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I'm having trouble with a circuit I designed and I'm hoping someone here may help me fix it.

I've successfully used the circuit below a few times. It was proposed by Russel McMahon in this answer. It let's the user power up a circuit by closing SW1 (Go!) long enough so that the MCU then can keep the circuit powered by keeping the signal HIGH_TO_RUN high. To turn it off, the MCU just have to bring the same signal low.

schematic

simulate this circuit – Schematic created using CircuitLab

This time I'm using two NiMH AA cells (~2.5V) to power a boost DC/DC converter based on the MC33063 IC which in turn brings the voltage up to ~5V and deliver about 125mA to an ATmega328P circuit. The circuit is the following.

enter image description here

The problem is that the minimum input voltage on my MC33063 is about 2.5V and Q1 is dropping it below that, to about 2.0V.

So, my questions are:

  1. How can I modify the circuit so that I can deliver ~200mA@5V to my load from the same two NiMH AA cells?

  2. Can I replace Q1 with an P-Channel MOSFET (such as the IRF9530) to decrease the voltage drop on the transistor? What changes to the circuit should be made then?

Note: The linked datasheet says that the minimum input voltage for the MC33063 is 3.0V, but I'm pretty sure there are variants out there that take less voltage (1.8, 2.0 and 2.5V if I'm not mistaken). I'm pretty sure mine is the 2.5V variant (at least I've tested it).

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Ricardo
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    The P-Channel MOSFET will work fine, but you need a MOSFET with a lower \$V_{GS}\$ turnon. The IRF9530 can have up to a 4 V threshold, but will probably require 4.5 V to get for a 200 mA load. You may want an n-channel FET as well to pull the gate down more. Searching Digikey for low \$V_{GS}\$ parts looks promising. – caveman Jan 18 '15 at 19:52
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    PMOS can work if you select the right part. I don't think you need to make any other changes. You can even keep the BC337, I believe (doesn't need to be changed to NMOS). But if the boost needs 2.5V, then I think your system design is kind of marginal. I would suggest you search for a boost that works down to a lower voltage, like maybe 1.8 or so. – user57037 Jan 18 '15 at 20:08
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    "I'm using two NiMH AA cells" + "minimum input voltage on my MC33063 is about 2.5V" = No way this is going to work. Maybe you'll be able to use 10% of the batteries capacity, but, depending on the batteries and the load current, voltage may as well break down as soon as you turn it on. – JimmyB Jan 19 '15 at 13:58
  • @HannoBinder - I guess you and mkeith are right - my design is indeed marginal. The voltage breaks down after I try to draw more than 50mA... I wasn't noticing it because the ATmega328P is pretty robust regarding these low voltages. I now checked and I have the 3V MC34063AP, but I've already ordered a few MC34063E that is supposed to take 2.1V at the input. In any case, your feedback is useful and appreciated. – Ricardo Jan 19 '15 at 14:30
  • I think I was just having a hard time sourcing the right ICs down here. Ordered a few LT1111, LT1302 and MAX756 to play with, so things should improve in the near future. – Ricardo Jan 19 '15 at 14:31
  • Adding another NiMH isn't an option? This would give more headroom for operation and at the same time can provide more energy, i.e. longer run time. – JimmyB Jan 19 '15 at 15:18
  • @Hanno - yes, the idea of adding a 3rd NiMH crossed my mind, but that would weaken the gadget appeal :D I actually wanted to use just one cell, but that's really out of my reach. The original gadgets had 4 NiMH and no voltage regulator, so 3 AA cells is not that different anyway. – Ricardo Jan 19 '15 at 15:33

1 Answers1

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I would replace Q1 by an enhancement PMOS with low gate threshold and low RDSON value.

Then the changes in the circuit would be to replace R6 by a short circuit as it would not be necessary anymore.

There are two ways of connecting the MOSFET, shown in the schematics below. Option A is the classic one, or the Option B. The Option B advantage is that it provides protection against reverse polarity of the battery. You can read an explanation of its working mechanism in http://www.ti.com/lit/an/slva139/slva139.pdf (figure 3).

The disadvantage of using Option B in your circuit is that even when the channel is OFF, there is a diode voltage drop between the input and the output. And therefore in the output you would have battery-0,7 V (a diode drop).

Therefore in your particular circuit you would need to use Option A.

schematic

simulate this circuit – Schematic created using CircuitLab

Roger C.
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