Operational amplifiers are widely used in various types of electronic products to amplify or attenuate analog signals, so that the signal amplitude reaches a reasonable range for other circuits to compare or sample.

The differential amplifier has an advantage that ordinary amplifiers do not have: it can detect one or more signals that are not in common ground, and each measured signal or amplifier is not affected by the non-equipotential, so that each signal and the amplifier are not affected. continue to maintain the “isolation” feature. However, such a good advantage has not been valued by instrument manufacturers.

At present, most oscilloscopes cannot detect more than two signals that do not share the same ground at the same time, and even when only a single channel is used, they cannot directly measure non-isolated signals, such as 220V mains or 220V rectified voltage, because the probe’s The ground and the AC ground wire are connected, and a short circuit is a test. If the pre-stage sampling adopts the form of a differential amplifier circuit, this problem can be solved easily. However, Fluke’s oscilloscope does support the measurement of different ground signals, but whether it uses a differential amplifier circuit, I have not studied it.

The following figure is the sampling circuit of the rectifier voltage. According to the experience of the pioneers of science and technology, when the two input resistances are equal and the two feedback resistances are also equal (for the time being, the non-inverting terminal resistance is also called the feedback resistance), the amplification ratio of the circuit is RF/RI , the figure below is 10/1000, or 0.01 times, the attenuation circuit.

I can’t understand the formula derivation process in the textbook. It’s my fault that I didn’t learn mathematics well, but I believe that the formula is correct, because I have calculated it in my own way of understanding, and also experimented, and enlarged the ratio. It is indeed RF/RI. I will share my derivation method below, which is also the calculation method of each voltage point, but it should be noted that this calculation method is used when the measured signal and the amplifier do not share the same ground. The calculation method is different when it is on the ground, which I will talk about later.



In the figure, the tested voltage is 540VDC, the upper is positive and the lower is negative.

We know that when the op amp works in the amplification area, the voltages of the positive and negative input terminals are equal (the ideal state is exactly the same, there is a slight deviation in practice, and the deviation value is determined by the quality of the op amp), that is, the virtual short, the load of the measured signal The current can be equivalent to the figure on the right, from which we calculate the loop current of the signal under test, 540V/2000K=0.27MA, the red arrow is the current direction, OK.

We also know that the op amp also has a virtual-off characteristic, that is, the current at the positive and negative input terminals is almost 0, which can be ignored. Then we can conclude that the current flowing through the two input resistors is the same as the current flowing through the two feedback resistors. , that is, the currents of the four resistors are all 0.27MA. Therefore, we can calculate that the voltage of the feedback resistor is 0.27MA*10K=2.7V, and the current flows from bottom to top, so the voltage of the non-inverting input terminal is -2.7V, because of the virtual short, the voltage of the inverting input terminal is also -2.7V, negative feedback The resistance current flows from left to right, that is, the resistance voltage is high on the left and low on the right, and the amplitude is 2.7V, so the output voltage is 2.7V lower than the negative input terminal -2.7V, that is -5.4V.

If the formula is used to calculate, the magnification ratio is 0.01, 540*0.01=5.4V. Since it is an inverting circuit, a negative sign must be added, that is, the result calculated by the algorithm using the formula is the same as that calculated by my own algorithm, but I The algorithm is easier to understand (I personally think this), and can calculate the voltage of each point, which can quickly determine which point voltage is abnormal during maintenance, so as to find the fault point faster and improve the maintenance efficiency.

A friend said that the output voltage is equal to the sum of the two input voltages (-2.7+(-2.7)=-5.4). Is this an adder? This isn’t actually an adder, the result is just a coincidence, since both feedback resistors are the same in this circuit. In order to clear the doubts of friends, let’s take a look at the picture below.



I changed the negative feedback resistor from 10K to 1K, let’s see what happens? The current of the signal under test has not changed, and the voltage of the two input terminals has not changed, but the output has changed. Because the feedback resistance has become smaller, the voltage of the feedback resistance has changed from 2.7V to 0.27V (0.27MA*1K=0.27V), so The output becomes -2.7+(-0.27)=-2.97V.

Therefore, the output voltage is not necessarily equal to the sum of the voltages of the two input terminals. Of course, anyone who uses this kind of circuit does not easily change the circuit shape, and still maintains the two input resistances and the two feedback resistances are equal.

The above calculation method is only for the case where the signal under test and the op amp do not share the ground. If the signal under test has a direct relationship with the op amp, the calculation method is roughly the same as that of the ordinary amplifier. see the picture below



The picture is a current detection circuit. A small resistor is connected in series in the load loop, and a weak signal is obtained from both ends of the small resistor and sent to the differential amplifier for 10 times amplification.

Let’s first calculate the load current, 12V/5.1Ω=2.35A, the load resistance voltage is 2.35A*5Ω=11.76V (rounded up), the sampling resistance voltage is 12-11.76=0.24V, the inverting amplification is 10 times, the result should be For -2.4V, let’s verify it.

First come the voltage at the non-inverting terminal, the load resistance voltage is 11.76V, so we calculate that the voltage at the non-inverting terminal is 11.76V/110*100=10.69V. Due to the virtual short, the inverting terminal is also 10.69V, and the negative feedback current is (12- 10.69) V/10K=0.131MA, the negative feedback resistor voltage is 0.131MA*100K=13.1V, the left is high and the right is low, so the output voltage is 10.69-13.1=-2.41V (the extra 0.01 is due to the deviation of the previous rounding), The result of the calculation is the same as the result of applying the formula. Again, by measuring whether the voltage value of each point is normal or not, the fault point can be found quickly, which is the importance of theory.

We know that the load has an input resistance, and the power supply has an output resistance. Why does the voltage drop when the load is applied? It is the output resistance of the power supply that is at fault. The signal will also have an output resistance. Please see the image below.



The left of the figure is an inverter, which reversely amplifies the signal sent by the front stage, and the C pole obtains a signal with the same frequency as the input signal, opposite phase, and an amplitude of 12V.

But if a load is added to the output signal, and the resistance value is the same as the C-pole resistance, what will be the result? On the right of the figure, we can see that the frequency and phase of the output signal have not changed, but the amplitude has changed to 6V, which is a full half. This is the effect of the signal output resistance (C-pole resistance). In electronic circuits, it is often encountered such a situation that some signals (or voltages) are going to drive some loads, but their own output resistance factors cause insufficient driving capability, so they are unable to drive the subsequent loads. What should I do? At this time, it is necessary to add a first-level buffer between the driving source and the subsequent load, as shown in the following figure.



A triode is added between the driving source and the load, and the driving signal directly pulls the B pole. When the signal is at a high level, the B pole has a current flowing in, flowing out from the E pole, flowing through the load resistance, and finally reaching the ground, but the B pole current flows. The inflow of the triode makes the triode turn on quickly and generates the C-pole current. Because the triode has an amplifying effect, the C-pole current = B-base current * magnification, the C-pole current is very large, and the triode reaches a saturation state, and the E-pole voltage is about 11.4V , which is only 0.6V lower than the B pole voltage. If this voltage drop is ignored, it can be regarded as VB=VE, thus ensuring sufficient voltage amplitude for the load resistance and improving the driving ability. Because this only amplifies the current driving ability, not the voltage, VE changes with VB, that is, the emitter voltage changes with the base voltage, people call this triode circuit an emitter follower, which is also an emitter follower. originator. Later, when the op amp appeared, there was an upgraded version of the emitter follower, the triode version, the input and output signals had a voltage difference of 0.6V, but the voltage difference of the op amp version was much smaller. as shown below



Friends, do you think it’s strange that when we talk about differential amplifiers, we talk about emitter followers again? Don’t worry, please listen to me slowly.

The two differential amplifier circuits mentioned above, one is attenuated by 100 times, and the other is amplified by 10 times, so let’s have a circuit that neither amplifies nor attenuates, how? See the image below.



What does the circuit above do? The voltage amplification factor is 1, that is, no amplification. And the current driving ability must be enhanced, right, isn’t it the emitter follower? Yes, in fact, it is an emitter follower, which does not amplify the voltage, but only improves the current driving capability. Of course, it doesn’t matter what the circuit is called, it’s important to know how it works so that it can be helpful for repairability.

In fact, the differential amplifier also has non-inverting amplification and inverting amplification. The above are all inverting amplification. When the measured signal is connected to the positive side of the amplifier, it is non-inverting amplification, otherwise it is inverting amplification. We take the current detection circuit above Make some changes, see the image below.



I reversed the positive and negative of the sampled signal, and the circuit changed from inverting amplification to non-inverting amplification. After calculation, the output voltage value of the amplifier is still 2.4V, but the polarity is positive.

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