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I'm working with the DRV103 PWM driver chip; this chip generates an initial long pulse followed by PWM and is useful for driving coils/solenoids. The problem is, this chip operates at 8-32V, while the solenoids I'm driving with it operate at 48V.

Normally this would be simple amplification, but it's complicated by the fact that the DRV sinks rather than sourcing, with it's OUT pin pulled high to its supply voltage while 'off' and sinking low when 'on'. This means I have to change a PWM supply that sinks from ~20V to GND to one that sinks from 48 to GND.

I've come up with this circuit, is there a way to do it with a single transistor/component? The output must be active when DRV_OUT is low, to preserve the initial long pulse, and the solenoid is expected to draw 600-700mA. The PCB this circuit will be on only has a single 48V input, which I've regulated down to power the DRV.

enter image description here

The resistor divider is there because the cheap transistors I've got in there can only tolerate up to 20V at the gate (but up to 60V drain-to-source). DRV_OUT has a pullup resistor not shown here.

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  • \$\begingroup\$ Like specifying capacitors for at least double the expected voltage, I like to spec MOSFETS for at least twice their operating voltage. So I'd spec a >= 100v MOSFET for a 50v bus. Also consider that D101 will slow down the release period of the solenoid, if that is important. \$\endgroup\$ – rdtsc Oct 3 at 22:17
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Assuming that the only problem is the voltage — the DRV_OUT can otherwise handle the solenoid current, you could use the big brother of the common logic level conversion circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Otherwise, you'll have to use your 2-transistor circuit — but be sure to add a pullup to Q101's gate, either to the driver Vcc or to another voltage divider.

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  • \$\begingroup\$ That's perfect, I love the simplicity! There is a pullup on DRV_OUT in the main schematic, not shown here for confidentiality reasons, but I'll use your suggestion. Thanks :) \$\endgroup\$ – Isaac Middlemiss Oct 3 at 21:18
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This is a variation on Dave Tweed's circuit. If you use a BJT you can get much lower voltage loss in the common-base (or -gate) amplifier -- like less than one volt.

You probably want to look around for a "super beta" transistor, or maybe a Darlington, and you want to be careful about supplying adequate base voltage -- the 0.7V that we all think about applies to small-signal types. Ditto adequate base current, if you're using a resistive divider or voltage regulator (and this is why I'm suggesting a super beta transistor).

Don't just pull a 2N3055 or other 50 year old transistor type off the shelf -- Zetex made some awesome advances in BJT technology in the 1990's. They didn't gain enough traction to take power electronics back from FETs, but they left behind some transistors with astonishingly better capabilities than the "old" ones.

schematic

simulate this circuit – Schematic created using CircuitLab

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