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 R1 is close to R2 and they are both large compared to Rb, I should get around 2.5 V in the middle of the divider when the Arduino’s output is high. I can then pick Rb 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:
V1 = R1 i1
V2 = R2 i2
V2 = Rb ib + 0.6 V
Vin = V1 + V2
i1 = i2 + ib
I consider Ib and Vin to be known quantities. Fiddling with these 5 equations for 7 unknowns gives me:
Vin (R2/R1) + Ib(Rb - R2) - 0.6 V = 0
Have I missed an equation? Hm. Anyway, I am willing to use R1 ~ R2 = R to get closer to determining a value for Rb. To wit:
Rb = (Vin - 0.6 V - Ib R)/Ib
However, there is one thing that I haven’t used. The quantity in the numerator above must be positive, since Rb must be positive. This leads to:
R < (Vin - 0.6 V)/Ib
R < (5 V - 0.6 V)/(40 mA)
< 110 Ohm.
With R = 110 Ohm I find that Rb = 0. (If R is smaller, Rb > 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.