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Hello smart people of the internet. Please help me to solve a puzzle. I'm pretty sure there is something simple I'm missing. So here is what I have: I have a 2.2v battery that I want to discharge via R-BAL (see scheme below). I have a MOSFET and a simple switch that controls the gate. There is also a pull-down R2 resistor that keeps the MOSFET closed. Everything works nice here, but the issue that I have is that I want to have my MOSFET to be opened as maximum as possible. To achieve that I need to have Vgs at around 10v.

enter image description here

So, I added a MT3608 DC-DC converter (this one) to my circuit as shown below. Firstly, I do not connect VOUT+ to the gate and measure 10v as shown below. All looks great here.

enter image description here

However, as soon as I connect VOUT+ to the gate, the reading drops back to 2.2v

Here are 2 questions please

  1. Why does it drop back to 2.2v?
  2. I suspect the answer to the first question will be something like "you cannot do that because xxx". Therefore my second question is: is it any way for this circuit to have 10v on gate-source of the MOSFET?
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  • \$\begingroup\$ Check all your wiring, especially your MOSFET. Is it backwards? \$\endgroup\$ – DKNguyen Jun 9 at 1:29
  • \$\begingroup\$ Thank you for the comment. Does the scheme look good to you? Will try to double check the wiring, but I played with it a lot already... maybe will try to take another mosfet... \$\endgroup\$ – Sancho.RubyROID Jun 9 at 1:32
  • \$\begingroup\$ I see no obvious problem. What happens if you add a serires resistor at the gate? Maybe your battery is just too weak to supply more current? \$\endgroup\$ – DKNguyen Jun 9 at 1:33
  • \$\begingroup\$ Is R2 too low, perhaps? \$\endgroup\$ – Hearth Jun 9 at 2:06
  • \$\begingroup\$ If you are using MOSFET with metallic case that could be "body diode". \$\endgroup\$ – user2960039 Jun 9 at 2:34

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