Hi,

This one has me stumped and no doubt it’s due to my lack of knowledge about transistors.

**Background**: I bought a pager motor with a maximum rated voltage of 4 V (and 80 mA max current) which I intend to drive at 3.3 V. I already have an L239D dual H-bridge, but the supply voltage needs to be greater than 4.5 V and the motor supply voltage cannot be smaller than the logic voltage. The voltage sources at my disposal are 12 V and 5 V.

**Goal**: Use the Arduino’s 5 V digital output to switch 3.3 V supply for a motor.

**Stretch goal**: Use this solution to build my own H bridge from NPN and/or PNP transistors and diodes.

**Restrictions**: I want to do this using only one transistor, either NPN or PNP. I have resistors and other common components; I do not want to solve this by buying a different type of transistor, a relay, an H bridge IC or resorting to a fundamentally different approach. Perhaps someone will show me why one transistor is not really a good solution, and I’ll face the fact that I need to look for another way of turning on the motor. If that is the case, I’ll almost certainly look for an H bridge that can have a motor supply down to 3.3 V, and I’ll use a stiff voltage divider to bring the Arduino’s 5 V digital out to 3.3 V for the H bridge IC.

My attempt at a solution (on paper): I have drawn a schematic and attached the photo. Please suggest a better circuit if this one is off the mark. The idea behind this attempt is that if R_{1} is close to R_{2} and they are both large compared to R_{b}, I should get around 2.5 V in the middle of the divider when the Arduino’s output is high. I can then pick R_{b} to get enough current for saturation (say 40 mA for safety)… and that’s where I grind to a halt, because I feel I’m overlooking something.

Going back to basics, I have identified the following 5 equations at my disposal:

V_{1} = R_{1} i_{1}

V_{2} = R_{2} i_{2}

V_{2} = R_{b} i_{b} + 0.6 V

V_{in} = V_{1} + V_{2}

i_{1} = i_{2} + i_{b}

I consider I_{b} and V_{in} to be known quantities. Fiddling with these 5 equations for 7 unknowns gives me:

V_{in} (R_{2}/R_{1}) + I_{b}(R_{b} - R_{2}) - 0.6 V = 0

Have I missed an equation? Hm. Anyway, I am willing to use R_{1} ~ R_{2} = R to get closer to determining a value for R_{b}. To wit:

R_{b} = (V_{in} - 0.6 V - I_{b} R)/I_{b}

However, there is one thing that I haven’t used. The quantity in the numerator above must be positive, since R_{b} must be positive. This leads to:

R < (V_{in} - 0.6 V)/I_{b}

or

R < (5 V - 0.6 V)/(40 mA)

< 110 Ohm.

With R = 110 Ohm I find that R_{b} = 0. (If R is smaller, R_{b} > 0). Does this really mean that I can just make a 110/110 Ohm voltage divider, connect the base to the middle, and get 40 mA flowing into the base? This feels wrong.

And again, I feel that I am missing something. Could someone please point out the flaws in my reasoning or analysis? This is probably a very simple task to solve, but I lack the experience to know what design to choose. Thank you in advance.