Electronic – Powering a 14v incandescent bulb with Raspberry Pi & Transistors

You've chosen a good incandescent bulb with a relatively long life. Here's a datasheet I found at Farnell: Chicago Miniature Lighting. There, find the CM73 entry. You can infer from the data that it is operated at a lower filament temperature than most of the similar choices at \$14\:\text{V}\$. (MSCP relates to the "candle power" and it's less than most of the choices there and the expected lifetime is correspondingly higher, as well.)

The above might be important to consider because of the following information on Tungsten: Tungsten Relative Resistivity vs Temperature. Without knowing a lot about incandescents (I'm no expert), I do know that the filaments are typically operated at or below \$2900\:\text{K}\$ (sometimes as high as \$3300\:\text{K}\$ for photography lighting reasons related to the blackbody radiative distribution of wavelengths.) If I had to guess (and I do), your bulb is probably operated at or below \$2400\:\text{K}\$. If you look at the chart, you'll see that this implies about a 12:1 ratio. This suggests that the bulb will have about \$\frac1{12}\$\$^\text{th}\$ the resistance when cold, as when it reaches a stable and hot temperature.

The reason I said this might be important is that there are times when it's helpful to know these details when designing a circuit. Having to deal with such a wide range of operating currents can mean the difference of one approach over another. In the case of your BJT switch, the collector current is limited (to first approximation) by the base recombination current and the BJT's unique \$\beta\$ value. So there is a kind of self-limiting taking place with the BJT switch, though just how much limit there is will depend on the individual BJT itself.

An easy way to insert some negative feedback (opposition) into the circuit is to add an emitter resistor. If the current pulses really high, then the voltage drop across the emitter resistor will act to reduce the base-emitter voltage and that will reduce the collector current in response. Then, as the current requirements decline the resistor will have a smaller voltage drop across it, allowing the BJT to act more like a switch and supply the desired collector current. It's cheap and easy.

With perhaps 12:1 current variation as it warms up (assuming no managed limits were added to your circuit), adding this emitter resistor can do a good job of limiting the cold-start current without limiting the warm operating current.

So let's look at a circuit:

^{simulate this circuit – Schematic created using CircuitLab}

There are other considerations to get to (mostly about dissipation.) But let me walk you through my thinking process here:

1. I know that \$80\:\text{mA}\$ (operating at temperature) is a collector current that most small signal BJTs can handle, continuously. So this opens the door to a smaller BJT, which is nice.
2. For selecting an emitter resistor, I want one that won't drop too much voltage when running normally with \$80\:\text{mA}\$ but I also want one that will drop enough voltage at significantly higher currents that the BJT will be almost off. A small signal BJT in this context is almost off when the base-emitter voltage is about \$600\:\text{mV}\$ (though with larger BJTs that might be closer to \$500\:\text{mV}\$.) So let's use the smaller figure of \$500\:\text{mV}\$ for off.
3. The base recombination current is usually chosen to be \$\frac{1}{10}\$\$^\text{th}\$ the collector current so that the BJT acts very close to a switch behavior. (\$V_\text{CE}\le 200\:\text{mV}\$, not uncommonly.) You can almost always get by with less than that -- though again with larger BJTs designed for more current the situation is a little different. Given that you are using a Raspberry PI, it can probably handle the implied \$8\:\text{mA}\$. But there are port limitations, too, and I think we should design this for a base recombination current of \$5\:\text{mA}\$ to avoid other complications. It's likely to be sufficient because it uses \$\beta=\frac{80\:\text{mA}}{5\:\text{mA}}=16\$. Experience tells me this is probably fine.
4. At \$3.3\:\text{V}\$ on your I/O pin and an expected source current of \$5\:\text{mA}\$, I would expect to see something like \$3.3\:\text{V}-5\:\text{mA}\cdot 70\:\Omega\approx 3\:\text{V}\$ at the loaded I/O pin, itself.
5. I'd like to see the emitter resistor drop no more than about \$200\:\text{mV}\$ at an operating \$80\:\text{mA}\$ for the bulb. So this suggests \$\frac{200\:\text{mV}}{80\:\text{mA}}=2.5\:\Omega\$. Let's use a nearby (and slightly lower) \$R_1=2.2\:\Omega\$.
6. The base resistor should be \$R_2=\frac{3\:\text{V}-900\:\text{mV}-80\:\text{mA}\cdot 2.2\:\Omega}{5\:\text{mA}}\approx 390\:\Omega\$. The \$3\:\text{V}\$ comes from the calculation at step 4. The \$900\:\text{mV}\$ comes from my experience guessing about the base-emitter voltage required for a collector current of \$80\:\text{mA}\$ in small signal BJTs.

The remaining question is what happens when the bulb is cold and is it worth the bother putting that emitter resistor in place. The base recombination current as a function of lamp current is probably the best way to examine this circuit. I'll use a very simple, fixed value for the base-emitter voltage of \$V_\text{BE}\approx 700\:\text{mV}\$ as a broad estimate that covers a lot. (The base-emitter voltage will change by about \$60\:\text{mV}\$ for each 10-fold change in collector current and the 12:1 ratio for the lamp suggests that in the worst case we'd see perhaps \$60-70\:\text{mV}\$ variation of the base-emitter voltage over the entire range of currents possible for the lamp. So we can ignore it, as it's small.)

The formula for the base current can then be approximated with \$I_\text{B}=\frac{3\:\text{V}-700\:\text{mV}-I_\text{LAMP}\cdot 2.2\:\Omega}{390\:\Omega}\$ and assuming that \$I_\text{E}\approx I_\text{C}\$ for simplification purposes, we find that the transistor must provide \$\beta=\frac{I_\text{LAMP}\cdot 390\:\Omega}{2.3\:\text{V}-I_\text{LAMP}\cdot 2.2\:\Omega}\$. For a switch, it's almost certain that this value must be at or below about 50. (That's if you are lucky. It's likely even less than that.) This turns into \$I_\text{LAMP}=\frac{2.3\:\text{V}}{2.2\:\Omega+\frac{390\:\Omega}{\beta}}\$. Estimating a saturated \$\beta=50\$ gives \$I_\text{LAMP}\le 230\:\text{mA}\$. Estimating a saturated \$\beta=30\$ gives \$I_\text{LAMP}\le 150\:\text{mA}\$. And even if you happened to get a saturated BJT with an unlikely high \$\beta=100\$, that would only give \$I_\text{LAMP}\le 375\:\text{mA}\$.

So as you can see this does seem to perform nicely. It limits the cold start current while still allowing normal operation once the bulb has fully warmed. The slight downside is that this adds a little overhead to the BJT switch, leaving just a slightly smaller voltage left over for the lamp. I don't believe the loss of a few hundred millivolts is enough to be any problem with a \$14\:\text{V}\$ bulb, though. So I'd give that a go and see.