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Voltage inverter employs PWM

2014-12-12 11:49  
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Jeff Wilson, STMicroelectronics

Use two diodes and two capacitors to generate a negative voltage.

This Design Idea describes a circuit employing a small microcontroller-based sensor module with only three connections: 5V dc, an RS-232 transmit-data output, and ground. A dedicated single-voltage level shifter or a dc/dc converter would be too costly, but the design still needs to supply ±3V at 1 mA to drive the transmit-data pin. Because a spare PWM (pulse-width-modulator) output on the 5V microcontroller could drive ±5 mA at nearly 5V, a PWM-based voltage inverter using a BAT54S dual-series Schottky diode, two capacitors, and a limiting resistor would produce the negative voltage (Figure 1).

Voltage inverter employs PWM
Figure 1.The diodes and capacitors produce a negative voltage.

The microcontroller’s PWM output drives the inverter with a 1-kHz, 50%-duty-cycle, 0 to 5V waveform. When the PWM output is 5V, it charges C1. The lower diode in D1biases in a forward mode to connect the terminal to ground. When the PWM output is low, it transfers the charge in C1to C2by forward-biasing the upper diode in D1. Meanwhile, it inverts the charge by taking the positively charged terminal of C1nearly to ground potential. When the PWM output switches high again, the cycle repeats.

Due to D1’s minimum voltage drop of 0.2V, it is impossible to get to −5V from 5V, so the voltage output will be approximately −4.6V, with 0.2V loss in each phase. The design requires a limiting resistor, R1, only when the driving microcontroller is sensitive to the current transients when switching or if the switching transients disturb the analog inputs on the microcontroller.

The PWM’s timebase is 1 kHz, so component values must accommodate that frequency. If you need other frequencies, you must calculate new component values using the following equation:

where C is the value of C1and C2, F is the PWM switching frequency in hertz, and R is the total resistance of the PWM’s output-driving circuit.

When calculating the total resistance of the PWM output, you must take into account the drive rating of the digital output. A simple substitution for the value of R is V/A, where V is the drive voltage of the PWM’s output and A is the current drive of the output in amps. For example, the original values for this design

and

or 0.1 μF.

You can also use this circuit as a negative- voltage supply for ADC/DAC and op-amp dual supplies. For analog usage, you probably need to use additional filtering or micropower voltage regulators on the output to filter out the switching transients.

EDN


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