A differential amplifier frequent use is the amplification of the voltage difference between its inputs, while rejecting the common-mode level. However, the output common-mode level cannot be zero. The operational amplifier technological limitations, as well as the outside resistor tolerances let the common-mode voltage to make it to the amplifier output as an output error. As a consequence, the amplifier output voltage is the input signal difference times gain, plus the output common-mode voltage.

# common-mode

## The Differential Amplifier Common-Mode Error – Part 2

### Power Supply Output Current Measurement with a Differential Amplifier

When designing a differential amplifier, part of the art is to manage the errors affecting the precision of the circuit. In MasteringElectronicsDesign.com: The Differential Amplifier Common-Mode Error – Part 1 of this presentation I discussed the common-mode error of a differential amplifier. I also showed that, given the circuit in Figure 1, the common-mode voltage can be viewed as V2, when we consider V1-V2 as a signal that rides on top of V2. The same goes for V1, which can be considered the common-mode voltage of the differential amplifier when -(V1-V2) is the signal that rides on top of V1.

**Figure 1**

Most of the times, however, the input signals V1 and V2 would vary in time, whether there is an AC signal riding on top of a DC signal, or the input signals have a noise component as in Figure 2.

**Figure 2**

Because of that, it is customary to consider the common-mode voltage the average of the input signals, V1 and V2, as in Figure 3, so that the common-mode input signal lands in between V1 and V2.

Let’s note this signal with Vcm, and the difference V1-V2 with Vd.

_{} |
(1) |

From a signal difference point of view, each input will be referred to the common-mode voltage as shown in Figure 3. In this case, the difference signal Vd = (V1-V2) is split in two, so that the input R1 has a signal Vd/2 and the input R3 has a signal -Vd/2 as referred to the common-mode voltage Vcm.

What is the common-mode error in this case?

With these notations, I can express the input signals as in (2).

_{} |
(2) |

In MasteringElectronicsDesign.com: The Differential Amplifier Common-Mode Error – Part 1 I demonstrated that the output signal of the differential amplifier can be expressed as a function of V1-V2 and V2 as shown in (3).

_{} |
(3) |

By replacing V1 and V2 with the expressions (2), Vout becomes,

_{} |
(4) |

After calculations, the differential amplifier output becomes,

_{} |
(5) |

In equation (5), the first parenthesis is the differential gain and I will note it with Gd. The second parenthesis is the common-mode gain, noted with Gcm.

_{} |
(6) |

One can see that, if the resistor ratios are equal, Gcm is zero. We should note that this gain is the same as in the MasteringElectronicsDesign.com: The Differential Amplifier Common-Mode Error – Part 1, when the same expression multiplied V2. Indeed, this proves that, no matter the level of the common-mode voltage at the amplifier input, V2, Vcm or anything in between for that matter, the common-mode gain is the same.

Equation (5) also shows that the larger Vcm, the larger the common-mode error at the differential amplifier output. Since many times we cannot do anything about the common-mode voltage level, the electronics designer can only minimize the common-mode gain to reduce the error. This can be done by matching the resistor ratios.

One good example of using the differential amplifier is current measurement. One way is to measure the voltage drop across a small resistor. Another way is to measure the current inductively, with a magnetic probe.

Measuring the current through a network branch with a small resistor, called *sense resistor*, is preferred by many designers, because it can be very precise. Depending on the expected current level, the resistor value is chosen so that the voltage drop on this resistor is around a few hundred millivolts. A differential amplifier connected across the sense resistor amplifies the voltage drop to a manageable level, usually around 2.5V or 5V, so that an Analog to Digital Converter (ADC) can measure it with good resolution.

If the measurement has to be done at a power supply output (see Figure 4), the common mode voltage can be high, because it equals the power supply voltage level.

**Figure 4**

Let’s say that this is a 12V power supply that sources a nominal current of 5A to a system it powers. Based on the powered system functionality, the load current can vary in time and we need to monitor it. The voltage drop on Rsense has to be small enough so that the powered system still receives approximately 12V. If, instead of 12V, the system is powered at 11.9V, it can be good enough for most applications. Therefore, we can choose the drop on Rsense as 100mV. At 5A, the sense resistor has to be Rsense = 20 milliohms.

Also, let’s say we need to read the current with a microcontroller. For this, we need to use an ADC, with a reference voltage of 2.5V. We can design the differential amplifier resistors so that the nominal current of 5A means 2V at the amplifier output. This means that the nominal value is placed at 4/5^{th} of the ADC range, so that there is some room for positive or negative load current variation.

If the resistor ratios are equal, the differential gain, Gd is

_{} |
(7) |

The gain of the differential amplifier has to be

_{} |
(8) |

Let’s choose R2 = R4 = 20 kohms and R1 = R3 = 1 kohm. What tolerances should I select for these resistors? Resistors with 1% tolerance are quite common nowadays and they are not expensive. With 1% tolerance resistors, what is the common-mode error?

Since V2 = Vpower, let’s choose equation (3) to calculate the output voltage, for a nominal power supply current of 5A.

_{} |
(9) |

With the tolerance t = 0%, the output is the ideal nominal value Vout = 2V.

When the tolerance is t = 1%, and in the worst case in which the resistor values may be as follows,

_{} |
(10) |

the output voltage is Vout = 2.413V. The extra 0.413V is the common-mode error which is significant, as it represents 20.6% of the nominal value.

What if we use resistors with 0.1% tolerance? For the worst case scenario described above, the output becomes Vout = 2.042V. The error of 42mV means that the power source current is measured with an error of 2.1%. Depending on the application requirements, this measurement may be good enough, or may not be acceptable. If the error is too high, the designer has to choose either better matched resistors, or choose instrumentation amplifiers. Analog Devices’ AD620 can do the job with high precision.

There are some other questions that rise from this experiment:

Can the common-mode voltage damage the operational amplifier used for the differential amplifier circuit?

Is the sense resistor small enough so that the differential amplifier components do not modify its value and generate errors?

Is the offset voltage of the differential amplifier small enough so that the output offset does not appear as an error?

Are the operational amplifier input bias currents small enough, or their offset for that matter, so that there are no perceived errors at the amplifier output?

Is the temperature coefficient of the differential amplifier components small enough so that any temperature variation does not result in measurement errors?

I will discuss all these possible errors in future articles. Stay tuned.

## The Differential Amplifier Common-Mode Error – Part 1

The common-mode voltage can bring errors in the differential amplifier applications. What is the common-mode voltage? The common-mode voltage is the voltage level common to both inverting and non-inverting inputs of the differential amplifier. In many applications, the differential amplifier is used to amplify the difference between two voltages, for later processing, or to isolate a signal from common-mode noise, or to amplify a signal that rides on top of some large voltage level. If the common-mode voltage is not rejected, it appears as an error at the amplifier output.

It is customary to consider the common-mode error as being negligible, based on the high Common-Mode Rejection Ratio (CMRR) of the operational amplifiers. This is not always the case. Once the electronics designer connects resistors around this amplifier, in a differential configuration, the common-mode error starts to be significant.

The common-mode voltage Vcm and the differential voltage Vd are shown in the group of equations (1).

_{} |
(1) |

Why these expressions? How was Vcm defined like that and why? We will start by looking at the significance of each input voltage in the differential amplifier.

Looking at Figure 1, V1 is the input voltage between R1 and ground, while V2 is the input voltage between R3 and ground.

**Figure 1**

As we saw in **MasteringElectronicsDesign.com: The Differential Amplifier Transfer Function**, the signal at the amplifier output is as follows:

_{} |
(2) |

If we arrange this equation differently, as in (3),

_{} |
(3) |

one can see that, in the unique case in which

(4) |

the circuit amplifies the difference of the input signals, V1-V2. In other words,

_{} |
(5) |

So, which is the common-mode voltage? In order to give you an answer, let’s rearrange the input signals as in Figure 2.

**Figure 2**

It should be clear now that, when the ratio of the resistor pairs is equal, V2 contribution to the output signal is zero. This can also be seen from equation (2) written differently, as in (6). In equation (6), I grouped the terms so that two main signals are shown: the difference V1-V2 and V2.

(6) |

How did I arrive at this equation? It can be done in two ways: mathematically, using simple algebra methods, or, by using the Superposition Theorem.

Using the Superposition Theorem is easier, because we can consider that there are two voltage sources in the circuit in Figure 2. One source is V1-V2 and the other one is V2. Based on the Superposition Theorem if we take out one source, V2, and replace it with a wire, we find the first term of equation (6). Indeed, when R3 is connected to ground, the amplifier in Figure 2 becomes a non-inverting amplifier. As I showed in a previous article, **MasteringElectronicsDesign.com: The Differential Amplifier Transfer Function**, Vout1 is the voltage at the non-inverting input times the gain given by R4 and R3.

(7) |

** **

With Vout1 I noted the output voltage when V2 is zero.

By rearranging

we arrive at the first term of equation (6).

The second term of equation (6) is the output voltage when V1-V2 is made zero. In this case the amplifier in Figure 2 is a differential amplifier with the same voltage, V2, at both inputs. Hence, the second term of equation (6).

Equation (6) is important because it shows the common-mode error. Since the circuit amplifies the difference V1-V2, this signal appears as riding on top of V2. Hence, V2 can be seen as a common-mode voltage. If the resistor ratios are rigorously equal, the second term in equation (6) is zero. If they are not, the same term will show up at the amplifier output as an error. This is the common-mode voltage error.

How big is this error and why should the electronics designer be concerned about it?

Let’s consider that the ratio of the resistors is equal, as in equation (4), and that only R2 has a tolerance *t* which can be positive or negative, but smaller than 20%. In other words:

(8) |

For resistors, this is a practical assumption. Examples of usual resistor tolerances are 0.1%, 1%, 10%, 20%. In my example R1, R3 and R4 are ideal resistors, with 0 tolerance, while R2 has a tolerance of, say, 10% which I noted with *t*. This creates a mismatch in the resistor ratios R2/R1 and R4/R3 , so that the common-mode voltage V2 appears at the differential amplifier output, scaled by a factor dependent on the tolerance *t*. This voltage level is the common-mode error.

To calculate this error, let’s write the common-mode portion of the differential amplifier output by taking into consideration the tolerance *t* of resistor R2,

(9) |

where with Vocm I noted the common-mode voltage at the differential amplifier output. Since the signal of interest is the difference V1-V2, the common-mode error at the differential amplifier output is Vocm.

After calculations, and using (4), Vocm becomes

(10) |

We can consider that t·R2/R1 is small compared with the ratio R2/R1 which determines the gain of the amplifier. Also, for gains larger than 10, the value of 1 in the denominator can be neglected. Therefore, the common-mode error Vocm is

(11) |

Equation (10) shows that, if one resistor, R2, has a tolerance other than zero, there is a significant error at the differential amplifier output, which is approximately the common-mode voltage times that tolerance.

As an example, if V2 = 10V, V1 = 10.1V, and

the circuit in Figure 1 amplifies the difference between these two signals, so that the output is 2V.

However, if R2 has a tolerance of +10%, the error at the circuit output is Vocm = 10V·0.1 = 1V. As a result, the differential amplifier output will be the sum of the differential output of 2V and the error of 1V, which makes 3V. The error of 1V is significant.

If R2 has a tolerance of 0.1%, the error is 10mV, which can be considered negligible in some applications. That is why the usual recommendation is to have either highly matched resistors for the differential amplifier, or resistors with 0.1% or even 0.05% tolerance.

The same logic is valid for V1 that can be viewed as the common-mode voltage, while the circuit amplifies the negative difference -(V1-V2). In the next part I will show that the convention for the common-mode voltage is

and also the reason why this is the preferred method.