DC step-up regulator wiring

Question

I'm using a pololu step-up voltage regulator (http://www.pololu.com/catalog/product/2117) to go from an unregulated battery (VIN line from an Arduino, powered by a 3.7v LiPo cell) to 12V.

No matter what I do, I can't get a voltage to show across the VOUT. There's definitely a voltage applied to VIN, and it's within spec (above the 2.5V minimum for the regulator). The VOUT line just falls rapidly to 0 when I put a meter on it.

Seems like a simple enough circuit. What could I be doing wrong?

EDIT: The circuit below is for the final application. I tried simply supplying 5V on the VIN line to the regulator, and measuring voltage between VOUT and GND, with the same 0 value.

^{simulate this circuit – Schematic created using CircuitLab}

Here's another way I tried to wire it, with the same effect

^{simulate this circuit}

Answer 1

The current out of that 754410 is probably not enough to drive the boost converter. I would build the same circuit like so:

^{simulate this circuit – Schematic created using CircuitLab}

And 100 uF may not be enough. You may need 1000 uF! If so, you also want to add a few Ohms of resistance between boost-out and capacitor-plus, to avoid putting too much capacitive load on the boost converter all at once.

Answer 2

Some observations:

1. The minimum voltage for the SN754410 (according to a TI datasheet) is 4.5V. You're using the driver outside its normal range if you're feeding it from a 3.7V LiPo.

2. (Gross simplification) Best-case output current for the 754410 is 1A. Assuming perfect 3.7V, 1A output from the driver (which it won't be), that's a theoretical maximum of 3.7W. Assuming your little booster is 80% efficient (which it won't be) that's 2.96W available for your load, or around 246mA for the solenoid. Again, this is assuming lots of ideals - reality will be worse. You didn't specify the DC resistance of the solenoid control winding so I cannot predict how much current it will draw. (\Gross simplification)

3. The regulator should have \$ 33 \mu F \$ of capacitance on its input per the web link you cited. It's not shown on your drawing.

4. You haven't mentioned the A-h capacity of your 3.7V LiPo. The Arduino will be eating some of that energy too. A heavy load will likely discharge it fairly quickly - if the boost continues to receive input, it will try and draw more and more current to keep the output regulated (a switching converter shows a negative impedance load to the source.)

Answer 3

Power out is always less than power in. Stepping voltage up means you draw a much bigger current at input (in the same ratio).

A Solenoid (even a small one) takes a lot of current. The output from the half H drive is unlikely to be 3V3. To get 1.4A max current out you'll need to put in at least 5 times that. At 3V that's a whopping 7A, I think that rather exceeds the current ratings of the bridge, supply and the regulator.

With any luck your regulator has just shut down until you treat it with more respect.