How to amplify the rectangular wave with a frequency of several hundred KHz and a peak-to-peak value of 3.3V from a single-chip microcomputer by more than ten times (to more than thirty volts...

DevilMayCry

How to amplify the rectangular wave with a frequency of several hundred KHz and a peak-to-peak value of 3.3V from a single-chip microcomputer by more than ten times (to more than thirty volts... [Copy link]

 

As shown in the figure above, I want to use a single-chip microcomputer[/url] to generate the pulse or rectangular wave shown in the figure. First, the DAC built into the single-chip microcomputer forms a rough waveform (peak-to-peak value of about 3.3V). What method should be used to enlarge it to the shape shown in the figure? Or which great god has a better method to generate the waveform shown in the figure? I don't know much about electronic technology, thank you for your guidance!!!

chunyang

Your pulse is a positive and negative bidirectional pulse, which can only be realized using an op amp, but it is not easy to make the edge steep enough, so a high-speed op amp is required.

maychang

The signal generated by the microcontroller has only two values: either high or low. There are three values in your diagram: 0V, 15V, and 30V. Therefore, it is impossible to amplify the signal from the microcontroller.

lzwml

maychang posted on 2017-3-31 18:16 The signal generated by the microcontroller has only two values: either high level or low level. There are three values in your diagram: 0V, 15V, and 30V. Therefore, it is impossible to...

Adding a DC bias conversion circuit can be based on 15V. It is not difficult for the microcontroller to output a signal with a 10us period. As Chunyang said, the difficulty lies in the conversion circuit to ensure that the 0-30V conversion can ensure that the edge is straight. It is not easy. I am not good at device selection.

xujian2000

Go to TI company to find it.

xujian2000

Go to TI company to find it.

xujian2000

Go to TI company to find it.

xujian2000

Go to TI company to find it.

xujian2000

Go to TI company to find it.

maychang

lzwml posted on 2017-3-31 20:20 Adding a DC bias conversion circuit can be based on 15V. It is not difficult for the microcontroller to output a signal with a 10us cycle. It is different from what Chunyang said...

You said that "adding a DC bias conversion circuit" is not necessary. It has been explained before: the microcontroller output has only two levels. There are three in your curve. In addition, the microcontroller outputs a digital signal, not an analog signal, so there is no need to "amplify" it. The circuit you want is nothing more than a circuit that outputs 30V when the microcontroller outputs a high level, and outputs zero when the microcontroller outputs a low level, and the edge is steep.

littleshrimp

Look at the motor driver chip H bridge

chunyang

maychang posted on 2017-3-31 18:16 The signal generated by the microcontroller has only two values: either high level or low level. There are three values in your figure: 0V, 15V, 30V. Therefore, it is impossible to get the value from the single chip...

Just use DAC or PWM + low pass filter.

chunyang

lzwml posted on 2017-3-31 20:20 Adding a DC bias conversion circuit can make the 15V the benchmark. It's just that the microcontroller outputs a signal with a 10us cycle. It's not a difficult task, and chunyang said...

Adding DC bias is useless, this is a three-level circuit.

chunyang

There is another simpler way to achieve the original poster's requirements, which is lower in cost and has better edge characteristics at low cost: Use the two IOs of the MCU to drive a 74HC238 (38-wire data distributor/decoder), or use 74HC239 (dual 24-wire data distributor/decoder). The data distributor is added to prevent the subsequent circuit from losing control and short circuiting due to power-on transients and software runaway. Here, Y1 and Y2 are specially selected as outputs, and Y0 is not used to prevent accidental loss of control. When Y1 and Y2 are valid, the two IOs of the MCU must be in the opposite logical state, which is highly reliable. Then connect an electronic switch composed of an NPN and a PNP transistor after Y1 and Y2. The inputs of these two switches are connected to 30V and 15V voltages respectively, and the outputs are connected in parallel and connected to a resistor to the ground. In this way, when the control IO of the MCU is at 10 or 01, one of the electronic switches is turned on and the output is 30V or 15V respectively. When the control IO of the MCU is at 00 or 11, both electronic switches are turned off and the output is biased to 0V by the resistor.

DevilMayCry

Thanks for your help, I will try your method! :handshake