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With the widely available DC-DC boost ready built modules based on MT3608 chip can I control it by adding the N-ch MOSFET to the Vin- line like this: dc-dc boost application schematics

The module itself are pretty reference in design similar to this: dc-dc boost module reference design

Module load are COB LED +12V light. +Vbat are Li-Ion power which also powers the MCU through 3.3V LDO and I want to enable/disable module & LED from MCU.

PS: I know the DC-DC IC has the ENABLE pin, but it's hard wired to the Vin+ on board and I don't want to deal with cutting traces.

I'm also aware it's possible to control LED by adding MOSFET to the LED GND line but this way the module will be always powered up and I want to save battery power by turning it off completely when not used.

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    \$\begingroup\$ Only if there is no common ground and maybe not even then. You should use the enable pin, it is what it is there for. \$\endgroup\$ – Chris Stratton Apr 17 at 6:53
  • \$\begingroup\$ As I've already said the enable pin are hard wired. The module has the common ground between input & output side but the output load (LED) aren't connected to anything else except the module. I.e. both in&out sides of the module has only one ground connection point - through the MOSFET. I guess the problem here because the inductor on the DC-DC can give a spike burst back to the batter when it's cut off from the ground? \$\endgroup\$ – NStorm Apr 17 at 7:09
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Yes, it should work. But...

Note that your LEDs "GND" will not be the same as the GND of the rest of your ciruit, it will have an offset equal to, $$V_{DS} = R_{DS_{(ON)}} \cdot I_{DS}$$

Thus, a slightly "cleaner" way of doing this would be to have a P-channel MOSFET acting as a high-side switch. This would keep all GND the same and will save you some headaches. If you are not confortable driving the P-MOS directly (if the Vbat is to high for your MCU) you can use a circuit as shown bellow, where another transistor (N-channel) is used to drive the P-MOS. I did not chose the transistor carefully, don't pick the same (please find some that have Vgs compatible with your application).

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Thank you for your answer. Yes, I knew I could achieve this by P-ch & another driving transistor (yes, Vbat are too high for logic). But can you explain a bit more about the offset voltage on the ground with N-ch solution? \$\endgroup\$ – NStorm Apr 17 at 7:48
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    \$\begingroup\$ Your low side switching N-channel transistor, when activated, will start conducting electricity, right ? But it resistance is not quite 0Ohms. It equals Rds(on) (a.k.a the equivalent resistor of your transistor when fully activated), thus from Ohms law U=R I you will have a voltage drop across the transistor. This means that In- of your switching converter is at $$ R_{DS_{(ON)}} \cdot I$$ away from GND. Thus Vout- will also be offset from GND by the same amount. This offset will be small, so you might be able to get away with it, but it might not be the cleanest way of doing this. \$\endgroup\$ – benguru Apr 17 at 8:00
  • \$\begingroup\$ Thanks. Yes, the MOSFET I'm using has a very low Rds(on), which is ~0.035Ohms and the maximum current from the battery (at it's lowest charge level) would be ~0.6A. So it's only around 21mV which should be totally negligible I guess. \$\endgroup\$ – NStorm Apr 17 at 8:45
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If the MT3608 module's ground ( = IN- = OUT-, these pins are shorted on the module) is kept separate from the rest of the circuit then: yes you can.

The schematic should look like this:

schematic

simulate this circuit – Schematic created using CircuitLab

I have added a current limiting resistor R1, because the 12 V LED COB modules I have seen all need a constant current, not a constant voltage. Maybe your LEDs are different. I do not know, but I just mention this to warn you of potential issues that occur if you apply 12 V directly to LED modules which actually need a current instead of a voltage.

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  • \$\begingroup\$ Yes, you are right. In- and Out- are shorted on the module. The Out- side are NOT connected separately to anything except LED "cathode". My LED COB module has already built-in series resistance adjusted for optimal current from +12VDC voltage source. So your schematics are basically equivalent to my LED module. \$\endgroup\$ – NStorm Apr 17 at 7:33
  • \$\begingroup\$ I've updated my schematics to draw a led & series resistance to resemble actual "COB LED module" I was referring to equivalent circuit. \$\endgroup\$ – NStorm Apr 17 at 7:40

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