Using a 2N7000 to light a LED

I plan on using a 2N7000 (as I have plenty of them) to power the LED's from the outputs of a 74HC595N.

My understanding of this is the gate can connect to the 595 with no resisters but how do I work out the resistor value in series with the LED's. Im using 2 LED's in series, each LED is 3.3V 20mA with a supply of 12V.

I would have gone for a 270 Ohm if I wasn't using the FET but wasn't sure on the effects of the FET in the circuit as not used FET's before so thought Id ask here first.

You should be posting this in the electronics section.

The FET has no voltage drop as far as I know, but with 12V I would use a larger resistor. (470 Ohm for example).

The FET has an Rds of ~3 ohm with Vgs of 5V and 50mA current draw. You want typically a max of 20mA thru an LED, so with same Rds the voltage Vds will be 0.02 * 3 = 0.06V.

So: (12V - 3.3V - 0.06V )/.02A = R of 432 ohm. 470 ohm is standard value I think and will work well.

So if I'm using 2 LED's in series with each FET, each 3.3V 20mA would that work out to 270 ohms. Thanks for the .pdf CrossRoads.

CrossRoads: The FET has an Rds of ~3 ohm with Vgs of 5V and 50mA current draw.

I must be thick because I can't see how you find/decipher this information from the .pdf

Page 2
On characteristics (See Note 1)
RDS(ON), Static Drain-Source On-Resistance -
VGS = 5.0 V, ID = 50 mA, TJ =125°C, Typical Rds = 2.8 to Max 5 ohm.


  1. Pulse Test: Pulse Width < 300µs, Duty Cycle < 2.0%.

So figuring more steady state operation vs pulse, temps could be elevated, I used the higher resistance in the table.

Could be lower if you look at Figure 2.
The 0.06V in the calculation would drop then, but has marginal impact, and 270 would work well as you figured:
You might want to connect a couple up and see what the Vf of the LEDs actually is and adjust your R from there before soldering very much up.
(12 - 3.3 -3.3)/.02 = 270.