Help-Transistor hFE Explanation

JDR04

Joined May 5, 2011
356
Hi All. I'm planning a small project using a BC337 transistor.

So my LED requires 130mA and I'm trying to calculate what BASE current I will need. As far as I understand it, required mA divided by tansistor hFE or gain should give me the required mA to switch transistor on.

But when I look at the datasheet for BC337 I see hFE all over the place. Which one do I use?

Have attached the datasheet.

Thanks Guys - JDR04

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tshuck

Joined Oct 18, 2012
3,527
For driving the transistor for use as a switch, assume a hFE of 10 , or so...

#12

Joined Nov 30, 2010
18,076
If you can't afford 13 ma in your drive side, use 2 transistors to make a Darlington or a double inverting circuit driven by 1.3 ma.

crutschow

Joined Mar 14, 2008
23,159
The hFE in the data sheet is for operating the transistor in the linear region, not when using it as a switch. For that you should use an hFE of 10 or so, as tshuck noted, to insure that the transistor is fully on.

JDR04

Joined May 5, 2011
356
Thanks Guys, so for most NPN transistors I can safely use 10 as the hFE.

Thanks a lot - John

alexfreed

Joined Oct 8, 2012
72
That would be a very conservative estimate. Most low power transistors are much better in this respect. In particular BC337 has a MINIMUM of 100 at Ic=100 mA.

Indeed saturation requires more base current, but that is for 500 mA, not 130.

INHO 50 would be a safe bet. Assuming you want to switch 130 mA, not to regulate it. If you want to use the transistor as a current regulator you should use a topology that does not rely on a particular value of hFE. BTW at 130 mA it will get hot. Much better to use power MOSFETs. For a few pennies more you get a lot less power converted into heat.

crutschow

Joined Mar 14, 2008
23,159
......................
INHO 50 would be a safe bet. Assuming you want to switch 130 mA, not to regulate it. If you want to use the transistor as a current regulator you should use a topology that does not rely on a particular value of hFE. BTW at 130 mA it will get hot. ...................
It may get hot as a linear regulator at 130mA but not as a switch if it's well saturated.

alexfreed

Joined Oct 8, 2012
72
The saturated voltage drop will be about 0.5V. At 130 mA that's 65 mW. Sure it won't burn anybody's fingers but warm it will be.

SgtWookie

Joined Jul 17, 2007
22,201
Here is ON Semi's datasheet for the BC337: http://www.onsemi.com/pub_link/Collateral/BC337-D.PDF
Look on page 3, figure 5, "On" Voltages, at the bottom plot; it's spec'ed with Ib=Ic/10; with Ic being 200mA, Vce will be less than 0.1V.

The BC337 is guaranteed a minimum hFE of 100 @ Ic=100mA, but that's with Vce=1v; certainly not for use as a saturated switch. We try to make certain that our members get their circuits working as easily as possible; "fudging" the numbers too much could be asking for trouble.

MikeML

Joined Oct 2, 2009
5,444
For driving a LED, the transistor should be operated as a linear constant current sink; not a switch, therefore not saturated. As such, it will be dissipating some power, so will need to be mounted on a heatsink.

Here is an example: suppose you have 0V and 5V from a PIC port pin. You have six high-power LEDs. You would like to drive 130mA through the LEDs from a nominal 24V supply when the port pin is at 5V; otherwise turn off the LEDs.

Here is how to do it using a BC337. Note the emitter resistor. It sets the LED current when the base is at ~5V. In the simulation, I vary the supply voltage V1 from 15V to 25V to see what effect that has on the LED current I(D1) Green trace. Once the supply voltage is >20V, it is a good current sink as shown by the nearly constant LED current.

Note the power dissipation in Q1 Red trace and in R1 Blue Trace. That is why Q1 should be on a heatsink.

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WBahn

Joined Mar 31, 2012
24,581
Why should it be operated as a constant current sink to drive an LED? The OP has given no indication that a specific constant current is of any importance all -- he just wants to turn on an LED. Use a transistor as a switch and a resistor to set the current. If something more sophisticated than that is needed, THEN use something more sophisticated.

MikeML

Joined Oct 2, 2009
5,444
Why should it be operated as a constant current sink to drive an LED? The OP has given no indication that a specific constant current is of any importance all -- he just wants to turn on an LED. ....
Nope, reread the OP's first post. He specifically says that he wants to operate his LED at 130mA.

alexfreed

Joined Oct 8, 2012
72
Look on page 3, figure 5, "On" Voltages, at the bottom plot; it's spec'ed with Ib=Ic/10; with Ic being 200mA, Vce will be less than 0.1V.
Interesting. I looked at the Fairchild spec that stated CE saturation voltage 0.7 V at 500 mA. So I extrapolated to 0.5 at 100 mA.

So I just tested a no-name "major brands" BC337. At 130 mA I got exactly 0.1 V CE drop while forcing 12 mA into the base.

At 2.2 mA base current I still got only 0.14 V CE.

#12

Joined Nov 30, 2010
18,076
A real measurement is worth a dozen theories.

WBahn

Joined Mar 31, 2012
24,581
Nope, reread the OP's first post. He specifically says that he wants to operate his LED at 130mA.
Well, okay. So when someone says that they have an LED that needs 10mA, we need to conclude that they need a constant current sink for it?

I could be wrong, but if someone gives a current for an LED without further stipulation, I figure that they are saying that this particular LED wants about that current to give off enough light. I suspect that if you were to search through all of the threads you would find (if you could track down the needed info) that that was what was meant for more often then that they need a controlled and regulated current of that amount.

But it's certainly a fair point to get clarification on. So, to the OP, what are your requirements. Do you need exactly 130mA (or within a tight tolerance of 130mA), or are you just trying to get something in the vicinity of 130mA through the LED to turn in on?

MikeML

Joined Oct 2, 2009
5,444
Its one thing to say "I want about 10mA to light my indicator LED", and another thing to say " I want to light a 130mA LED lamp".

If you use an NPN as a saturated switch with current-limiting resistor in series with the LED, you still need to dissipate exactly the same amount of power as in the circuit I posted with a resistor in the emitter of the transistor. The parts count is the same. My way does a much better job of regulating the current through the LED.

JDR04

Joined May 5, 2011
356
Thanks for the info so far guys. All I'm trying to do is switch an IR LED at 38Khz. I'm trying to use the IR LED as efficiently as possible in order to maximise the distance. Its going to be part of a IR break beam system.

Sorry for the lack of clarity on my behalf.

Thanks again - JDR04

MikeML

Joined Oct 2, 2009
5,444
Thanks for the info so far guys. All I'm trying to do is switch an IR LED at 38Khz. I'm trying to use the IR LED as efficiently as possible in order to maximise the distance. Its going to be part of a IR break beam system.

Sorry for the lack of clarity on my behalf.

Thanks again - JDR04
So how will you limit the current through the LED?
What supply voltage is available?
What is the forward voltage drop across the LED at 130mA?

WBahn

Joined Mar 31, 2012
24,581
Its one thing to say "I want about 10mA to light my indicator LED", and another thing to say " I want to light a 130mA LED lamp".

If you use an NPN as a saturated switch with current-limiting resistor in series with the LED, you still need to dissipate exactly the same amount of power as in the circuit I posted with a resistor in the emitter of the transistor. The parts count is the same. My way does a much better job of regulating the current through the LED.
Better performance almost always comes at a price. There are two practical differences for the current source relative to the switch configuration. First, you are dissipating more power in the transistor and less in the resistor, which will tend to raise the parts cost even though you are dissipating the same total power. Second, your control voltage must be higher that it otherwise could be. In any particular situation it's quite possible that neither of these matter and so use the current source (and you can even eliminate the base resistor). But in other situations one or both of those might be deciding factors.

JDR04

Joined May 5, 2011
356
My plan was to limit the current throught the IR LED with a resistor in series. The power supply will be a standard 9V battery dropped down to 5V. Forward voltage drop of LED is 1.3 to 1.5v.

Hope this helps and thanks. JDR04