Hi , working on building an analog tach using the "current" circuit recommended on the National Semiconductor datasheet. The circuit has a diode connected to pin 11. The signal I need to count goes from about +1 volt to about +13 volts. Using the diode I'm thinking will require the input go below .7 volts to count. The datasheet says I can set the comparator input level but can't figure out how it's done. Would I want to set the comparator to about +6 volts and how could I do that? Any suggestions?

 

I guess you're talking about this schematic?

The voltage divider on the input to pin 1 will reduce the amplitude of the signal by 1/3 as well as "de-bounce" it somewhat due to the capacitor and RC time. If you used that exact input scheme, your 1v to 13v signal will wind up being 0.66v to 8.66v.

The 10k resistor and diode on pin 11 cause pin 11 to be biased at about 0.32v to 0.7v, depending on the diode used.
A 1N5817 would give about 0.32v. A 1N914/1N4148 would measure about 0.63v.

However, your average voltage at pin 1 will be roughly 4.66v, so a diode on pin 11 won't really be adequate to trigger the comparator. Changing the diode to a 4.7k or 5.1k resistor would be close enough, you'd then have pin 11 biased at about 4.47v or 4.73v. 14v * 4.7k/(10k+4.7k) ~= 4.47v. 14v * 5.1/(10k+5.1k) ~= 4.73v.

 

Thanks! Now I understand. If the B+ varies though could I use the regulated approximate 8 volts from pin 9 and use a voltage divider?

 

Yes, two equal resistors; 3.9k to 39k - anywhere in that range. 10k would be a good compromise between low current and stable reference voltage.

 

Thanks again!

 

Two more posts of you and we could have had them typed in proper size and color!

OK, next time then...