Chris said:
I'm not sure how I would do this, but I need to take a 5 volt power
source and drop it down to 0.7 volts and match the impedance to 75
ohms.
That's a strange thing to ask for. Are we talking 5V DC here? That's my usual
assumption when talking 5V power. Do you need .7 volts open load, or looking
into a 75 ohm load?
And what kind of specs do you need for accuracy?
I'd say swing two silicon diodes (1N4148 style) with a 200R resistor in series,
you get 1.4V at the junction of the diode and resistor. Now, take a 75 ohm
resistor in series with the load and when you attach it to the load you'll get
0.7V across the load, for a power dissipation of 7mW in each resistor. Slap a
100nF cap across the diodes.
There you go. Not the most efficient or elegant thing, but cheap.
5V
+
|
|
<
| 75R
+----V^V^V^---- to LOAD
|
+--- cap to ground
|
V
V two forward biased diodes
|
~
Oh dear that looks like crap. But you get the idea. Get rid of one of the diodes
if you want 0.7V open load. Play with the type of diode and read the datasheet
to figure out the resistor value for getting closer to 0.7V. But this should get
you started, and should be pretty close to .7v at the 75 ohm load.
