# TTL input circuit to drive diode

Discussion in 'Electronic Basics' started by Gary H, Nov 23, 2006.

1. ### Gary HGuest

Hi. I need a little help with something that should be pretty basic. I
have a control board that has a TTL output to turn on and off a laser. The
laser is nothing more than a diode that consumes about 60mA. So, I want to
be able to turn the diode on or off with the TTL output. I don't really
know much about TTL requirements. But, would a NPN transistor with resistor
between the base and the TTL Output work? (Collector to + side of diode,
emitter to GND.)? If not, what would be the best thing to do?

Thanks!
Gary

2. ### DJ DelorieGuest

That's what I do. Or an N-channel MOSFET, which requires less math.

For a BJT (NPN) transistor, the math goes like this:

Consider the NPN with Rb between its base and the TTL output, and Rc
in series with the diode load, between Vcc and the collector. The
emitter is simply grounded.

The goal current is Ic. There will be a voltage drop across the
transistor (Vce) and across the diode (Vf) at current Ic. Subtract
those from Vcc to get the voltage across Rc (Vrc). Compute R given Ic
and Vrc (V=IR R=V/I, so Rc = Vrc/Ic).

Now, look up Hfe for the transitor. This is the "gain" of the
transistor. Divide Ic by Hfe to get Ib. This is a minimum Ib, of
course. Now, there will be a voltage drop across the transistor again
- Vbe. Subtract that from the TTL Voh to get the voltage drop across
Rb, and calculate Rb from Vrb and Ib. This resistance is a maximum,
you can use smaller values if you want.

For a MOSFET, the math is easier. The voltage drop across the MOSFET
(Vds) is negligible, you can just calculate Rc (er, Rd now) by (Vcc -
Vf)/Ic. As for the base, any medium resistor is sufficient, since the
gate is capacitive. You probably could omit the resistor completely.

3. ### ChrisGuest

Hi, Gary. If you actually have a TTL output, you might not know that
they sink a lot more current than they can source. That means you
might want to use a PNP transistor, with the output active low (the
diode goes on when the TTL output is low) like this (view in fixed font

|
| VCC
| +
| |
| .-. VCC
| | | +
| | | |
| '-' |
| |\ ___ | |<
| -| >O--|___|-o-|
| |/ |\
| |
| .-.
| | |
| | |
| '-'
| |
| V ~
| - ~
| |
| ===
| GND
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

Thus was it done in days of yore when TTL ruled the earth.

As is mentioned in a prior post, if you're stuck with an active high
output (diode must go on when the TTL output is high), you might want
to go with a logic level MOSFET. If you do, make sure to use a pullup
at the TTL output to ensure the high voltage goes over the typical 3.5V
you'll see as a TTL high. That might not fully turn on your MOSFET.

Good luck
Chris

4. ### Gary HGuest

Thanks. I appreciate you taking time to show the equations and how they
work. It makes sense.

Gary

5. ### RobinGuest

If you need to switch the diode fast, a transistor arrangement might be
too slow for you.

If your TTL output comes from a package that has other unused gates,
you could parallel them together to provide the extra drive?

Cheers
Robin

6. ### MartinGuest

I suspect we are in violent agreement since you say:
"you can use smaller values if you want"
But I think that is not quite strong enough ...

Good design practice is to ensure the transistor is well saturated.
Do this by dividing the Rb above by 10
(thus increasing the base current by a factor of 10)

<snip>

Martin