Driving transistor from HC273

Do I need a resistor when driving a transistor from a HC273? It will
be on for 1ms and off for 10ms.

Transistor will be a 2N2222 type driving an LED at a maximum of 60ma or
so.

On my breadboard this works fine, but I don't want to make any stupid
engineering mistakes for the final [email protected]! It involves 900-1800 of
the buggers!

 

10% duty.

I'm trying to imagine a topology that uses a BJT and shouldn't use a
resistor. How do you have it hooked up now? Are you pulling the base
around with an HC output, directly?

Jon

 

Right now the output of the HC273 is going directly to the base of the
2N2222.

I just thought about using a 74ABT273. I need to confirm something.

http://www.standardics.philips.com/products/abt/pdf/74abt273a.pdf

or

http://focus.ti.com/lit/ds/scbs185b/scbs185b.pdf

When it says that IOL is 64ma, that means it can sink 64ma to ALL
outputs at the same time? The output pin going to a resistor going to
the cathode, and finally +5?

The link to TI's datasheet says "Latch-Up Performance Exceeds 500 mA
Per JEDEC Standard JESD-17"...this means that one chip could sink 8
outputs at 64ma without issue? There could be the situation where the
chip has to drive all outputs at 60ma 100% of the time.

I was wrong above when I listed duty cycles for the transistor.
Consider the possibility that it could be as low as 10% or as high as
100%

Is using a transistor my best bet, or will this work well (well in that
the final design will cost a few hundred in parts!)

 

You should be using a series R to limit the base current.

Graham

 

I'll let someone better informed about all that answer. I don't have
time tonight to look it up and come to my own opinion.

I probably would first consider using an emitter resistor. Leave the
base directly connected, put the LED between the collector and the +
supply, and put the resistor between the emitter and ground. This
way, you can program the setup to provide a known current. If you are
using 5V, then the resistor would be about 4.1V/60mA or say 68 ohms. I
don't like _only_ using a base resistor here because it won't limit
the current or, if it does, it will do so only by depending on the
BJT's beta. Better, I think, if you are stuck adding only one
resistor, to put it in the emitter leg. At least, that's what I think
this late at night.

Jon

 

The short answer for a BJT is yes.

As you are driving the base of a bipolar, yes. As the bipolar is
sinking LED current then the resistor should be roughly (Vcc-0.6)/6mA.
For 3.3V systems, that would give ~ 4.7k (for a standard size) in
series with the base. You should, of course, have a collector resistor
to limit the LED current.
For the base resistor, I didn't take into account outputs being less
than Vcc due to the output current.

Are you using the 2N2222 because you have them? In such a situation (to
minimize parts) I would use a small FET such as the venerable VN2222LL
for through hole, 2N7002 surface mount. No gate resistor required as
the '273 is always driving it's outputs. Of course, some use gate
resistors to limit the charging current in the FET, but the value is
not critical (somewhere between 2k and 39k would probably work just
fine). In this case, that really shouldn't be an issue.

On your breadboard, it's probably working fine because the '273 outputs
are being limited to about 0.7 ~ 0.8V out (in the high state) and the
internal drivers can (so far) handle that current (and abuse). Don't
count on it lasting too long, though. Without looking at the datasheet,
I can't say whether it will damage the '273, but it's not a good
practise.

On the LED, 60mA is *very* bright. I use LEDs for panel information,
and I don't run those over 10mA. In either case, you will need series
resistors in the LED current path.

Cheers

PeteS

 

As somebody already said, you should use a resistor to limit the
current into the base, and also to limit the current in the LED. It is
far better to be safe than sorry, especially in a production run.
Besides, resistors cost pennies which comapred to the cost of a field
failure is negligible

As a general rule of thumb, you determine how much collector / emitter
current you want to have flow. Then calculate the base current
required, assuming the minimum Beta of the transistor. Then multiply
this figure by ten for a 10x overdrive. Given this figure, you can
calculate a base resistor necessary to provide this amount of current
according to your supply. Assuming the emitter is grounded, you would
have Vsource - .7 / 10x base current = resistor.

 

Why would 6mA of *base current* be required John ?

Graham

 

---
If he's planning on using a common-emitter connected transistor to
drive an LED at 60 mA, forcing beta to 10 (to make sure the
transistor is running in saturation) yields:

Ic 60mA
beta = ---- = ------ = 10
Ib 6mA

 

60 mA ? through the LED ? Did I miss that somewhere ?

For a typical small signal transistor in common emitter I'd not over drive it
that much though. I'd probably select a series R for a beta of ~ 50.

Graham

 

Apparently. ;-) The OP specified 60mA at ~10% duty cycle using a
2n2222.

That might be a tad optimistic. The Fairchild datasheet for the 2n2222
shows a minimum hfe of 35, albeit at a much lower Ic.

 

---
Yes. Look above for:

"Transistor will be a 2N2222 type driving an LED at a maximum of
60ma or so.
---

 

I wondered about that.

The 2N2222 seems to be a bad choice then. I really can never understand why
these old parts keep re-appearing ad nauseam. The modern 'general purpose'
small signal transistors I use bottom out with a beta of about 200.

Graham

 

Ok. Given the 10% duty cycle that makes sense.


I fail to see the advantage here. The led currrent is limited by a series R. If the
transistor doesn't saturate to the last millivolt it really doesn't matter, it's not
a real 'hard switching' application.

The extra base mA will result in wasted power consumption though.

Using say a BC548 I'd set Ib @ ~ 1mA.

Graham

 

Follow up. 1800 led drivers @ 10 % duty cycle > 1800 * 10% * ( 6-1 mA ) saving on the
PSU.

That's nearly an AMP !

Graham

 

---
So what?

he's going to be using about ten amps to drive the LED's, he didn't
say anything about a power problem, and he wants to use 2N2222's, so
that's how it goes.

 

Off of the 11A continuously drawn by the LEDs. Man....this thing is
really gonna put out some heat.

 

Yup. A year ago, I bought 1000 to-92 pn2222a for $8. Less than 1
cent each. Cheap.

Jon