Path: utzoo!attcan!uunet!dino!sharkey!bnlux0!geller
From: geller@bnlux0.bnl.gov (joseph geller)
Newsgroups: sci.electronics
Subject: Re: How can I turn on this bulb with TTL?
Summary: a correction to my original reply
Keywords: NPN, TTL, bulb, 12 volts, bias
Message-ID: <1383@bnlux0.bnl.gov>
Date: 22 Jul 89 04:00:00 GMT
References: <4363@merlin.usc.edu>
Reply-To: geller@bnlux0.UUCP (joseph geller)
Organization: Brookhaven National Lab., Upton, N.Y.
Lines: 54


	The article from tomb@hplsla.HP.COM offers a real nice solution
to the original TTL - light bulb question.  I'd just like to correct an
error I made in my reply.

	I stated that the maximum high level output for a standard TTL
output is 400 uA.  This is the maximum 1 level steady state output current 
you should design for, but as others have mentioned, the totem pole output
is capable of much higher 1 level currents.  

	The higher currents are specifically intended to charge circuit
capacitance to give the 10 - 20 nano second TTL transitions.  To design a
circuit which uses the 1 level output current (the totem pole sourcing I)
to directly drive a transistor is iffy at best.

	The TTL totem pole output stage looks like this : 


                        +5
	             |---|                The voltage drop across Ric goes up
                     Rib Ric         as the one level current (Ioh) does. 
                     |   |           The power dissipation, or chip heating,
                     |   C           also goes up with Ioh.
     internal chip-- | -B   NPN           At Ioh max, 400 uA, the power 
                         E           dissipated by Ric may be as low as
                         |           micro watts at very low switching speed.
                        \/           As the frequency of the switching goes up
                        --           so does the average power, because of the 
                         |           load circuit C's.
                         |--(TTL OUTPUT PIN)
                         |                The real iffy part is to try to say
                         C           what an acceptable Ioh above the 400 uA 
     internal chip------B   NPN      might be.  If a circuit draws 10 mA from
                         E           the output (Voh would be about 2.2 V), the
                         |           power dissipation in Ric is up to on the
                       common        order of 10 milli watts steady state.
                                          Remember that Ric and the NPN's are
                                     integrated components on the chip.

	Any way my rough calculations for that original circuit show that 
several mA might have been available; my comment that 400 uA was the limit was
incorrect.  Thanks to brianr@sbs.uucp for his correction by email.

	Perhaps integrated drivers are not the best answer when a junk box
full of transistors is available.  I would recommend using the PNP solution 
first.  It's certainly got the big advantage that when the PNP driver is off 
you are not dumping power into a resistor.   If your not on battery power
a pull up resistor from totem pole or open collector TTL, which then biases
an output NPN when the TTL output is high, also works fine.

	 Component failure due to heating is usually one of the first failure
modes in a circuit were ratings were exceeded.  I suppose there are cases where
it might be OK to exceed mfgr max values but, my experience is that there are
plenty of "gremlins" to deal with even when those max values are not exceeded.