Working on a project for a friend that involves remotely illuminating three small LED strips (I’d guess about 80mA each, but final lights have not been chosen). The whole controller and power source needs to be battery powered and carried on a person (i.e. in a pocket). The LEDs will be used very infrequently for short periods of time, so battery size and life shouldn’t be an issue
I chose the Moteino (very small, low power with built in wireless) and I’ve got the comms and the software working. I’ve read a great deal about swtiching high power loads and seen lots of Darlngton arrays, TIP120, etc. but nearly all examples are with 5v and hardwired source. I read someone was having issues when using the TP120 with a 3.3v battery powerwed source because of a 1.7v voltage drop (were very few details).
So, I’m lookign for a suitable to switch the LED’s. The idea is that I’ll have a 6v/12v battery source (or whatever is required for the LEDs). Drop that to 3.3v for the moteino and drive the LEDs being switched on digital ouptut pins. Anyone got any suggestions?
One issue with a 3.3V source is the Vforward of LEDS. Typical values are:
red ~ 1.4V
green ~ 1.5 - 1.8V
White (and any that use phosphorescence) ~ 3-4V
It is those that require a higher Vf that are the problem.
If you use an Darlinton BJT then voltage drop from collector to emitter will be over 1V and can be 2V. Then there is not enough Voltage across the LED to turn it on. Look up the data sheet for the TIP120 and check the Voltage drop at the current you need.
A MOSFET is a better choice since many can have very low Voltage drops but 3.3V still may not be enough the turn on some LEDs.
Also, look up the data sheet for the Vf spec of the LEDs you want to use.
Now, if you really need to use 3.3V then get a DC-DC boost converted with a 5V output.
Also, there are some LED driver chips made to solve to issue you have.
Check Linear Tech’s web site for possible LEDs drive solutions.
waltr:
One issue with a 3.3V source is the Vforward of LEDS. Typical values are:
red ~ 1.4V
green ~ 1.5 - 1.8V
White (and any that use phosphorescence) ~ 3-4V
It is those that require a higher Vf that are the problem.
If you use an Darlinton BJT then voltage drop from collector to emitter will be over 1V and can be 2V. Then there is not enough Voltage across the LED to turn it on. Look up the data sheet for the TIP120 and check the Voltage drop at the current you need.
A MOSFET is a better choice since many can have very low Voltage drops but 3.3V still may not be enough the turn on some LEDs.
Also, look up the data sheet for the Vf spec of the LEDs you want to use.
Now, if you really need to use 3.3V then get a DC-DC boost converted with a 5V output.
Also, there are some LED driver chips made to solve to issue you have.
Check Linear Tech’s web site for possible LEDs drive solutions.
Hi Waltr
Thanks for the response. I’m not whiz at electronics, so maybe I misunderstood something. I’m not sure the relavence of the vForward of the LEDs and the 3.3v. I thought the way the circuit would work is that the power supply would drive the LEDS (enough V and A) and the digital output on the 3.3v would just need enough current to cause the switching transisitor to switch the main supply to the LEDS
LEDs are diodes and have a minimum forward voltage (Vf) to start conducting. Each type (color) of LED has a different Vf. Then it is the current through the LED that determines their brightness.
For a simple LED circuit one needs a Voltage source (power supply or battery) and a current limiting (setting) resistor.
So for a typical low cost Red LED that has a Vf of 1.4V and using a voltage source of 3.3V.
Lets set the current to 10mA (about half of typical max current).
Th LED drops 1.4V so awe need to drop 3.3 - 1.4V across the resistor with 10mA flowing through the resistor and LED.
So (3.3-1.4)/0.01 = 190 Ohm (Ohms Law). The LED lights and all is ok.
Now lets use a White LED with a Vf of 4V (same source).
Well even without a resistor the LED never starts to conduct, 3.3V < 4V so the LED will never light.
We can run into the same problem if we try to switch on the Red LED with a Darlington BJT.
Assume a 1.7 V drop on the BJT. So 3.3V - 1.7V = 1.6V which is just over the LEDs conduction voltage.
The resistor would be (1.6-1.4)/0.01 = 20 Ohm.
Hope that helps. Now go google LEDs to look at data sheets (the Vf spec) and look at drive circuits.