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I plan to use the MAX1555 chip

The chip output a logic bit to indicate if current is pull from the charger ( POK pin- Chip 1551 , CHG pin - chip 1555)

The datasheet talk about the low state, being around 150-300 mV.

What is the HIGH level voltage? I need to know because I implement a little circuit that ligh led 1 when current is pull and light led 2 when the device is done charging.

That HIGH level will affect my inverter and MOSFET choice.

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  • \$\begingroup\$ You realize that only the '1551 has nPOK and the '1555 only has nCHG, right? \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 6 '15 at 21:27
  • \$\begingroup\$ @IgnacioVazquez-Abrams Thanks for the precision, I will add it to the question \$\endgroup\$ – MathieuL Aug 6 '15 at 21:32
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The POK pin is open-drain which means you can select the voltage it is when the output is high. Refer to this picture.

enter image description here

The chip drives a MOSFET if the battery is being charged, making the output low. Otherwise the output is high impedance. If you pull up the signal with a resistor you can choose its high state to be whatever you want (Vcc in the picture).

The datasheet says to limit the voltage on the POK pin to +7V which means the MOSFET inside the chip can't handle more than 7V of potential across it.

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  • \$\begingroup\$ R1 being the pull resistance if i understand correctly \$\endgroup\$ – MathieuL Aug 6 '15 at 21:31
  • \$\begingroup\$ @MathieuL Yes that is correct. Choose it so it is high enough where not a lot of current will flow through M1 when active, and low enough that it switches fast enough for your application. 1k - 10k are good starts. \$\endgroup\$ – ACD Aug 7 '15 at 12:24
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From the datasheet:

The MAX1551’s nPOK is an active-low, open-drain output...

...

The MAX1555’s CHG is an active-low, open-drain charge status indicator.

Its high therefore is whatever you pull it up to. Note that pulling it above +7V may damage the chip though.

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