I'm building a boost converter and want to implement a shutdown feature. The particular chip I'm using, LM3478, has a shutdown feature which is activated by a high level on the frequency adjust pin. In this case, I'm assuming a high level would be 12V since the chip is powered by 12V. The control signal I will be using will be from an MCU which has an output level of 3.3V.

I cannot simply use a P-channel MOSFET since 3.3V at the gate is not enough to turn the MOSFET off if the source voltage is 12V. I could use a PNP transistor since it is current controlled but that would require the control signal to be open-collector (tri-stated to turn off, low to pull up).

I'm thinking I could maybe combine a PNP transistor and a small MOSFET to do this. The PNP emitter is connected to 12V and the collector is connected to the pin on the chip. The MOSFET drain is connected to the transistor base through a resistor and the MOSFET source is connected to ground. I would then control the whole thing by a signal on the MOSFET gate.

Is there a better way to do this?

  • \$\begingroup\$ Are you in frequency adjust mode? \$\endgroup\$ – markrages Sep 14 '12 at 16:42

enter image description here

3.3 volts > 1.4 volts. So 3.3 volts should keep it shut off.

But for more confusion see also note 5. The two sentences seem contradictory to me.

EDIT: Here's a pic of what I'm talking about. Push-pull totem output is internal to the microprocessor. You should reduce \$R_{FA}\$ by the output resistance of the totem pole, which is in the vicinity of 25 ohms. It's in the micro's datasheet.

enter image description here

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  • \$\begingroup\$ But this would still require an open collector output since I can't be pulling the pin low when not in shutdown. Maybe I can use a diode to prevent that, though? When pulled high, the diode is forward biased but when pulled low, the diode is reverse biased essentially taking the shutdown signal out of the circuit? \$\endgroup\$ – Emil Eriksson Sep 14 '12 at 20:38
  • \$\begingroup\$ Why can't you pull the pin low when not in shutdown? \$\endgroup\$ – markrages Sep 15 '12 at 23:29
  • \$\begingroup\$ Because the frequency when not in shutdown is set by the external resistor to ground on the frequency/shutdown pin. If the pin is pulled to ground, that's the same as no resistance. \$\endgroup\$ – Emil Eriksson Sep 16 '12 at 19:12
  • \$\begingroup\$ Connect the resistor to uC output? I draw picture... \$\endgroup\$ – markrages Sep 16 '12 at 19:45

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