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I am using the UCC27423DGN dual low-side MOSFET driver as a current sink in order to drive some piezo elements. In general the gate drivers I am using are inverted, so on a high pulse on the Pulse_A/B lines, the outputs Pulse_A/B_Out should drop from VDD to GND.

In the design, VDD is 10.5 V from a switching regulator, and Pulse_En, Pulse_A, and Pulse_B are all 3.3 V signals from the microcontroller.

With no output connected (no piezo element attached), on circuit startup I am seeing that the gate driver is consistently shorting the VDD signal to GND once VDD is applied, seemingly permanently damaging the gate driver. Even after removing the IC from the board, a short can be seen between VDD and GND on the TSSOP pins.

The small resistor and capacitor on the output are meant to act as a low-pass filter. As far as I know, the voltages should be within spec and the circuit fails before a load is attached. Any help would be appreciated, as I am not sure what I am missing.

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ocrdu
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md-raz
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1 Answers1

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You're simply operating the gate driver way outside of its absolute maximum ratings.

According to the datasheet, the driver can drive up to 4.5A pulses for at most 0.5µs. It can only drive 0.2A continuously. Its output resistance is typically 2.2 Ohms.

When the driver switches its output low, a current of around 4.5A will flow through it as the huge 3.3µF capacitor discharges. Unfortunately, an RC low-pass with 2.2 Ohms and 3.3µF has a time constant of around 7µs, and the current only decays to a safe level after 3 tau. As a result, you're whacking the driver with an excessive current for more than 20µs, while it's only specified to survive this current for 0.5µs.

In other words, you're exceeding the safe operating area by a factor of 40. It's no wonder that the driver dies. You simply can't drive such huge capacitors with this MOSFET driver.

Jonathan S.
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  • Thanks for swift reply, I did see the 5us requirement but did not know if that would be the reason why It was failing, as there was no output connected. In light of this, I will most likely have to use a different IC, as the load will probably have even higher capacitance. Thanks again. – md-raz Mar 21 '23 at 17:07
  • @md-raz Or increase the 0.1 Ohm resistor to create a much lower current? – Mr47 Mar 22 '23 at 09:37