A dedicated switching IC for driving up to 5 LEDs is the Texas Instruments TPS61060/61/62 series of devices.

 

hi stryker,

So just to clarify, you have selected to send 1ma to the BASE which turns on. This NPN has a gain of 50 so the current going through the collector-emmiter will be 50ma. Is that correct? If it is, i have several questions stemming from this:
Its not the base current I have set at 1mA, its the collector current

- How do you know that 1ma will be going to the base? All we know is that the voltage drop from base-emmiter will be 0.7V and 1A goes through R2.
- If we know that 50ma is going through R1, we can calculate what the voltage drop would be and consequently what the new BASE FET voltage is correct? and we know what voltage we need to give a certain current from the FET collector-emmiter, and this way we can find R1 right?
There never could be 50mA flowing thru R1, use ohms law.

Also, one other thing that just popped in my head. I looked at the specs for the MOSFET, and if a 6+ volt is applied , the emmiter-collector current is 100 A?! If i had 3 leds and i need about 12V to supply them, and i need 1A (needs a 2V gate), where would the leftover 10V go??
The FET current can never get higher than 1Amp with the values shown.
Lookup negative feedback, from the NPN, the current in the FET is a Constant Current of 1Amp.


The voltage difference between the supply and the voltage drop across the LEDs [in SERIES] would be across the FET.

In a case like that the 10V at 1amp [considering just one LED] would generate a heating effect of 10Watts, you would need a heat sink on the FET.

You must NOT connect the LED's in parallel.

Are we there yet..

__________________
Eric
"Good enough is Perfect"

PIC tutorials:
Gramo's: www.digital-diy.net/
Bill's: www.blueroomelectronics.com/

 

So you selected the 1ma based on the BASE EMMITER ON voltage VS collector chart on the NPN specs sheet? Basically you can get the collector current based off the on voltage to the base... On the NPN i am looking at i would have about 10mA of current with the 0.7V base voltage.

Yes that makes sense, without the NPN negative feedback there the current would go to 100A, but as soon as it rises to 1A, the NPN turns on and brings the current to stabilize...


And to select R1. I look at the specs of the MOSFET such as the FQP50N06L. To get 1A drain current, the gate voltage must be ~2.4V . Knowing this, we want to select R1 to give that voltage to the FET. Using a power supply of 6V, we need a 6-2.4V = 3.6V drop. So R= V/I = 3.6/10mA = 360 ohm?

Am i getting it haha? I think im starting to understand it better lol.

 

For reference: this is the source which i got the circuit from

http://www.instructables.com/id/Powe...stant-current/

 

To elaborate on my previous comments about bipolar transistors: The base-emitter voltage is always around 0.6-0.7V (typical of a foward biased diode) when the transistor is on. It's the base current that determines the collector current. The base voltage will vary slight with base current of course (again similar to a diode) but it's the current that is the determining factor.

If you look at the gain curves for a transistor it will typically peak over some collector current range and droop at the high and low current levels. Typically you want to bias the transistor so its collector current is somewhere in the peak range of gain, in this case 1mA was selected.

In your calculation for R1 you used 10mA, it should be 1mA. Also the 2.4V gate voltage is actually gate-to-source voltage. Thus you need to add the 0.7V source voltage (across R2) to the 2.4V gate-source voltage to give 3.1V required at the gate (to common). Thus R1 = (6V-3.1V) / 1mA = 2.9k ohm.

 

Here is a spec sheet im looking at for the NPN:
http://www.fairchildsemi.com/ds/2N%2F2N4123.pdf

There is a graph called base emmiter on vs collector current. Is taht the one i look at? In order to get 1ma @ 25C i would need a ~.68V at the base correct? If i wanted 10ma, i would need ~.72V . So if i set my R2 resistor @ 1A to drop .6V, i should be getting 0.1mA would i not?

Yes you are correct about the 3.1V needed because of the 0.6V drop.

Also, is the method i chose before correct. By looking at the FET specs and looking what type of output current is given for a given gate voltage? On some of the FETS i was considering (power mosfets), the graphs they give do not typically show the region of 1A. If they do show it, it is at the leftmost part of the graph and its hard to read what type of Voltage i need. Do you have any suggestions on how to get an accurate voltage i need to apply to the gate for it to output 1A? Are there emperical equations that you can use to calculate it rather than trying to read it from a small graph?

 

 

Ok, this makes sense now and i finally know how to calculate the values i need now. Thank you everyone for all your help, i will post a topic once i order the parts i need and build a circuit!!!!