C & L Impedance

Capacitor Impedance ZC=1jωC=−jωCZ_C = \frac{1}{j\omega C} = -\frac{j}{\omega C}

Inductor Impedance ZL=jωLZ_L = j\omega L

Amplifiers

I will generalize all single stage operational amplifiers using the following two-input system.

The basic function of an op-amp is to ensure V+=V−V_+ = V_-. In terms of feedback systems, you can think of the output VoV_o as the control signal meant to steer V−V_- towards V+V_+ to minimize error, while VinV_{in} is a disturbance signal. For the following examples, we assume infinite input impedance (no current flowing into our inputs) and zero output impedance. With these assumptions, we get the following:

V−=Z2Z1+Z2(Vin+Vo)+VoV_- = \frac{Z_2}{Z_1 + Z_2} (V_{in} + V_o) + V_o =Z2Z1+Z2Vin+(1−Z2Z1+Z2)Vo= \frac{Z_2}{Z_1 + Z_2} V_{in} + (1 - \frac{Z_2}{Z_1 + Z_2})V_o =Z2Z1+Z2Vin+Z1Z1+Z2Vo= \frac{Z_2}{Z_1 + Z_2} V_{in} + \frac{Z_1}{Z_1 + Z_2}V_o

Then, using using V+=V−V_+ = V_-, we have the following equation:

V+=Z2Z1+Z2Vin+Z1Z1+Z2VoV_+ = \frac{Z_2}{Z_1 + Z_2} V_{in} + \frac{Z_1}{Z_1 + Z_2}V_o Z1Z1+Z2Vo=V+−Z2Z1+Z2Vin\frac{Z_1}{Z_1 + Z_2}V_o = V_+ - \frac{Z_2}{Z_1 + Z_2} V_{in} Vo=(Z1+Z2Z3)V+−(Z1+Z2Z1)(Z2Z1+Z2)VinV_o = (\frac{Z_1 + Z_2}{Z_3})V_+ - (\frac{Z_1 + Z_2}{Z_1})(\frac{Z_2}{Z_1 + Z_2}) V_{in} =Z1+Z2Z1V++−Z2Z1Vin= \frac{Z_1 + Z_2}{Z_1}V_+ + -\frac{Z_2}{Z_1}V_{in}

For a basic inverting amplifier, let us ground our non-inverting input, V+=0VV_+ = 0V. Then, using the above equations, we get:

Vo=−Z2Z1VinV_o= -\frac{Z_2}{Z_1}V_{in}

Giving us a voltage-gain of VoVin=−Z2Z1\frac{V_o}{V_{in}} = -\frac{Z_2}{Z_1}

For a basic non-inverting amplifier, let us consider the input to be V+V_+ and ground VinV_{in}. Now, we have

Vo=Z1+Z2Z1V+V_o = \frac{Z_1 + Z_2}{Z_1}V_+

Giving us a voltage-gain of VoV+=Z1+Z2Z1\frac{V_o}{V_+} = \frac{Z_1 + Z_2}{Z_1}

The usefulness of this representation comes from being able to subtract our inputs VinV_{in} and V+V_+. Consider the simple example of adding a dc offset to an ac input: we can set V+V_+ to a dc voltage and get a final dc offset of Vdc=Z1+Z2Z1V+V_{dc} = \frac{Z_1 + Z_2}{Z_1}V_+. (This can also be accomplished using a adder using passives, but the point still stands that this circuit is capable of adding voltages with a certain amount of gain).

Difference Amplifier

Note in the previous example, the inputs are scaled first before being subtracted. If we want to instead subtract with a uniform gain or subtract with unity gain, we need to scale our non-inverting input. A very simple way of scaling our input is using a voltage divider, giving us a circuit known as a difference amplifier.

Vo=(Z3+Z4Z3)(Z2Z1+Z2)Vin+−(Z4Z3)Vin−V_o = (\frac{Z_3 + Z_4}{Z_3})(\frac{Z_2}{Z_1 + Z_2}) V_{in_+} - (\frac{Z_4}{Z_3}) V_{in_-}

Now, if to get the same gain as Vin−V_{in_-}, we simply set

(Z3+Z4Z3)(Z2Z1+Z2)=(Z4Z3)(\frac{Z_3 + Z_4}{Z_3})(\frac{Z_2}{Z_1 + Z_2}) = (\frac{Z_4}{Z_3}) Z1+Z2Z2=Z3+Z4Z4\frac{Z_1 + Z_2}{Z_2} = \frac{Z_3 + Z_4}{Z_4} Z1Z2=Z3Z4\frac{Z_1}{Z_2} = \frac{Z_3}{Z_4}

Now, we can easily set Z1=Z3Z_1 = Z_3 and Z2=Z4Z_2 = Z_4 to subtract the signal scaled by Z4Z3\frac{Z_4}{Z_3}

Instrumentation Amplifier

A common amplifier is the instrumentation amplifier or INA. Before we discuss the advantages of an INA, let us look at the circuit and derive the equation for the output.

I hope that you can appreciate that the lower half of the circuit is in the same configuration as a difference amplifier, so let us focus on the upper portion.

Looking back at the basic op-amp circuit example, we can say that the output of the left op-amp takes V2V_2 and the voltage at node 2 as inputs, while the right op-amp takes V1V_1 and the voltage at node 1 as inputs.

Thus, the output of the left op-amp is

VL=Rg+R1RgV2−R1RgVn2V_L = \frac{R_g + R_1}{R_g}V_2 -\frac{R_1}{R_g}V_{n_2}

and the output of the right op-amp is

VR=Rg+R1RgV1−R1RgVn1V_R = \frac{R_g + R_1}{R_g}V_1 -\frac{R_1}{R_g}V_{n_1}

Knowing this, we also know that the final output is

Vo=R3R2[Rg+R1Rg(V2−V1)−R1Rg(Vn2−Vn1)]V_o = \frac{R_3}{R_2}[\frac{R_g + R_1}{R_g}(V_2 - V_1) - \frac{R_1}{R_g} (V_{n_2} - V_{n_1})]

Now, note that due to function of the op-amps (makes the non-inverting and inverting inputs equal) and the short from nodes 1 and 2 to the inverting inputs of the op-amp, the following must hold true:

Vn1=V2,Vn2=V1V_{n_1} = V_2, \quad V_{n_2} = V_1

Therefore,

Vo=R3R2[Rg+R1Rg(V2−V1)−R1Rg(V1−V2)]V_o = \frac{R_3}{R_2}[\frac{R_g + R_1}{R_g}(V_2 - V_1) - \frac{R_1}{R_g} (V_1 - V_2)]

=R3R2(Rg+R1Rg+R1Rg)(V2−V1)= \frac{R_3}{R_2}(\frac{R_g + R_1}{R_g} + \frac{R_1}{R_g})(V_2 - V_1) =R3R2(1+2R1Rg)(V2−V1)= \frac{R_3}{R_2}(1 + \frac{2R_1}{R_g})(V_2 - V_1)

Finally, we have a gain of

VoV2−V1=R3R2(1+2R1Rg)\frac{V_o}{V_2 - V_1} = \frac{R_3}{R_2}(1 + \frac{2R_1}{R_g})

Now the obvious question is, why use an INA when a difference amplifier does the same job? The answer lies within the imperfections and tolerances of real-world components.

Going back to the difference amplifier, let us consider what happens when R1=aR3R_1 = aR_3 and R2=bR4R_2 = bR_4 to account for imbalances in resistor values. Then, returning to this equation

Vo=(R3+R4R3)(R2R1+R2)Vin+−(R4R3)Vin−V_o = (\frac{R_3 + R_4}{R_3})(\frac{R_2}{R_1 + R_2}) V_{in_+} - (\frac{R_4}{R_3}) V_{in_-}

We get the following equation with adjusted values

Vo=(R3+R4R3)(bR4aR3+bR4)Vin+−(R4R3)Vin−V_o = (\frac{R_3 + R_4}{R_3})(\frac{bR_4}{aR_3 + bR_4}) V_{in_+} - (\frac{R_4}{R_3}) V_{in_-} =(R3+R4R3)(bR4aR3+bR4)Vin+−(R4R3)Vin−= (\frac{R_3 + R_4}{R_3})(\frac{bR_4}{aR_3 + bR_4}) V_{in_+} - (\frac{R_4}{R_3}) V_{in_-}

If we consider VCOM=Vin++Vin−2V_{COM} = \frac{V_{in_+} + V_{in_-}}{2} and Vd=Vin+−Vin−V_d = V_{in_+} - V_{in_-} such that Vin+=VCOM+Vd2V_{in_+} = V_{COM} + \frac{V_d}{2} and Vin+=VCOM−Vd2V_{in_+} = V_{COM} - \frac{V_d}{2}, then we can rewrite the equation above:

Vo=(R3+R4R3)(bR4aR3+bR4)(VCOM+Vd2)−(R4R3)(VCOM−Vd2)V_o = (\frac{R_3 + R_4}{R_3})(\frac{bR_4}{aR_3 + bR_4}) (V_{COM} + \frac{V_d}{2}) - (\frac{R_4}{R_3}) (V_{COM} - \frac{V_d}{2}) =VCOM[(R3+R4R3)(bR4aR3+bR4)−(R4R3)]−Vd2[(R3+R4R3)(bR4aR3+bR4)+(R4R3)]= V_{COM} [(\frac{R_3 + R_4}{R_3})(\frac{bR_4}{aR_3 + bR_4}) - (\frac{R_4}{R_3})] - \frac{V_d}{2}[(\frac{R_3 + R_4}{R_3})(\frac{bR_4}{aR_3 + bR_4}) + (\frac{R_4}{R_3})]

While ideally, we would only see VdV_d in the output, when the resistors are not perfectly balanced, the common-mode term gets amplified by a factor of

ACOM=G−HG+HA_{COM} = \frac{G - H}{G + H}

while the difference term gets amplified by a factor of

Ad=(1+H)G+2G2H+GA_{d} = \frac{(1+H)G + 2G^2}{H+G}

where G=R4R3G = \frac{R_4}{R_3} and H=abH = \frac{a}{b}.

From this, we can get common-mode rejection ratio, or CMRR, typically expressed in dB.

CMRRdB=20log⁡10(∣ACOMAd∣)CMRR_{dB} = 20 \log_{10} (|\frac{A_{COM}}{A_{d}}|)

Line Level

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Citations

  1. Horowitz, P., & Hill, W., (2015) The Art of Electronics
  2. Self, D. (2010) Small Signal Audio Design