IC current drive interface: BJT or MOSFET? - Electronic Circuits Projects Diagrams Free

solis365 · 20th March 2009, 02:04 AM

Hi, I have made a few posts on here about a project I'm working on where I am PWMing some high-power LEDs. (The thread is here). Since the LEDs are high power (350-700mA), I needed some sort of way to drive them from a microcontroller PWM output pin. As seen in the simple schematic I supplied on that thread, I had planned on using MOSFETs.

However, now I am beginning to think that BJT's may be easier to implement. Here is the plan:

NPNs will take the place of the FETs in the schematic; however the emitters will go directly to ground. Given that base-emitter voltage (Vbe) varies very little, you can approximate it to be constant (lets say 0.7V). Thus, your base voltage Vb will be 0.7V if the emitter goes to ground. If the IC pin's logical high is 5V, then you will need to place a series resistor (Rb) between it and the base of the NPN that will drop 4.7V. This resistor will limit the base current (Ib) by ohms law:

Ib = (5V - Vb) / Rb

Since the Emitter current (Ie) of a BJT is β*Ib (or hfe * Ib) and collector current (the current drawn through the LEDs) is

Ie - Ib = Ic

So by using Rb to set Ib, Ie = β*Ib , --> Ic = (β - 1)*Ib

By selecting NPNs with a decent β, and then selecting the resistors, I can just control the amount of current I pull through the LEDs.

I have access to large amounts of NPNs so I can just go through and test them and select ones with similar β-values.
Seems easier to work with than FETs, or is it just me?
The fact that Vbe remains almost constant makes it a lot easier to work with.

Any opinions?
And thanks for all the help

audioguru · 20th March 2009, 03:16 AM

The beta of a transistor applies only when it has plenty of collector to emitter voltage. All transistors have a "max saturation voltage" spec when the base current is 1/10th the collector current.

You cannot use the beta to control the conduction of transistors because the beta increases as the transistor heats.

The max allowed output from a microcontroller pin is 25mA. Then a transistor will saturate properly when the collector current is a max of only 250mA.
You can test many transistors to find a few that saturate well with a higher collector current, or use darlington transistors or Mosfets.

You need a current-limiting resistor in series with the LEDs.

Thread Tools
Show Printable Version Email this Page
Display Modes
Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

Similar
Title	Starter	Forum	Replies	Latest
sink, source current and drive current	ervinzhou	General Electronics Chat	0	4th October 2008 01:52 PM
BJT and MOSFET- Switching and Drive	dknguyen	General Electronics Chat	12	21st January 2008 10:09 AM
Mosfet gate drive	wael_sal	Electronic Projects Design/Ideas/Reviews	1	2nd November 2006 09:59 PM
IDE drive interface USB	cwtech	Electronic Projects Design/Ideas/Reviews	2	12th January 2004 07:22 AM
Power Mosfet drive techniques	Optikon	General Electronics Chat	8	21st November 2003 12:50 AM

20th March 2009, 02:04 AM	#1
solis365	IC current drive interface: BJT or MOSFET? Hi, I have made a few posts on here about a project I'm working on where I am PWMing some high-power LEDs. (The thread is here). Since the LEDs are high power (350-700mA), I needed some sort of way to drive them from a microcontroller PWM output pin. As seen in the simple schematic I supplied on that thread, I had planned on using MOSFETs. However, now I am beginning to think that BJT's may be easier to implement. Here is the plan: NPNs will take the place of the FETs in the schematic; however the emitters will go directly to ground. Given that base-emitter voltage (Vbe) varies very little, you can approximate it to be constant (lets say 0.7V). Thus, your base voltage Vb will be 0.7V if the emitter goes to ground. If the IC pin's logical high is 5V, then you will need to place a series resistor (Rb) between it and the base of the NPN that will drop 4.7V. This resistor will limit the base current (Ib) by ohms law: Ib = (5V - Vb) / Rb Since the Emitter current (Ie) of a BJT is βIb (or hfe Ib) and collector current (the current drawn through the LEDs) is Ie - Ib = Ic So by using Rb to set Ib, Ie = βIb , --> Ic = (β - 1)Ib By selecting NPNs with a decent β, and then selecting the resistors, I can just control the amount of current I pull through the LEDs. I have access to large amounts of NPNs so I can just go through and test them and select ones with similar β-values. Seems easier to work with than FETs, or is it just me? The fact that Vbe remains almost constant makes it a lot easier to work with. Any opinions? And thanks for all the help

20th March 2009, 03:16 AM	#2
audioguru	The beta of a transistor applies only when it has plenty of collector to emitter voltage. All transistors have a "max saturation voltage" spec when the base current is 1/10th the collector current. You cannot use the beta to control the conduction of transistors because the beta increases as the transistor heats. The max allowed output from a microcontroller pin is 25mA. Then a transistor will saturate properly when the collector current is a max of only 250mA. You can test many transistors to find a few that saturate well with a higher collector current, or use darlington transistors or Mosfets. You need a current-limiting resistor in series with the LEDs. __________________ Uncle $crooge