a_drunken_dwarf said:couldn't I just hook up a lot of resistors in series for a high voltage drop? How is an attenuator different from doing this? Does it introduce less noise? Thanks!
Dean Huster said:It's two resistors in series. Let's put them down on paper in schematic form in a vertical line, R1 on top and R2 on bottom. The bottom of R2 is grounded. The top of R1 goes to your voltage source to be dropped or attenuated, in this case, the output of your TTL chip. The junction of the two resistors is the output with respect to ground, i.e., the output is across R2.
Two resistors are used in the design like this rather than just a "bunch of resistors in series" so that you have more control if the load resistance bounces around a bit.
Choose a value out of thin air for R2. You want this resistance to be about 4 to 10 times lower than the load resistance (the petrie dish). We'll pick 100 ohms just for the heck of it. Divide your desired output voltage by the resistance value you've chosen for R2 to get the current through R2 and the rest of the series circuit. So, 10mv divided by 100 ohms is 100µa. Let's assume that the TTL output of the chip is 4v p-p. The voltage across R1 will be the source voltage minus the voltage across R2 or 4 - 10mv which is 3.99 v. The value of R1 will be the voltage across R1 divided by the current through R1 or 3.99 divided by 100µa which is 39,900 ohms. Nearest standard 5% value for this will be 39K ohms. You'll have to use a little trial and error here since we're making a few assumptions, such as the actual output voltage of the chip, etc. But that's the general way do figure the thing. Dropping 5v down to 10mv is a bit of a drastic change, so there's going to be a lot of slop in this circuit.
Dean
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