I built the following circuit to charge a large capacitor from a low power source before switching the load:


simulate this circuit – Schematic created using CircuitLab

MAX9064, MCH3484, MCH3383.

The MAX9064 has an internal REF of 0.2V and the resistor network should be triggering it on at 1.5V and off at 1.1V.

I think I may have messed up with the MOSFETs, as they are both Power MOSFETs? In any case, the circuit doesn't work properly. I get 0.5V at Vout with C1 removed. With C1 in place it charges way too slowly, but I'm still seeing a slowing rising mV value at Vout instead of 0V like it should be.

I'm thinking that either 1) I've used the Power MOSFETs incorrectly or 2) There just isn't enough current for it to function (but these are all super low current devices so it should?)

If it is the Power MOSFETs, is there a way I can make them work in this circuit or do I need to get different ones?

Note: I'm using the MAX9064 which has push-pull output. At 1.8V it is supposed to supply Vcc - 0.15 to 0.23. So it's output at the switch on voltage should be around 1.3V. M1 only requires a drive of 0.9V which it should be supplying.

  • \$\begingroup\$ Could you test M2 by short its gate to ground directly? To eliminate a possible cause... \$\endgroup\$ – Huisman Mar 31 '19 at 12:47
  • \$\begingroup\$ Can you further measure pin 4 of the MAX9046 wrt GND (the multimeter's internal resistance can be neglected wrt to the 350 kohm) while applying an adjustable voltage source to pin 1 of the MAX9046 increasing it to 1.8V and decreasing it again to 1.0V. When passing the 1.5V with increasing voltage you should notice an increase at pin 4 of 40mV as the ouput of the MAX9064 contributes through R3. Same when decreasing the power supply. I think it should see a drop at about 1.2V instead of 1.1V. Please verify it this works correctly. \$\endgroup\$ – Huisman Mar 31 '19 at 12:52
  • 3
    \$\begingroup\$ M2 needs a gate to source resistor to make it turn off. \$\endgroup\$ – EinarA Apr 1 '19 at 2:59
  • \$\begingroup\$ I think I might actually have a faulty component. I need to do some more testing and will update... \$\endgroup\$ – hekete Apr 1 '19 at 9:18
  • \$\begingroup\$ Of course @EinarA, +1. \$\endgroup\$ – Huisman Apr 1 '19 at 11:29

First of all:

The MAX9060 and MAX9061 have open-drain outputs

so it will never drive M1. You need to add a pull-up resistor.

As OP indicates, the MAX9064 is used which has a push-pull output.

My first thought was M1 has too much leakage current, pulling down the output of MAX9064.
However, the Gate to Source Leakage Current of the MCH3484 is at most 10 uA at \$V_{GS}\$=4V, and with lower \$V_{GS}\$, it will be even less. According the MAX9064 datasheet, at (test conditions) \$V_{CC}\$ = 1.2V and \$I_{SOURCE}\$ = 40μA the ouput voltage is typically 1.12V or min 1.0V. So with a lower \$I_{SOURCE}\$, it will indeed be enough to drive the MAX9064.

Regarding measuring an output while switched off
You probably measure a voltage due to the Zero-Gate Voltage Drain Current of M2. It is in the order of uA (at VDS=−12V, VGS=0V it is −10 μA) which will cause a voltage across the measurement resistor (which is in the order of MOhm) of the multimeter.
Apply a 1kOhm resistor as load and you should measure nothing.

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  • \$\begingroup\$ I'm using the MAX9064 which has push-pull output \$\endgroup\$ – hekete Mar 31 '19 at 7:22
  • \$\begingroup\$ The datasheet is a bit confusing because it has so many different versions of the chip in it and only one of them has push-pull. I will update the question to make it clearer. \$\endgroup\$ – hekete Mar 31 '19 at 7:37

Okay, so there are/were two problems.

1) My input source circuit had died at some point and was not delivering the expected voltage!

2) As @EinarA pointed out a Gate to Source resistor was required on M2 or it never turns off!

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