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I'm playing around with an old MagSafe charger and designing a circuit to be powered off it.

From what I've read, the charger's safety mechanism first provides 6V with a current limit of around 100uA. The load is expected to apply a resistive load that pulls that voltage down to around 1.7V for a second or so. This seems to be some sort of protection against shorts (i.e. a short would have pulled the voltage close to 0V). Only after one second and the charger is happy there is no fault does it then send the full 14-20V down. It makes sense to do this since the 85W of power the charger is able to deliver is something that can seriously ruin your day if shorted.

Unfortunately, that means extra complications if I'm designing a circuit to be powered entirely off the charger. In particular, I don't want to provide power to the rest of my circuit until the charger is delivering full power.

I came up with a scheme to achieve this by taking advantage of a N-channel MOSFET with a high threshold voltage to do this.

schematic

simulate this circuit – Schematic created using CircuitLab

When VIN is at 6V, M1 will not turn on and R1 will be the only load being driven (which will pull down the voltage as per the spec). When the charger starts delivering the full voltage (around 14-20V), VIN will be above the threshold voltage of M1 which will deliver power to the rest of the circuit.

Now, in practice, I understand that a MOSFET is not really completely off if a lower-than-threshold-voltage is applied to the gate. Normally this isn't a problem but in this case, I can't really leak more than 100uA into the rest of the circuit lest the safety circuit thinks there's a short circuit.

Will this actually be an issue? What parameter should I look for when shopping for MOSFETs? Is the idea sound or have I missed something blatantly obvious?

tangrs
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  • You don't need to worry about the current when the gate voltage of the MOSFET is low, it will be less than a uA and even lower when Vgs = 0. Aayway, you will not find a MOSFET that is still off when Vgs = 6V, forget about that. What you need is another resistor from gate to ground and a 6 V zener diode between Vin and gate. The zener + resistor will give Vgs = 0 when Vin = 6V and Vgs = more than 8 V when Vin = 14V That is what you need. – Bimpelrekkie Jun 04 '15 at 12:59
  • If you want to power a circuit with the charger you should open the charger and keep only its power unit. And I am not sure you can find a mos that fits your spec, I highly doubt it actually. 20V of Vgs is a lot. – Vladimir Cravero Jun 04 '15 at 13:00
  • This will not work, there is no separate power unit and charger. The charger unit will be in the laptop. It must be because it needs to monitor the temperature of the battery-pack ! Remember exploding laptop/phone batteries ? That's why ! People call it a charger but it's actually only a power adapter, in this case switching to 14 V when a proper load is detected. – Bimpelrekkie Jun 04 '15 at 13:07
  • One last thing, (pun intended) Apple is notorious for implementing stuff in non-standard ways. So it might be that there is more needed to make the adapter supply 16V. I see there's a "charge control pin" on the Magsafe connector. Maybe that requires some data sequence for the adapter to supply 16 V. Just guessing, I have no personal experience with Apple products. – Bimpelrekkie Jun 04 '15 at 13:17
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    @Rimpelbekkie Ahh, I see how the zener + resistor idea could work. Although, I just realised that in practice, Vgs will never be 6V because R1 is supposed to pull the input down to ~1.7V when the charger is supplying 6V. – tangrs Jun 04 '15 at 13:22
  • @Rimpelbekkie, if by "non-standard" you mean adding safety & usability features, then yeah. And it's all described in the article the OP linked to. – Anthony May Jun 04 '15 at 14:00

1 Answers1

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Expanding on your idea here is something that might work:

schematic

simulate this circuit – Schematic created using CircuitLab

You should choose R1 and R2 so that their sum is less than \$17k\Omega\$, but their ratio can be pretty much anything: the constraing on the mosfet maximum gate source voltage is (quite) relaxed. You can possibly even use a relay to dissipate even less power, but you need to be sure its coil can be powered through R1.

Vladimir Cravero
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  • So essentially, we're picking R1 and R2 to divide the input in a way that when Vin = 6V, Vgs is as close to 0V as possible (while still maintaining Vgs >= threshold voltage when Vin = 14-20V)? – tangrs Jun 04 '15 at 13:27
  • Yep. You actually want to fully saturate the transistor, check the datasheet to see how high you can go while remaining low enough. I think some 6V should do the job, so something like 4k7 for R2 and 10k for R1 might be ok. – Vladimir Cravero Jun 04 '15 at 13:29