Electrical – Choosing a transistor for the Raspberry Pi 5v fan

I thought your question was pretty well-framed. You have PNP BJTs, it seems, but no NPNs? Not so good. But it can be done. At least, I think so. The main worry I have is that a single I/O pin might have to sink some significant current. But I think the Pi can safely handle perhaps \$15\:\text{mA}\$ on a single I/O pin. So perhaps it is okay.

Here's the simple circuit to try:

^{simulate this circuit – Schematic created using CircuitLab}

The basic idea is:

1. When the I/O pin is LOW, it can sink enough current to include the needed base current for \$Q_1\$ (plus whatever \$R_1\$ is supplying, then.)
2. When the I/O pin is HIGH, it still sinks some current but only the current needed for \$R_1\$ because there won't be enough voltage for the base-emitter of \$Q_1\$ to activate it.

In general, if the base-emitter voltage falls below about \$V_{\text{BE}_\text{OFF}}\le 500\:\text{mV}\$, \$Q_1\$ will be turned off. So that's the goal for turning off the load. And for the \$120\:\text{mA}\$ you are considering for \$Q_1\$ when turned on, I'd want a base current of at least \$I_{\text{B}_\text{ON}}=6\:\text{mA}\$ and also a base-emitter voltage that is at least \$V_{\text{BE}_\text{ON}}\ge 800\:\text{mV}\$.

(Now, this is a bit of a compromise. I'd prefer still more base current. But, unfortunately, I'm also trapped by some limits for your I/O pin. So this is one of those compromise decisions that are sometimes required to balance competing issues.)

There is another worry to deal with. The I/O pin itself has some resistance. I find that assuming \$100\:\Omega\$ is conservative enough for most circumstances. (You could measure it, too.) Often, it's lower than that, but sometimes it may be a little higher. So that's the number I'm going with.

This provides two equations and two unknowns (the resistors.)

$$\begin{align*} \left(I_{\text{B}_\text{ON}}+\frac{5.0\:\text{V}}{R_1}\right)\cdot\left(R_1\:\vert\vert \:R_2\right)&\le 5.0\:\text{V}-V_{\text{BE}_\text{ON}}\\\\ \frac{3.3\:\text{V}\cdot R_1+5.0\:\text{V}\cdot R_2}{R_1+R_2}&\ge 5.0\:\text{V}-V_{\text{BE}_\text{OFF}} \end{align*}$$

Once I solve for \$R_2\$, I then have to subtract off the I/O pin resistance. So from that fact, I arrived at the given resistor values above. The peak current required from your I/O pin is about \$11\:\text{mA}\$ and all the power dissipation figures work out to reasonable values. So it probably will be fine, as shown.

In case it helps, here are the resulting equations. I'll use \$V_\text{CC}=3.3\:\text{V}\$ and \$V_\text{DD}=5.0\:\text{V}\$ for these purposes. Then, highlighting the common factor on the left:

$$\begin{align*} R_1&=\left\{\frac{V_{\text{BE}_\text{ON}}}{I_{\text{B}_\text{ON}}}\left[V_\text{CC}-V_\text{DD}\left(1-\frac{V_{\text{BE}_\text{OFF}}}{V_{\text{BE}_\text{ON}}}\right)\right]\right\}\cdot\frac{1}{V_\text{DD}-V_\text{CC}-V_{\text{BE}_\text{OFF}}}\\\\ R_2&=\left\{\frac{V_{\text{BE}_\text{ON}}}{I_{\text{B}_\text{ON}}}\left[V_\text{CC}-V_\text{DD}\left(1-\frac{V_{\text{BE}_\text{OFF}}}{V_{\text{BE}_\text{ON}}}\right)\right]\right\}\cdot\frac{1}{V_{\text{BE}_\text{OFF}}} \end{align*}$$

We could take this one more step and add another PNP BJT to the chain and operate it similarly, as above, except that the first BJT (nearest the I/O pin) would serve to short-out the following BJT to turn it off. And this would relieve the I/O pin of some of the drive current requirements above.

Something on the order of this:

^{simulate this circuit}

The earlier equations can be used to compute the values for \$R_3\$ and \$R_4\$ in this new circuit, where this time \$I_{\text{B}_\text{ON}}\approx 500\:\mu\text{A}\$, perhaps? (Because the load now is perhaps at worst under \$10\:\text{mA}\$.) \$R_2\$ then just has to be sufficient to turn \$Q_1\$ on and \$R_1\$ is more of a pull-up now and much less critical in value.

I/O pin current will now be a lot less. Probably under \$1\:\text{mA}\$.