# What do I need to make this circuit work?

I’m trying to light two 2.2V 20mA LEDs using a 9 volt battery, a PIR motion sensor and a transistor.

Wired as shown with 9 volts applied at the collector and the 3.3 volt signal from the PIR applied at the base of the transistor, the output is only about 2.6 volts which will only light one LED.

My idiot brain thought the PIR would output a signal when motion is present, then the transistor would act as a switch and I would get approximately 9 volts at the emitter. I could then use a resistor to knock it down to about 4 volts to light the two LEDs.

Apparently there is a large voltage drop at the transistor resulting in a 2.6 volt output.

• Does anyone know why I only get 2.6 volts coming out of the transistor?
• Any idea how I can correct this to get what I want?

The emitter voltage is 0.7V below the base voltage.

So $$\3.3-0.7=2.6V\$$. Place the diodes in the collector and a resistor in the emitter as shown. Select the resistor for a current that will give the desired brightness.

For example if 1 mA is desired then $$\R_1=2.6/.001=2.6k\Omega\$$. The nearest 5% value is $$\2.7k\Omega\$$.

simulate this circuit – Schematic created using CircuitLab

• Most LEDs will not be very bright with only 1ma of current.
– Nedd
Commented Jan 15, 2023 at 10:11
• RussellH, Thanks but can you clarify a couple of things for my pea brain? First my LEDs are 20mA so am I correct in calculating a 130 Ohm resister would work? Also having the LEDs on the collector side would the 9 Volts not fry the 2.2 Volt LEDs? Commented Jan 15, 2023 at 14:54
• GodJihyo, Thanks but how do I calculate what resistor to add between the motion sensor and base? Commented Jan 15, 2023 at 14:56
• @GodJihyo: The base current is fixed by the collector current and the current gain. Adding a base resistor will not change the base current for a given emitter voltage. Commented Jan 15, 2023 at 15:12
• @SamSpade: The collector appears as a current source that supplies the desired current. In this case the LEDs use only the voltage that they require. The rest is dropped across the transistor and emitter resistor. Commented Jan 15, 2023 at 15:19