Rep: 23
2
I’ve cut the plug off, but no voltage on the two output wires? Do I need to apply a load? Give feedback to the adapter? Etc.
Questa è una buona domanda?
Rep: 782,5k
1,1k
893
2,3k
@robertdunn60451 The tiny center pin is a charge control (Adapter Sense) pin that assists with changing the LED color, and also assists with switching the adapter off. There should be low power output (3V-6V) from any of the other ports until contact is made with the center charge control pin. This pin is also used to transfer additional Information like the power type and serial number of the power supply. it is my understanding the Magsafe uses 3 volts to power a power management chip inside whatever it is connected to. Ones it communicates and verifies with that PMIC about charge compatibility, it will supply the full voltage. If you can't measure it like that, apply a 40 kΩ resistor between the MagSafe's output and ground which will then enable the full voltage.
For more info check sites like this
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Thanks very much. I put a 36k resistor across the output, plugged it in, and immediately got 16.5 V.
Rep: 34,7k
10
22
44
if you cut the plug off you can't really expect to get power, so i don't exactly get what your asking. yes you need to apply power to get power
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Daniel, thanks for your reply. Please see my comment above.
I want to cut off the DC out plug (the MagSafe plug), and take the straight DC from the adapter. But I don’t get anything out after I cut off the MagSafe plug.
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if you mean can you connect the wires to a dc bench supply then yes, between 18-20v at around 3a
da daniel
Yes, this is what I want, to make the adapter into a DC power supply. I’ve cut the MagSafe plug off, but don’t get any voltage from the newly exposed wires. I’m assuming that the outer shield and the one inner wire are DC out.
da Robert Dunn