I have the Schmitt trigger circuit below that I read the output as 1.6V for the logic High. I wonder how can I increase the output voltage of the comparator below without changing the threshold of the Schmitt trigger?
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\$\begingroup\$ What op-amp are you using, and can its output swing to both rails? How about the inputs: is the input range from rail to rail? \$\endgroup\$– KazMay 5, 2013 at 3:58
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\$\begingroup\$ I am using lmc6772 comparator. the datasheet is on ti.com/lit/ds/symlink/lmc6772.pdf The input voltage is very close to 10V and 0V as I measure \$\endgroup\$– svenMay 5, 2013 at 4:07
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\$\begingroup\$ Warning flag: open drain output. That means that to get an output voltage, you have to provide a pull-up resistor, etc. \$\endgroup\$– KazMay 5, 2013 at 4:44
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1\$\begingroup\$ The resistors have a slightly low value, which will result in high dissipation (they get warm). There should be no problem replacing them with 12k 47k and 56k. \$\endgroup\$– jippieMay 5, 2013 at 7:53
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1\$\begingroup\$ Please justify the 1.6V for logic high - I agree with jippie that a logic high output will produce about 9.9V and you should see this BUT if the input is logic high the output may be round about 1V because of the 17mA current taken by the open-drain output - maybe this is the 1.6V you are seeing. What are you trying to achieve? \$\endgroup\$– Andy akaMay 5, 2013 at 10:47
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1 Answer
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This might not be exactly right, but because the comparator has an open-drain output, you have to think of it as being something resembling the device surrounded by the dashed box:
simulate this circuit – Schematic created using CircuitLab
When M1 is off (high output), the voltage at OUT is determined by the R1/R2 voltage divider. An open drain output does not drive a high voltage. Instead, it shuts off to a high impedance state, and the high voltage is created by a pull-up resistor (either supplied by you, or perhaps one that is provided in internally ("weak pullup" type feature)).
If you look at the hysteresis example in the datasheet, it works differently. The output is bootstrapped from the input. If we draw it with our explicit open drain model, it looks like this:
So here, if OA1 cuts off the output transistor, then OUT is at the same voltage as IN. It is pulled up to IN by the two resistors. And of course if the transistor is fully on, then OUT is pulled to ground.
