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In article <4363@merlin.usc.edu> cyamamot@nunki.usc.edu (Cliff Yamamoto) writes:
>Greetings,
>	I know this is a simple circuit, but I can't seem to get the bias
>right or something.  The circuit is as follows :
>
>		  C----bulb---> + 12 volts
>TTL---1 Kohm----B				NPN transistor
>		  E---+
>		      |
>		     GND
>
>The problem is when the TTL is high, the bulb barely lights up.  This circuit
>is for automotive use so the bulb is just an instrument panel bulb (not a high
>powered spot light).  Isn't 2N2222 or 2N3904 good enough for this?  How does
>one go about selecting a transistor for this?  And lastly, how does one figure
>out the bias needed to fully turn it on?
>
>Thanks for all your EE101 advice!  :)
>Cliff Yamamoto



	To interface from anything (such as TTL) to the outside world (as 
a lamp) you  must consider two questions first.  1)  What is the drive 
signal ?  2) What are the current and voltage requirements of what is being
controlled ?

	1) Your drive signal is the output of a TTL gate.  There are many
TTL families (S, LS, F, etc.) but for the moment lets assume the original
TTL, as in 7400, etc.  TTL devices are specified by fan out.  A typical logic
element, as the 7400, has a fan out of 10.

	When one TTL gate drives another it supplies a one or a zero to the 
input of the following gate.  Since TTL is made of transistor stages you must
consider voltage and current at this point.  It is not good enough to simply 
sat zero is zero volts and one is 5 volts.

	When you design a TTL circuit that does not exceed the fan out limit,
the voltages are well defined.  A zero may vary from 0 volts to .8 volts and 
a one may vary from 1.8 volts to 5 volts  (the .8 and 1.8 may not be exact
for standard TTL, but there are spec'd numbers such as these).  

	If the TTL output of the driving stage is a one, it is actually back
biasing the input transistor on the following stage.  Since the load is
actually off, with a one input, the driver supplies very little current, or 
about 40 microamps.  With a fan out of 10 a standard TTL chip is rated to
source only 400 microamps.  The voltage is typically on the order of 4 volts.

	When the driver is sending a zero, it really turns on the input stage
of the following gate.  Now the current is much higher, on the order of one
and half milliamps.  So a typical TTL gate can sink over 15 mA.
	
	So ... your drive signal can supply 400 uA at about 4 volts.  The 
limit in your circuit is not the transistor or the resistor.  It is that you
simply cannot draw 4 mA from the standard TTL gate at a one output.

	There are many many ways to interface to TTL.  For typical LED or
small incandescent lights drawing less than 30 mA, many engineers opt for the 
standard 7406 or 7407 open collector outputs.  To use the output you wire the
drive power supply to your load (if it is an LED you MUST use a series 
resistor) and the other side of the load to the chip output pin.

	For higher current loads the Motorola ULN chips or the Sprague chips,
as the UDN series are fine.

	You can also build a power FET stage, or a darlington stage, but,
since this sounds like a first TTL interface project I would recommend one
of the packaged drivers.

	In a pinch, you can use the 7407 or 7406 to sink the current from a 
resistor tied to Vcc when its output is on, and when its output is off use 
the same current to bias your 2N2222.  Since the 06 and 07 have opposite 
logic, one or the other will light your light at the right time.    

	I don't want this to be too long, so just quickly lets look at your 
load, the light. If it draws 500 mA at 12V, the control transistor must have a 
maximum collector current of at least 500 mA (and usually more for engineering
margin).  The 2N2222 is about 800 mA, so it is close but, OK.  If the 2N2222
has a voltage drop of about .2 V when it is full on, you are asking for .2 * 
.5 or about 100 milliwatts of power dissipation, also OK.