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I'm designing a self sustaining guitar that has 6 amplifiers onboard and runs on a li-ion pack. Power draw is a about 2.5A when on "full power" and I decided to make my own step up converter to get the 3.7v from the battery pack to 6v.

I went with a MAX1771 controlling an IRF630 mosfet and just followed the reference example circuit from the datasheet, the bootstrapped version as apparently it's better for a low Vin.

Without load the circuit works fine but when plugging it to the amp board the output goes back to Vin through the diode, the controller's duty cycle is 90% ON at 50Khz -max power I assume- and can't see any voltage spikes on the inductor from the switching.

Now the mystery: Why if I connect a resistive load to the output I can easily draw 500mA from the output and with the controller being quite relaxed on a low duty cycle, while if I connect the amps board the Vout goes back to battery level even if they're not using any power and the current draw is just 50mA?

one note: I noticed that when trying to power the amp board, SW out of the controller to the mosfet is 3.5v that seems a bit low to make the mosfet fully conductive. Though this signal can't be more than Vd and if Vd=Vout=Vin how is it supposed to start the system?

Any ideas?

Thank you in advance!

Cheers

Circuit diagram from the datasheet

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    \$\begingroup\$ Which inductor are you using? What's the saturation current? \$\endgroup\$
    – John D
    Commented Nov 30, 2020 at 23:13
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    \$\begingroup\$ What does your implementation look like? Did you design a PCB, and if so, what is the layout? Or is the boost converter assembled on a breadboard? A picture might help shed some light on this. \$\endgroup\$
    – marcelm
    Commented Dec 1, 2020 at 0:03
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    \$\begingroup\$ IRF630 seems like a poor choice in your application. It's Vgs(th) is somewhere between 2V and 4V, and that's the point where it just barely starts to conduct. The 'Typical Output Characteristics' graphs on page 3 of the datasheet also looks pretty sad at low voltages. If you compare to the MOSFETs in the MAX1771's application circuit you'll see that they have a Vgs(th) of 1V to 2V and much more impressive low-voltage operation characteristics. \$\endgroup\$
    – brhans
    Commented Dec 1, 2020 at 1:29

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I noticed that when trying to power the amp board, SW out of the controller to the mosfet is 3.5v that seems a bit low to make the mosfet fully conductive. Though this signal can't be more than Vd and if Vd=Vout=Vin how is it supposed to start the system?

IRF630 has RDSon of 0.3 Ω with 10 V Gate drive, and needs ~6 V to turn on fully. 3.5 V is indeed too low to turn the FET on properly. But even if it did turn on fully its Drain-Source resistance would be a problem because at 2.5 A output the input current must be > 2.5 * 6 / 3.7 = 4 A, but the FET would drop > 4 * 0.3 = 1.2V so it would need to draw even more current to get the required output (which would drop more voltage, which...).

Bottom line is you need a FET with lower Gate drive voltage and lower RDSon.

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  • \$\begingroup\$ Thank you Bruce for the explaination, I went ahead and read a bit about logic mosfets and ordered a IRLZ34NPBF (RDS(on) = 0.035Ω, Vgs 1 to 2V), Cheers! \$\endgroup\$
    – Alex P
    Commented Dec 1, 2020 at 19:45

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