Differential Amplifier

Differential amplifier is a basic circuit applied in lot of integrated linear and other types of amplifiers. In general, this circuit configuration is structure of transistors directly connected with their emitters (in case of bipolar junction transistors) or sources (in case of junction FET transistors). The differential amplifier has a big amplification of the difference of the signals leaded to its inputs (differential amplification Ad) and a relatively small amplification of the average value of these signals (synphase amplification or amplification of the common signal Acm). This circuit is also used and for amplifying of DC signals. The upper limit of its working frequency range is in the order of several MHz. The circuit configuration of the differential amplifier with bipolar transistors is shown on Picture 1. The power supply for this circuit has two voltage sources, Vcc = 12 V DC and Vee = -12 V DC. All other components have some casual values, just for simulating purposes.

Picture 1: Differential amplifier with bipolar transistors


The differential amplifier can have a symmetric output if the output signal is taken between the collectors of the transistors or a non-symmetric output if the output signal is taken from one collector.

Now, let's take a look of same basic theory approaches for the differential amplifier (DA). The basic relations for the DA can be derived under assumption that both transistors have identical parameters, which means that the amplification for the both inputs is identical, but with opposite sign. Using these assumptions, for the output voltage we can write:

Vo = Ad x Vi1 - Ad x Vi2 = Ad x (vi1 - vi2)

Every real DA doesn't amplifies only the differential input signal, but also and the signal of the average value of the inputs, so called common signal. Therefore, by introducing the variables:

Vid = Vi1 - Vi2 ---> as differential input signal, and

Vicm = (Vi1 + Vi2)/2 ---> as common input signal,


the output voltage for the real DA can be represented as:


Vo = Ad x Vid + Acm x Vicm


Here, the differential amplification is defined as:

Ad = Vo/Vd | vi1 = -Vi2 = Vid/2;


and the synphase amplification or the amplification of the common signal can be represented as:

Acm = Vo/Vcm | Vi1 = Vi2;


In other words, the differential amplification Ad is equal to the output voltage Vo divided by differential input voltage Vd in case when Vi1 is equal to Vi2 but with opposite sign (Vi1 = -Vi2) and that's equal to Vid/2. From the other hand, the synphase amplification is equal to the output voltage Vo divided by common input signal Vcm in case when both input signals are equal to each other (Vi1 = Vi2).

Important parameter which defines the quality of the DA is the ratio between these two amplifications Ad and Acm, so called CMMR - Common Mode Rejection Ratio. The CMMR is actually a measure for the asymmetry of the real differential amplifier:

CMMR = Ad/Acm


Well, that was just a basic theory approach for the DA. Here, we will not go in further deep math analysis for all parameters of the DA, like input/output impedance, the amplifications expressed through the transistor parameters and so on.


Time-domain analysis

As we defined some basic aspects in the theory approach for this circuit, now we can measure that for this circuit configuration example using time-domain analysis in LT Spice and we will see if our circuit is well configured or not. First, we will measure the amplification of the common signal or the synphase amplification Acm. After that, we will measure the differential amplification for our circuit and finally we can calculate the Common mode rejection ratio (CMMR) for the circuit.

Picture 2: Transient analysis - output voltages and current wave forms (time-domain) for common signal input mode


A. Common signal input mode

For the common signal input mode we need to have equal signals on both inputs of the DA. Since we talk about AC signals, by equal we mean that the signals need to have equal amplitude, frequency and phase. For that purpose, we choose Vi1 and Vi2 as voltage sources with sinusoidal form, with amplitude of 100 mV, frequency of 1 kHz and phase of 0 degrees for both:

Vi1 = +/- 100 mV at f = 1 kHz;
Vi2 = +/- 100 mV at f = 1 kHz;

The results of this time-domain simulation are shown on Picture 2. As we can see from the picture, the output voltages Vo1 and Vo2 are equal (that's why we do not see the green trace - Vo1 in the plot diagram since the blue trace - Vo2 covers it all). On the same picture also is shown and the output current which flows through the load Rl, Il. This current actually has zero value, and that's good, because DA should not amplify the common signal. As we can see from the diagram, the waveform of the current it's like a noise signal. The amplitude of the output voltage is 45 mV:

Vo1 = Vo2 ---> min = 6.907 V; max = 6.997 V; => |Vo| = 90 mV / 2 = 45 mV; - the amplitude of Vo

Il ~ 0; - is at order of 0.001 fA;

Acm = Vo/Vcm | Vi1 = Vi2 ---> Vcm = (Vi1 + Vi2)/2 = Vi1 = Vi2; Vi1 = Vi2 = 100 mV; Vo = 45 mV; => Acm = 45/90 = 0.5

So, according to the results from the simulation, our calculated value for the synphase amplification is Acm = 0.5, which is relatively ok, but not so good, since it should be near 0 - that's case with current amplification for this circuit, since the output current is ~ 0.

Picture 3: Transient analysis - output voltages and current wave forms (time-domain) for differential signal input mode


B. Differential signal input mode

For the differential signal input mode we need to have equal signals on both inputs of the DA, but with opposite phases. That means that the signals need to have equal amplitude and frequency, but different phase for 180 degrees one from another. For that purpose, we choose Vi1 and Vi2 as voltage sources with sinusoidal form, with amplitude of 100 mV, frequency of 1 kHz and phase of 0 degrees for Vi1 and 180 degrees for Vi2:

Vi1 = +/- 100 mV at f = 1 kHz;
Vi2 = +/- 100 mV at f = 1 kHz with Phi = 180 degrees;

The results of this time-domain simulation are shown on Picture 3. This time we plot just the difference between outputs Vo1 - Vo2. We can view on this difference as a output voltage Vo which is applied to the load Rl. The amplitude of Vo is 477 mV (here, we are looking only at AC component of the output voltage):

Vo1 - Vo2 = Vo ---> min = -477 mV; max = + 477 mV;

Il = +/- 47.7 uA;

Ad = Vo/Vd | vi1 = -Vi2 = Vid/2; ---> Vid = Vi1 - Vi2 = 200 mV; Vo = 477 mV; => Ad = 477/200 = 2.385


So, according to the results from the simulation, our calculated value for the differential amplification is Ad = 2.385, which is not big enough, but that depends on what do we want from our circuit. Here, our goal is just to see how we can do a simple analysis of the circuit. And finally, when we calculated the both amplification Acm and Ad, we can calculate the CMMR parameter for this circuit configuration as:

CMMR = Ad/Acm = 2.385/0.5 = 4.77;

The CMMR of 4.77 is not big enough, but also it's not very bad at all, since it's positive and it's bigger than one, which means our circuit amplifies the differential signal and suppresses the common signal on the inputs from it's output, and actually that is the main function of this amplifier configuration.

Just to mention, the circuit configuration is made just for analysis purposes. As we can see from the numbers, we have small differential amplification and the synphase voltage amplification is not close to zero, but that is the result of the values of each component in the circuit. If we choose other values we can achieve better results and bigger differential amplification, but our goal here was to see the basic principle of work for this type of amplifier.

This amplifier configuration can also be used and for classic amplifying of one single input signal, but then it will not have it's basic function as differential amplifier. Namely, if we want to use this amplifier for amplifying just one single input signal we need to re-configure the inputs. If we lead our input to input 1 - Vi1, then we should remove signal voltage Vi2 and connect the Ri2 resistor to the ground. Also, If we lead our input to input 2 - Vi2, then we should remove signal voltage Vi1 and connect the Ri1 resistor to the ground. With these modifications we will have amplified output signal on the load Rl. In both cases, we will have the same apsolut amount of amplification on the output, but the output signals will be different in phase for 180 degrees one from another - that's why one of the inputs of this circuit is called non-inverting and the other inverting. However, since the nature of this circuit is to amplify the difference between the two input signals we should use it that way. It's very useful in control applications. If we have same amplitudes of both inputs, we have no output signal. If one of the amplitudes change it value, the output signal will appear. Since the output is amplified, we can use it for controlling small changes between two inputs.