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This is my third revision of this supply switching circuit. Do you guys see any problem with this revision?

This is a circuit to switch from disposable lithium battery to USB power when the device is plugged into a computer. Q1 and Q2 are PFETs. When there's no USB power, both FETs Vgs is -3.6v. When USB is plugged in, the USB interface IC sends 3.3v to the FETs gates, sufficiently turning Q1 and Q2 off. VBUS is a regulated voltage from USB. The diode to is keep from back feeding the regulator when there's no USB power.

I can't use an ORing diode circuit, I cant afford the parasitic losses. The losses from the diode on VBUS are fine since it won't be from the battery.

Thanks guys

Schematic

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    \$\begingroup\$ I don't see what Q2 is bringing to the party. Have you tried simulating the circuit? If not, why not? \$\endgroup\$ – Andy aka Dec 17 '19 at 14:31
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    \$\begingroup\$ @Andyaka Q2 prevents the USB 5V from directly (over)charging the battery. \$\endgroup\$ – Cristobol Polychronopolis Dec 17 '19 at 14:49
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    \$\begingroup\$ @Andyaka Cristobol Polychronopolis is exactly right. Without Q2, VBUS voltage would conduct through the body diode of Q1 and charge the non rechargeable battery causing a hazard. \$\endgroup\$ – James Pie Dec 17 '19 at 15:20
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    \$\begingroup\$ So, what does Q1 bring to the party? \$\endgroup\$ – Andy aka Dec 17 '19 at 16:00
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    \$\begingroup\$ I didn't mean that the diode direction would change, but if you drive the gate of Q2 high, say 3.3V, then in the presence of VBUS (also 3.3V), Q2 is off but its body diode may still leak current from the battery to the load. That is the case because the voltage at the anode of the body diode is Vbat ~3.6V and the voltage at its cathode is VBUS-VD1= ~2.8V, so the body diode is forward biased. In this condition, you do have a leakage current flowing from the battery, through Q2's body diode, to the load. All of this in the absence of Q1. That's why I said that I could then see the need for Q1. \$\endgroup\$ – Big6 Dec 17 '19 at 20:11
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If you want to make the turning-off voltage of the PMOSFETs "independent" of their rated \$V_{TH}\$, you can use a circuit similar to the following:

Circuit Waveforms Transition is made after 1ms. You can reduce the forward voltage drop of the diode by using a schottky one. An output capacitor might be necessary if you want to overcome, or at least reduce the voltage drop which occurs during the transition.

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  • \$\begingroup\$ Thank you for the reply! However, this is unnecessary for the given application and adds too many components to the design. But I do love this simulator! The one I'm using isn't this advanced. What is it? \$\endgroup\$ – James Pie Dec 17 '19 at 15:30
  • \$\begingroup\$ It is a free software called LTspice (analog.com/en/design-center/design-tools-and-calculators/…) \$\endgroup\$ – vtolentino Dec 17 '19 at 15:57

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