TTL driving 50 ohm termination
Danny
I'm working on a school project where I'm supposed to generate a TTL
signal with different frequencies (between 1 Hz to 500 kHz) depending
on the user input. I'm generating this TTL signal using a Motorola
68hc12 microcontroller. My project specifies that this TTL signal
should be connected to a BNC connector and should be capable of
driving 50 ohm termination. Now I cannot get 100 mA out of the
microcontroller.
I tried using a 2N4401 transistor with the signal connected to the
base, collector connected to Vcc, and the emitter connected to the
BNC. The problem is that the emitter voltage (the output of the BNC)
would be at most 4.2V. Also, I looked into using a TTL hex buffer, but
its maximum current rating is like 20 mA! You should also now that the
output resistance of the microcontroller is 50 ohm.
I need a circuit that would be able to supply me with 100 mA without
decreasing my output voltage, any ideas?. Also, any idea how to kill
the negative spikes that appear on the scope screen?
Thanks
Tim Hubberstey
You don't need either 100 mA or 5V. Your specification says you need a
_TTL_ signal into 50 ohms. TTL requires an output voltage of 2.4V min
for a '1' and 0.8V max for a '0'. Your 2N4401 connection should do
exactly what you want. You might want to add a resistor (1k would be
fine) from the emitter to ground so you can see the output on a scope
without the BNC hooked up but this isn't necessary for operation.
--
Tim Hubberstey, P.Eng. . . . . . Hardware/Software Consulting Engineer
Marmot Engineering . . . . . . . VHDL, ASICs, FPGAs, embedded systems
Vancouver, BC, Canada . . . . . . . . . . . http://www.marmot-eng.com
Christopher R. Carlen
>
> You don't need either 100 mA or 5V. Your specification says you need a
> _TTL_ signal into 50 ohms. TTL requires an output voltage of 2.4V min
> for a '1' and 0.8V max for a '0'. Your 2N4401 connection should do
> exactly what you want. You might want to add a resistor (1k would be
> fine) from the emitter to ground so you can see the output on a scope
> without the BNC hooked up but this isn't necessary for operation.
=>2.0V for a high, <=0.8V for low.
The problem with the 2N4401 as described is it will toast if the output
is shorted. A resistor in the collector will help.
It also has an asymmetrical drive impedance, so I doubt the waveform
will look pretty if the end of the cable is unterminated.
I prefer to drive 50R coax with a symmetrical driver tailored for mostly
resistive 50ohm source impedance, so that the waveform at the end of the
cable looks pretty no matter if it is terminated or not.
Good day!
--
____________________________________
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA
crc...@sandia.gov
Christopher R. Carlen
The nicest way I've found to drive 50ohm coax is to use 4 HC CMOS
inverter gates in parallel driven by another single inverter. You can
use one gate of an HC14 hex Schmitt trigger inverter as an input noise
eliminator (use HCT instead if you want input level compatibility with
the output of LSTTL or similar bipolar TTL gates), and this also loads
your circuitry with only one gate as well as making the whole inverter
assembly function as a buffer, since the output will get inverted once more.
Drive the gang of 4 parallel gates with the single inverter. The
parallel 4 should all be gates of the same package. Leave one gate unused.
Now connect a 33 ohm resistor to the output of the parallel gates, and
the other end can drive the cable. This resistor, in series with the
internal output impedance of the parallel gates, forms a driver with
total output impedance very close to that of the cable. This is said to
be "back-terminated." It will drive 5V into a cable with no
termination, with clean waveforms at the end of the cable. There is a
reflection, but it is absorbed at the source end, and so the destination
end sees only a clean edge snap to 5V. Anything in the middle of the
cable will see an ugly waveform though, so this doesn't work if the
cable must have any "Ts". But if you terminate the cable with 50 ohms,
then the waveforms will be clean through the whole cable, but the output
will now be 2.5V. This is fine for driving real TTL. It is *not* fine
for driving any CMOS inputs.
That is why I always design digital inputs that see the outside world to
use TTL compatible input levels, by using LS, AS, F, HCT, or ACT logic, etc.
Note, this parallel 4 HC gate driver is not technically short circuit
proof. The HC gates are rated for 50mA continuous max current through
VCC or GND. Thus, about 100mA will flow if you short the output (after
the 33 ohm resistor). This doesn't exceed the individual gate maximum,
only the total package current rating. However, the parallel gates will
output about 3.8V before the 33 ohm resistor, so the total package
dissipation is only about 120mW, which is unlikely to cause problems. I
wouldn't do this for something that would be manufactured in large
quantities, but it's OK for R&D purposes.
If you want a totally short circuit bulletproof design, you can use
AC(T) logic, which can handle 200mA total package current. This can
also allow you to reduce to only 2-3 gates in parallel. But AC is more
tricky since it is godawful fast, so the edges at the end of the cable
can wind up more imperfect unless you fuss a great deal with bypassing
and transmission line issues.
Finally, another easy but slightly more expensive approach would be to
use a driver like TC4427. If you must preserve pulse widths, use
TC4427A, which still will distort them a little, but not as bad as
TC4427. You can use these in parallel also, and use about 44 or 46 ohms
output resistance.
Good luck.
John Jardine
Danny <dani...@yahoo.com> wrote in message
news:f814939b.03040...@posting.google.com...
The circuit is fast and will supply lots of current. It is usual to drive a
50ohm termination from a 50ohm 'source' (the 50ohm loads gets 50% of the
drive signal). Short the 47ohm out if a low output impedance is needed.
5V
===
|
|
|/
,-----| NPN
| |>
frequency in | | ___
0-5V --+ +-|___|-o to BNC
| | 47ohm
| |<
'-----| PNP
|\
|
|
===
0V
Transistors, 'small signal' types
regards
john