InspectorGadget
- Joined Nov 5, 2010
- 215
Do not short out R2. A boost converter shorts the inductor to ground in the ON cycle and if you destabilize the system you can burn out the drive transistor. Shorting the FB to ground will make the converter think it's always on the low side and try to drive it too hard.
Going back over the equations I see I made a terrible mistake. The voltage divider ratio is actually R2 / (R1 + R2). The output voltage equations are actually:
Vout = Vref * (R1 + R2) / R2
Vout = Vref + Vref * R1 / R2
So it's R1 you need to bridge. You can bridge R1 with the 200K pot without removing it, which will give you less chance of damaging the board.
Since it's a boost regulator, you can't get any lower than the input voltage. it won't do any good to short out R1. I believe it will just never turn on the transistor switch and you'll just get the input voltage passed through to the output through the inductor.
And as I said above, don't short out R2 because that would make it try to drive the output too high.
Here's the boost configuration schematic. As you decrease R1 in the feedback (not pictured), the boost control circuit decreases the boost "charging" of L by turning on the transistor switch for smaller and smaller amounts of time (lowering duty cycle). When you get R1 low enough, the switch will just stop switching at all and the input voltage will pass through L and D to the output.
Going back over the equations I see I made a terrible mistake. The voltage divider ratio is actually R2 / (R1 + R2). The output voltage equations are actually:
Vout = Vref * (R1 + R2) / R2
Vout = Vref + Vref * R1 / R2
So it's R1 you need to bridge. You can bridge R1 with the 200K pot without removing it, which will give you less chance of damaging the board.
Since it's a boost regulator, you can't get any lower than the input voltage. it won't do any good to short out R1. I believe it will just never turn on the transistor switch and you'll just get the input voltage passed through to the output through the inductor.
And as I said above, don't short out R2 because that would make it try to drive the output too high.
Here's the boost configuration schematic. As you decrease R1 in the feedback (not pictured), the boost control circuit decreases the boost "charging" of L by turning on the transistor switch for smaller and smaller amounts of time (lowering duty cycle). When you get R1 low enough, the switch will just stop switching at all and the input voltage will pass through L and D to the output.
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