# Understanding voltage drop on a transistor

I have a circuit like this:

(Imagine the base goes to a microcontroller and is connected properly.)

After the red LEDs, the voltage will drop to 3.2V.

Will the transistor (2N2222) drop any more voltage?

If this circuit goes to ground (remove the transistor) I've calculated about 15mA goes to each LED. If there was more voltage drop, would I need to reduce the resistance?

If the transistor base current is adequate the transistor will be driven into saturation and there will be about 0.2 V between the collector and emitter. This is small enough that it can be ignored in many calculations such as this.

That leaves about 3 V across the resistor so I = 3/75 = 40 mA shared between the three LEDs. Generally direct paralleling of the LEDs is avoided because the currents will vary quite a lot depending on their individual forward voltages, Vf.

See what I've written in Variations in Vf and binning for more on the topic.

• Is 0.2V at full saturation a standard for all transistors? (similar to forward voltage of leds) – user248702 Jan 24 at 16:06
• @Niall895 It depends on the transistor and on the collector current. It's usually around 0.1 or 0.2 volts for a silicon bjt, but it's typically much higher for a Darlington pair. – Hearth Jan 24 at 16:07
• @Niall895 This is one of the metrics you can find in a transistor's datasheet; it's referred to as the saturation voltage, symbol $V_{CE,sat}$. – Hearth Jan 24 at 16:10
• @Transistor, your " LED binning" and related articles are very educational to my current Micky Mouse toy project messing around with RGBYW 1W power LEDs for smarting lighting my home. It was only a week ago when plotting the I-V curves of colour LEDs then I surprisingly discovered why I cannot easily balance the RGB lighting. Your suggestion on CCS (Constant Current Source) and examples helps me greatly getting started. Actually I am going to dip my toes wet on CCS this morning! Many thanks. Cheers. – tlfong01 Jan 25 at 0:45

Does it matter at all if the LEDs get less than 15mA?

The 2N2222 has a saturation voltage of 0.3 V at 150mA. It may be less at 15mA.

That means there will always be a voltage drop across the transistor.

Taking the typical forward voltage of 1.7V for a red LED, a 5V supply, and a 75 ohm resistor, I get a current of 44 mA. ($$\I_{C} = \frac {5V-1.7V} {75 ohms} = 44mA \$$ - that's 14.67mA per LED.)

Now assuming 0.3V across the collector-emitter junction, and using the rest of the numbers from my example, I get 40 mA. ($$\I_{C} = \frac {5V-1.7V-0.3V} {75 ohms} = 40mA \$$ - that's 13.3mA per LED.)

That's a difference of just over 1mA for each LED.

Does that difference matter? No, not for a typical LED indicator circuit.

It especially doesn't matter for your circuit.

You would normally use one resistor for each LED. Using one resistor will potentially allow one LED to "snarf up all the juice" - one LED may light and the others not, or they could all receive different amounts of current.

LEDs aren't identical (though they can be very close to identical if they come from the same production batch.) Small differences can cause the LEDs to behave just a bit differently. If you are lucky, your LEDs will be a little differently bright. If you are unlucky, one will light up (and maybe burn out) and the others will be dark.

It therefore really doesn't matter much if you get a milliampere more or less - the LEDs won't be divying it up equally, anyway.

You should do something like this:

simulate this circuit – Schematic created using CircuitLab

Each LED has its own resistor, and each gets about $$\I_{C} = \frac {5V-1.7V-0.3V} {225 ohms} = 13.3mA \$$ - and each resistor will get its 13mA, pretty much regardless of what the other LEDs do.

• Thank you for that explanation, to clarify: my main concern wasn't for losing a milliamp or two but if the voltage drop on Vce was more than a volt, which would mean each led is getting <6mA each. – user248702 Jan 24 at 16:46