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Over the past few days, some of our readers have stumbled upon a familiar error message regarding the common mode error of a differential amplifier from Mastering Electronic Design. This problem can arise for several reasons. Now we will deal with them.
Common mode voltage can cause slippage in differential amplifiers. What was the common mode voltage? In many applications, a differential amplifier is randomly used to amplify the voltages between differential connections for further processing, or to separate the signal from common mode noise, or ultimately to amplify a signal that depends on the first fundamental voltage level. If the common-mode voltage does not deviate altogether, the entire amplifier output is faulty ela.
Common mode error is generally considered to be negligible based on the best common mode rejection ratio (CMRR) of detailed amplifiers. Isn’t this always a special case? Resistors somewhere around this amplifier, in differential configuration, in common mode, the particular error becomes significant.
Common-mode voltage Vcm and differential voltage Vd are displayed in a group with (1) equations.
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<p> Why is gif “> one of these terms? How was Vcm defined and even why? Let’s start with the value of each input voltage of the differential amplifier. </p>
<p> In Figure 1, V1 is injected between R1 but also ground, and V2 is injected between R3 and ground. </p>
<p> As we saw on MasteringElectronicsDesign.com: Differential amplifier transfer function, the entire signal to the output amplifier is done like this: </p>
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 (2) |
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(3) |
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(4) |
The
circuit increases the difference between the input characters V1-V2. In other words,
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(5) |
So what is the common-mode voltage? To answer, let’s rearrange the input signals as shown in Figure 2.
It should now be clear that if the ratio of the resistance pairs can be the same, then the contribution of V2 to the output computer code is zero. This can also be found from equation (2), written differently, in the role of (6). In equation (6), I have grouped the terms so that two main signals are usually displayed: the difference V1-V2 and V2.
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(6) |
How did I get this equation? This can be done There are two ways: using simple mathematical methods or using the superposition theorem.
The superposition theorem is easier to use because you can imagine there are two voltage sources in the circuit shown in Figure 2. One source is V1-V2 and the other is V2. Based on the superposition theorem, if we remove the information source V2 and replace it with a cable, we find the first term in Figure (6). If R3 is connected to ground, the amplifier in Figure 2 can become a non-inverting amplifier. As I showed in a previous article, MasteringElectronicsDesign.com: Differential Amplifier Transfer Function, Vout1 is the voltage that causes the input to be inverted. with the time savings provided by R4 and R3.
 Vout1 |
(7) |
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I noticed the output voltage when V2 is zero.
Equation (6) associated with the second term is the output voltage when V1-V2 is set to zero. In this combination, the amplifier shown in fig. 2 is a realdifferential amplifier with the same voltage V2 on both inputs. Hence the equation for the second semester (6).
Equation (6) is important because it indicates the common mode error. Since most of the circuit amplifies the V1-V2 difference, this connection appears to be driving V2 with an incredible common-mode voltage. If the drag coefficients are indeed the same, the second term in scenario (6) is zero. If not necessary, the same value will be displayed as amplifier output error. This is a common mode voltage error.
How big is this mistake and why should a digital designer care?
Note that the resistance ratio is straight, as in equation (4), and that the last R2 has a t tolerance, which can of course be positive or negative, but less than 20%. In other words:
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(8) |
This is useful for resistors. For example, common resistance tolerances can be 0.1%, 1%, 10%, 20%. In my exampleR1, R3 and R4 are ideal zero tolerance resistors. while R2 has a tolerance of, for example, 10%, which is denoted by both i and t. This results in a mismatch between the resistance ratios R2 / R1 and R4 / R3 now that a common-mode voltage V2 appears at the output of the differential amplifier, scaled by a factor dependent on the tolerance t. This voltage level is common mode error.
To measure this error, we record the common-mode component of the output of the differential amplifier, taking into account the tolerance t, including the resistance R2,
 C |
(9) |
where Vocm I marked the common-mode voltage at the output of the differential amplifier. Since the computer code of interest is the difference V1-V2, this common-mode error in the final result of the differential amplifier is Vocm.
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(10) |
We assume that t · R2 / R1 small is compared based on the ratio R2 / R1, which determines the gain of the amplifier. In addition, for gains always greater than 10, the value of 1 in the denominator of the container can be neglected. So the common mode error is Vocm
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(11) |
Equation (10) shows that if R2 has a tolerance other than three, there will be a significant error at the output of the main differential amplifier, approximately equal to most of the common-mode voltage of that tolerance.
As an exercise: if V2 = 10V, V1 corresponds to 10.1V and
The circuit in Figure 1 increases the difference between the two, i.e. when the output voltage is 2 V.
However, if R2 gives a tolerance of + 10%, the error at each output of the Vocm circuit is 10 V * 0.1 = 1 V. As a result, the difference at the output of the current amplifier is the sum, which is the sum of the difference in the output. connectit is 2V, and I would say the error is 1V, which is 3V. The 1V error is important.
If R2 provides a 0.1% tolerance, the deficit is 10 mV, which can be considered negligible in some applications. For this reason, it is generally recommended to use commonly used differential amplifier resistors, or perhaps even resistors with a 0.1% or even 0.05% tolerance.
Exact logic is applied to V1 and in particular can be viewed as a common-mode voltage, and the circuit also amplifies the negative alternative – (V1-V2). In the next part I show very well that the convention for the existing general mode
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