user 37518
Well-known member
As the others, I am not sure what you are trying to achieve.
The general Schmitt trigger is like this
You start by determining which are the VTL and VTH voltages that you need, those are your thresholds for trigger low and trigger high of the input signal. Then, select R1 and calculate R2 like this: R2 = [(Vcc-VTL)/VTH]*R1, R1 shouldn't be small, something like 100K could work fine to reduce power dissipation. Afterwards, select a small value for R5 like 1K and make R4 much larger than R5 like 100K, then, calculate R3 like this: R3 = [(VTH-VTL)/Vcc]*R4.
Example: Say you want VTL = 0.7V, VTH =1V, and you are using Vcc = 5V;
select R1 = 100K, then R2 = ([5-0.7)/1]*100K = 430K, both resistors are commercial values.
Then, select R5 = 1K, R4 = 300K; both commercial values. Then R3 = [(1-0.7)/5)*300K = 180K, also a commercial value.
Let us know if that worked....
If you add a cap at the input, you will have to modify the circuit because this circuit assumes that the source is DC coupled. You could add a small resistor from the left leg of R3 to ground, just after the cap, it should be quite smaller than R3, so the threshold voltages don't get affected very much, but, that resistor will essentially become your input impedance and if its very small, your cap will have to be very large, so maybe you will have to scale the entire circuit so R3 is large. In the example I gave above, a 2K resistor from the left leg of R3 to ground might work, if you then add a 100uF cap it will produce a pole near 1 Hz. Just make sure the "square wave" can drive the 2K load.
The general Schmitt trigger is like this
You start by determining which are the VTL and VTH voltages that you need, those are your thresholds for trigger low and trigger high of the input signal. Then, select R1 and calculate R2 like this: R2 = [(Vcc-VTL)/VTH]*R1, R1 shouldn't be small, something like 100K could work fine to reduce power dissipation. Afterwards, select a small value for R5 like 1K and make R4 much larger than R5 like 100K, then, calculate R3 like this: R3 = [(VTH-VTL)/Vcc]*R4.
Example: Say you want VTL = 0.7V, VTH =1V, and you are using Vcc = 5V;
select R1 = 100K, then R2 = ([5-0.7)/1]*100K = 430K, both resistors are commercial values.
Then, select R5 = 1K, R4 = 300K; both commercial values. Then R3 = [(1-0.7)/5)*300K = 180K, also a commercial value.
Let us know if that worked....
If you add a cap at the input, you will have to modify the circuit because this circuit assumes that the source is DC coupled. You could add a small resistor from the left leg of R3 to ground, just after the cap, it should be quite smaller than R3, so the threshold voltages don't get affected very much, but, that resistor will essentially become your input impedance and if its very small, your cap will have to be very large, so maybe you will have to scale the entire circuit so R3 is large. In the example I gave above, a 2K resistor from the left leg of R3 to ground might work, if you then add a 100uF cap it will produce a pole near 1 Hz. Just make sure the "square wave" can drive the 2K load.
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