bias offset current.

Input Offset Voltage

The second rule of the perfect Op Amp is that the voltage difference between the input terminals is zero (the Op Amp voltage rule).

For the typical inverting Op Amp you can imagine that a voltage difference exists at terminal [math]V_1[/math].

Let

[math]V_{io}[/math] = input offset voltage source on terminal[math] V_1[/math].

If [math] V_{in} =0[/math]

Then the inverting amplifier looks like a voltage divider with[math] V_{out}[/math] being stepped down to [math]V_{io}[/math] through the resistor [math]R_2[/math].

[math]V_{out} -I_(R_1+R_2) =0 \Rightarrow I = \frac{V_{out}}{R_1 + R_2}[/math]

[math]V_{io} - I_R1= 0 \Rightarrow V_{out} = I R_1 = \frac{V_{out}}{R_1 + R_2} R_1[/math]

[math]V_{out} = \frac{R_2}{R_1+R_2} V_{io} = \left ( 1 + \frac{R_2}{R_1} \right ) V_{io}[/math]

A direct measurement of [math]V_{io}[/math] can be made if you set [math]R_2[/math] to give you a large gain and have [math]V_{in}[/math] = 0 (no input). What you measure at [math]V_{out}[/math] will be [math]V_{io}[/math].

Vout

For a non-perfect Op Amp

[math]V_{out} = - \frac{A_0 R_2}{R_1(A_0+1) + R_2} V_{in} + I_{B1}R_2 + \left ( 1 + \frac{R_2}{R_1} \right ) V_{io}[/math]

[math]\approx - \left( \frac{R_2}{R_1} \right)V_{in} + I_{B1}R_2 + \left ( 1 + \frac{R_2}{R_1} \right ) V_{io}[/math]

V_{io} can be either > 0 or < 0.