Raspberry Pi Pico GPIO amplification

Question

I'm trying to control a handful of LEDs via Raspberry Pi Pico, on a single GPIO pin.

I've wired up 14 LEDs in parallel, each with it's own resistor (330 Ohm, 0.5 Watt). Combined, the LEDs need to pull 420mA, with input of 3.3V and a forward voltage drop of 2.65V.

I want to use only [email protected] (no greater than this) from a single GPIO pin for a control (base) input to a transistor, with the power source (collector) connected to +5V. These requirements are non-negotiable, but I have no idea what other components I will require to make this happen (I'm trying to learn how to pick the correct, presumably NPN, transistor and appropriate resistor for between the power source and the collector).

I don't care about, or require, filtering as the load being driven is LEDs.

I'd appreciate either a specific transistor/resistor pair recommendation, or at least a concise explanation as to how to determine these parts. I'm okay at reading data sheets, but there's more to transistors than my understanding at this time.

link to the LED part from digikey: https://www.digikey.com/en/products/detail/kingbright/WP710A10LVBC-D/5177415

Answer 1

With collector tied to any positive supply (over 3.3V, such as 5V or 12V even), and with load at the emitter, that configuration is an emitter follower, and the output potential \$V_{OUT}\$ (at the emitter) will be \$V_{OUT} = V_{IN}-0.7V\$.

This means that your GPIO potential of 3.3V can never produce more than \$V_{OUT} = +3.3V - 0.7V = 2.6V\$. That is also non-negotiable, and this configuration cannot possibly work for your LEDs, regardless of what resistor you use.

You must use common emitter, with emitter tied to ground.

With that configuration, and a saturated transistor dropping \$V_{CE}=0.2V\$ (worst case), each resistor/LED has \$5V - 0.2V = 4.8V\$ across it. Using 330Ω resistors and 2.65V LEDs results in a collector current (for each) of \$I=\frac{4.8V-2.65V}{330\Omega}=6.5mA\$

The collector current for 14 LEDs will be \$14 \times 6.5mA \approx 90mA\$.

Any common signal transistor should do, but just to be sure we should consider current gain \$\beta\$. For a GPIO sourcing at most 3mA, you require a transistor with current gain \$\beta = \frac{I_C}{I_B} = \frac{90mA}{3mA} = 30\$ (or more).

A signal transistor's current gain \$\beta\$ is typically over 100, but we can assume a worst case of \$\beta=50\$. The minimum base current required to saturate the transistor will be \$I_B=\frac{I_C}{\beta}=\frac{90mA}{50}=1.8mA\$. That's less than your constraint of 3mA maximum, so we're good.

Also, for completeness, check the transistor's power dissipation, just in case: \$P=I\times V = 90mA \times 0.2V = 18mW\$, which is well within any transistor's capability, without heat-sinking.

Since the transistor's base will be clamped to 0.7V maximum, you can't tie the GPIO directly to it. When the GPIO output is high, there will be a difference of \$3.3V-0.7V= 2.6V\$ to span with a resistor, which should pass 1.8mA. That's a resistance of \$R=\frac{V}{I}=\frac{2.6V}{1.8mA}=1.4k\Omega\$. Erring on the side of caution (slightly more base current is OK), round down to the nearest E12 value, for a base resistance of 1.2kΩ. You could round up to 1.5kΩ, it's very unlikely that the transistor will have a gain of less than 100, so it's pretty safe to work with slightly less base current too.

All that implemented:

^{simulate this circuit – Schematic created using CircuitLab}