Driving transistor from HC273

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

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

or

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

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

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

drive it

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

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

or

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

will

stupid

of

(Vcc-0.6)/6mA.

it

If the

it's not

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

That's nearly an AMP !

Graham

will

stupid

of

(Vcc-0.6)/6mA.

it

If the

it's not

on the

mA ) saving on the

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