Electronics/Op-Amps/Linear Configurations

From Wikibooks, open books for an open world
Jump to: navigation, search

Inverting Op Amp[edit]

Opampinverting.png

The closed loop gain "mon then" of an Inverting Op Amp is

R_1=R_{in}
A_f=-\frac {R_f}{R_1} (i)

Input Impedance of this configuration is Z_{in} = R_{in} (because V_{-} is a virtual ground, no current flows into the Op Amp ideally.)

To get formula (i) we take a KVL loop with V_{in}, R_1 and the inputs of the Op Amp. This gives

V_{in}=i_{in}R_1+v_d \,

Where v_d is v_{+} -v_{-} the voltage between the non-inverting and inverting inputs. But for ideal Op Amps v_d is approximately zero. v_d is zero because the input impedance is infinite, which means the current through the impedance must be zero by Ohms law. The zero current means that there is no voltage drop across the impedance. This gives:

i_{in}=\frac {V_{in}}{R_1} (5)

Using this idea.

i_{f}=\frac {V_{out}}{R_f}(6)

If we take KCL at the inverting input then

i_{in}=i_{d}-i_{f} \,

For an ideal Op Amp there is no input current because there is infinite resistance. So using equations 5 and 6.

-\frac {V_{in}}{R_1}=\frac {V_{out}}{R_f}

Since

A_f=\frac {V_{out}}{V_{in}}=-\frac {R_f}{R_1}

Example 1

Design an Inverting Amplifier to amplify a 100mV signal to 1V signal. Assume: No source impedance.

Solution:

Step 1: Work out required gain.

| A_f |=\frac {1}{0.1}=10

Step 2: Select an R_f value.

Choose R_f = 100k \Omega.

Step 3: Work out the required value of R_1 using equation (i).

R_1=\frac {100k}{10}=10k \Omega

Example 2

Design an Inverting Amplifier to amplify a 10mV signal to a 1V signal. The signal has a 100 \Omega source impedance. The amplifier must not invert the signal.

Solution:

Since the voltage cannot be inverted then there must be an even number of stages. For simplicity let us choose two stages. Assume an ideal Op Amp.

Step 1: Work out the required gain.

A_{f_{tot}}= \frac {1}{0.01}=100

Step 2: Choose the required gain for each stage.

The gain will be 10 for both stages. But in the first stage we must worry about loading.

Step 3: Choose a value of input impedance i.e. choose R_1.

Choose 10k \Omega. We can now calculate the voltage that input to the Op Amp by voltage divider.
V_{in} = \frac {R_1V_s}{R_1+R_s}= \frac {(10k)(10mV)}{10k+100}=9.9mV

We want the output of this stage to be 100mV.

A_{f1}=\frac {100}{9.9}=10.1

Step 4: Use A_{f1} to work out R_{f1}.

Using equation (i) R_f=A_{f1}R_1=101k \Omega

Step 5: Choose a R_f for the second stage.

Choose 100k.

Step 6: Calculate R_1 using equation (i).

R_1=10k \Omega

Non-Inverting Op Amp[edit]

The closed loop gain of a Non Inverting Op Amp is

A_f=1 + \frac {R_2}{R_1} (ii)

The input impedance of this configuration is Zin = ∞ (realistically, the input impedance of the op-amp itself, 1 MΩ to 10T Ω).

An op-amp connected in the non-inverting amplifier configuration

Ideal Op Amp Derivation[edit]

Take a KVL with the inputs of the Op Amp and R1.

V_{in}=v_d+V_{R1} \,

But v_d is zero since the Op Amp is ideal. Therefore

V_{in}=V_{R1} \,(3)

According to voltage divider rule

V_{R1}=\frac {V_{out}R_1}{R_1+R_2}(4)

Substitute equation 4 into 3.

V_{in}= \frac {V_{out}R_1}{R_1+R_2}

Thus

A_f=\frac {V_{out}}{V_{in}}=1 + \frac {R_2}{R_1}

Feedback Analysis Derivation[edit]

If the output is connected to the inverting input, after being scaled by a voltage divider K = R1 / (R1 + R2), then:

V+ = Vin
V = K Vout
Vout = G(Vin − K Vout)

Solving for Vout / Vin, we see that the result is a linear amplifier with gain:

Vout / Vin = G / (1 + G K)

If G is very large, Vout / Vin comes close to 1 / K, which equals 1 + (R2 / R1).

This negative feedback connection is the most typical use of an op-amp, but many different configurations are possible, making it one of the most versatile of all electronic building blocks.

When connected in a negative feedback configuration, the op-amp will tend to output whatever voltage is necessary to make the input voltages equal. This, and the high input impedance, are sometimes called the two "golden rules" of op-amp design (for circuits that use feedback):

  1. No current will flow into the inputs
  2. The input voltages will be equal to each other

The exception is if the voltage required is greater than the op-amp's supply, in which case the output signal stops near the power supply rails, VS+ or VS−.


Example 3

Design a Non-Inverting Amplifier to amplify a 100mV signal to 1V signal. Assume: No source impedance.

Step 1: Work out required gain.

A_f=\frac {1}{0.1}=10

Step 2: Select an R_2 value.

Choose R_2 = 90k \Omega.

Step 3: Work out the required value of R_1 using equation (ii).

10=1+\frac {90k}{R_1}
R_1=\frac {90k}{9}=10k \Omega

Example 4 (a quick design procedure):

I want to amplify a signal A with a gain of 8. We want an output swing of at least -3 to +3 V.

We have a 5V and -5V power supply handy, so we can use that.

1. gain of 8 for A.

2. Ground gain = 1 - (8) = -7.

3. feedback-resistor value: Rf = 100 kiloOhms.

4. resistor values for each input:

  • RA = 100 kΩ / 8 = 12.5 kΩ
  • Rground = 100 kΩ / | -7 | = 14.3 kΩ

Since A has a positive gain, connect its resistor to V+. Since ground has a negative gain, connect its resistor to V


Example 5

Design a two stage Non-Inverting Amplifier to amplify a 10mV signal to 1V signal. The signal has a source impedance of 100 Ω.

Solution Assuming an ideal Op Amp. Since this configuration has the input impendance of the Op Amp itself. We do not have to worry about loading since the input impedance is infinite.

Step 1: Work out the required gain.

A_f=\frac {1}{0.01}=100

Step 2: Choose the gain for each stage.

Choose 10 for both stages.

Step 3: Choose a value for the R_2 resistors in both stages.

Choose 90kΩ for both.

Step 4: Work out the value of R_1.

Using equation (ii) R_1=10k \Omega.

Voltage Follower[edit]

This configuration is also known as the unity gain Buffer. Since it can be used to counter the effects of loading of the source.

Opampfollowing.png

This configuration provides an input impedance even higher than a normal Non-Inverting since the gain reduces that input impedance. The gain is given by equation (ii). But R_2 is short circuited and R_1 is an open circuit.

A_f=1 + \frac {R_2}{R_1}=1+ \frac {0}{\infty }=1

Difference amplifier[edit]

Opampdifferencing.png

This configuration is just an Inverting and a Non-Inverting configuration connected simultaneously. Resistors R2 and R4 are a voltage divider. Consider the situation when R4 is open circuited and R2 is short circuited. Now from equations (i) and (ii) we know that the gain of V1 is

A_{f1}=- \frac {R_3}{R_1}

and the gain of V2 is

A_{f2}= 1+ \frac {R_3}{R_1}

Now if we set A_{f1} to -10 then A_{f2} will be 11. This means that Vout will be

V_{out}= V_1A_{f1} + V_2A_{f2}=V_2(11)-(10)V_1

This means that if V_1=V_2 that Vout will be V2. This is not very useful for the most because mathematically we would like the answer to be zero. But if we make the voltage at the non-inverting input equal to 10/11 then when the voltages are equal we will have zero.

When R4 and R2 are connected the gain of V2 is

A_{f2}= \frac {(1+ \frac {R_3}{R_1})R_4}{R_4+R_2} (x)

The gain of V1 is

A_{f1}=- \frac {R_3}{R_1} (y)

If we want A_{f2}=| A_{f1} |

\frac {R_3}{R_1}=\frac {(R_1+ R_3)R_4}{(R_4+R_2)R_1}

We just set R_4=R_3 and R_2=R_1.

This configuration has a low input impedance. The input impedance seen by V1 is R1 as in the Inverting Amplifier. The input impedance seen by V2 is R2 + R4.


Example 6 (a quick design procedure):

We often want to subtract one signal A from another signal B, and amplify the difference by 10. We want an output swing at least -6V to +6V. Oh, and for safety reasons, A and B each have a source impedance of 8 kΩ.

Our 5V and -5V power supply isn't adequate, so we pick a +12V and -12V power supply.

1.

  • gain of +10 for B.
  • gain of -10 for A.

2. Ground gain = 1 - ( +10 + -10 ) = +1.

3. feedback-resistor value: Rf = 100 kiloOhms.

4. resistor values for each input:

  • RA = 100 kΩ / | -10 | = 10 kΩ
  • RB = 100 kΩ / | +10 | = 10 kΩ
  • Rground = 100 kΩ / | +1 | = 100 kΩ

So we connect the 100 kΩ Rf from Vout to V, another 100 kΩ from ground to V+. Then we connect a 2 kΩ from RA to V, and another 2 kΩ from RB to V+.

If the input has a source impedance, the source impedance is part of the circuit. The 8 kΩ source impedance, plus the 2 kΩ physical resistors we added, give us a total of 10 kΩ between the ideal voltage source and the op amp input.



Inverting summing amplifier[edit]

Opampsumming.png

This is merely an Inverting Amplifier with extra inputs. The analysis is nearly identical but we have many currents equal to the feedback current. If we take a KCL at the Inverting input.

i_n+ \cdots + i_2 +i_1=-i_f

The value of the currents can be determined by Ohm's Law using the fact that vd is zero for an ideal Op Amp.

\frac {V_n}{R_n}+ \cdots + \frac {V_2}{R_2} +\frac {V_1}{R_1}=-\frac {V_{out}}{R_f}

If R_n=\cdots=R_2=R_1 then

\frac {1}{R_1} (V_n+ \cdots +V_2+V_1)=-\frac {V_{out}}{R_f}
A_f=-\frac {V_{out}}{V_{sum}}=-\frac {R_f}{R_1}

Just as it is for the Inverting Amplifier.

Note: There is also a Summing Amplifier made using the Non-Inverting Amplifier configuration. The configuration is a bit more complicated and harder to use, since it requires an understanding of Superposition.

Ideal integrator[edit]

Opampintegrating.png


The configuration is an Inverting Amplifier with the feedback resistor a Capacitor. The derivation proceeds the same.

i=-i_f
\frac {V_{in}}{R}= - C \frac {\; dV_{out}}{\; dt}
\frac {V_{in}}{RC}= - \frac {\; dV_{out}}{\; dt}

Integrate both sides with respect to

V_{out}=  \int_{t_0}^{0} -  \frac {V_{in}}{RC}\, dt + V_{initial}

Practically a Resistor is often connected in parallel with the feedback capacitor. This means that there is not infinite gain at very low frequencies, which makes the Real integrator much more stable.

Ideal differentiator[edit]

Opampdifferentiating.png


The configuration is an Inverting Amplifier with a Capacitor as Resistor one so the derivation proceeds the same as before.

i=-i_f \,
C \frac {\; dV_{in}}{\; dt}=- \frac {V_{out}}{R}
V_{out}=-RC \frac {\; dV_{in}}{\; dt}


This configuration is unstable for several reasons. The higher frequency inputs are going to have higher derivatives. Which means that circuit acts like a low pass filter, but more importantly this means that it will just saturate if a high frequency signal is put into the differentiator. This is also seen through the gain.

A_f =- \frac {R}{1/j \omega C}

This means high frequencies mean high gain and thus saturation.

Practically a Resistor is often connected in series with the capacitor.