Introduction: A Precision Rectification Experiment
I have recently done an experiment on a precision rectification circuit and got some rough conclusions. Considering that the precision rectifier circuit is a common circuit, the results of this experiment can provide some reference information.
The experimental circuit is as follows. The operational amplifier is AD8048, the main parameters are: large signal bandwidth of 160MHz, slew rate of 1000V / us. The diode is an SD101, Schottky diode with a reverse recovery time of 1ns. All resistor values are determined by reference to the AD8048 data sheet.
Step 1:
The first step of the experiment: disconnect D2 in the above circuit, short circuit D1, and detect the large signal frequency response of the operational amplifier itself. The input signal peak is kept at around 1V, the frequency is changed from 1MHz to 100MHz, the input and output amplitudes are measured with an oscilloscope, and the voltage gain is calculated. The results are as follows:
In the frequency range of 1M to 100M, the waveform has no observable significant distortion.
The gain changes are as follows: 1M-1.02, 10M-1.02, 35M-1.06, 50M-1.06, 70M-1.04, 100M-0.79.
It can be seen that the large signal closed-loop 3 dB cutoff frequency of this op amp is about a little more than 100 MHz. This result is basically in line with the large signal frequency response curve given in the AD8048 manual.
Step 2:
In the second step of the experiment, two diodes SD101A were added. The input signal amplitude remains at around 1V peak while measuring the input and output. After observing the output waveform, the oscilloscope's measurement function is also used to measure the effective value of the input signal and the period average of the output signal, and calculate their ratio. The results are as follows (data is frequency, output mean mV, input rms mV, and their ratio: output average / input rms):
100kHz, 306, 673, 0.45
1MHz, 305, 686, 0.44
5MHz, 301, 679, 0.44
10MHz, 285, 682, 0.42
20MHz, 253, 694, 0.36
30MHz, 221, 692, 0.32
50MHz, 159, 690, 0.23
80MHz, 123, 702, 0.18
100MHz, 80, 710, 0.11
It can be seen that the circuit can achieve good rectification at low frequencies, but as the frequency increases, the rectification accuracy gradually decreases. If the output is based on 100 kHz, the output has dropped by 3 dB at approximately 30 MHz.
The large-signal unity gain bandwidth of the AD8048 op amp is 160MHz. The noise gain of this circuit is 2, so the closed-loop bandwidth is about 80MHz (described earlier, the actual experimental result is slightly larger than 100MHz). The average output of the rectified output drops by 3 dB, which is approximately 30 MHz, less than one-third of the closed-loop bandwidth of the circuit under test. In other words, if we want to make a precision rectifier circuit with a flatness of less than 3dB, the closed-loop bandwidth of the circuit should be at least three times higher than the highest frequency of the signal.
Below is the test waveform. The yellow waveform is the waveform of the input terminal vi, and the blue waveform is the waveform of the output terminal vo.
Step 3:
As the frequency increases, the signal period becomes smaller and smaller, and the gap accounts for an increasing proportion.
Step 4:
Observing the output of the op amp at this time (note that it is not vo) waveform, it can be found that the output waveform of the op amp has severe distortion before and after the output zero crossing. Below are the waveforms at the output of the op amp at 1MHz and 10MHz.
Step 5:
The previous waveform can be compared to the crossover distortion in the push-pull output circuit. An intuitive explanation is given below:
When the output voltage is high, the diode is fully turned on, at which point it has a substantially fixed tube voltage drop, and the output of the op amp is always one diode higher than the output voltage. At this point, the op amp works in a linear amplification state, so the output waveform is a good header wave.
At the moment the output signal crosses zero, one of the two diodes begins to pass from the conduction to the cutoff, while the other transitions from the off to the on. During this transition, the impedance of the diode is extremely large and can be approximated as an open circuit, so the op amp at this time does not work in a linear state, but close to the open loop. Under the input voltage, the op amp will change the output voltage at the maximum possible rate to bring the diode into conduction. However, the slew rate of the op amp is limited, and it is impossible to raise the output voltage to make the diode turn on in an instant. In addition, the diode has a transition time from on to off or from off to on. So there is a gap in the output voltage. From the waveform of the output of the op amp above, it can be seen how the operation of the zero-crossing of the output is "struggling" in an attempt to change the output voltage. Some materials, including textbooks, say that due to the deep negative feedback of the op amp, the nonlinearity of the diode is reduced to the original 1/AF. However, in fact, near the zero crossing of the output signal, since the op amp is close to the open loop, all the formulas for the negative feedback of the op amp are invalid, and the nonlinearity of the diode cannot be analyzed by the negative feedback principle.
If the signal frequency is further increased, not only is the slew rate problem, but the frequency response of the op amp itself is also degraded, so the output waveform becomes quite bad. The figure below shows the output waveform at a signal frequency of 50MHz.
Step 6:
The previous experiment was based on the op amp AD8048 and diode SD101. For comparison, I did an experiment to replace the device.
The results are as follows:
1. Replace the op amp with AD8047. The op amp's large signal bandwidth (130MHz) is slightly lower than the AD8048 (160MHz), the slew rate is also lower (750V/us, 8048 is 1000V/us), and the open-loop gain is about 1300, which is also lower than the 8048's 2400. .
The experimental results (frequency, output average, input rms, and the ratio of the two) are as follows:
1M, 320, 711, 0.45
10M, 280, 722, 0.39
20M, 210, 712, 0.29
30M, 152, 715, 0.21
It can be seen that its 3dB attenuation is less than a little at 20MHz. The closed-loop bandwidth of this circuit is about 65MHz, so the output average drop of 3dB is also less than one-third of the closed-loop bandwidth of the circuit.
2. Replace SD101 with 2AP9, 1N4148, etc., but the final results are similar, there is no substantial difference, so I will not repeat them here.
There is also a circuit that opens the D2 in the circuit as shown below.
Step 7:
The important difference between it and the circuit using two diodes (hereinafter referred to as the double-tube circuit) is that in the double-tube circuit, the operational amplifier is only in an approximately open-loop state near the zero crossing of the signal, and this circuit (hereinafter referred to as a single-tube circuit) The operation in the middle is in a completely open loop state for half of the signal period. So its nonlinearity is definitely much more serious than the double-tube circuit.
Below is the output waveform of this circuit:
100kHz, similar to a dual-tube circuit, also has a gap when the diode is turned on. There should be some bumps in the original place. The input signal is directly transmitted through two 200 ohm resistors. It can be avoided by slightly improving the circuit. It has nothing to do with the problems we will discuss below. It is 1MHz.
Step 8:
This waveform is clearly different from the dual tube circuit. The dual-tube circuit has a delay of about 40 ns at this frequency, and the delay of this single-tube circuit is 80 ns, and there is ringing. The reason is that the op amp is completely open-loop before the diode is turned on, and its output is close to the negative supply voltage, so some of its internal transistors must be in deep saturation or deep-off state. When the input crosses zero, the transistors that are in the "deep sleep" state are first "wake up", and then the output voltage is raised to the diode at the slew rate.
At lower frequencies, the rate of rise of the input signal is not high, so the effects of these processes are not shown (as is the case with 100k above), and after the frequency is high, the signal rate at the input is large, thus "waking up" the transistor. The excitation voltage or current will increase, which causes ringing.
Step 9:
5MHz. There is basically no rectification at this frequency.
Step 10: Conclusion
Based on the above experiments, the following conclusions can be drawn:
1. When the frequency is very low, the nonlinearity of the diode is eliminated by the negative feedback of the op amp depth, and any circuit can get a good rectification effect.
2. if you want to achieve higher frequency precision rectification, single-tube circuit is not acceptable.
3. even with dual-tube circuits, the slew rate and bandwidth of the op amp will seriously affect the rectification accuracy at higher frequencies. This experiment yields an empirical relationship under certain conditions: if the flatness of the output is required to be 3 dB, the closed-loop bandwidth of the circuit (not the GBW of the op amp) is at least three times greater than the highest signal frequency. Since the closed-loop bandwidth of the circuit is always less than or equal to the GBW of the op amp, the precision rectification of the high frequency signal requires a very high GBW op amp.
This is also a requirement for an output flatness of 3 dB. If higher output flatness is required in the input signal band, the frequency response of the op amp will be higher.
The above results were obtained only under the specific conditions of this experiment, and the slew rate of the op amp was not considered, and the slew rate is obviously a very important factor here. Therefore, whether this relationship is applicable under other conditions, the author does not dare to judge. How to consider the slew rate is also the next question to be discussed.
However, in the precision rectification circuit, the bandwidth of the op amp should be much larger than the highest frequency of the signal.
