Schmitt Trigger Design Tutorial

🔶Terms: ◽UTP - Upper threshold point (output is high if input reaches this voltage) ◽LTP - Lower threshold point (output goes low if input falls to this voltage) ◽Re - the resistor on the node common to the emitter of the two transistors ◽Q1 - leftmost transistor ◽Q2 - transistor on the right ◽R1 - resistor connected to the collector of Q1 (Left) ◽R2 - resistor connected to Q2 ( Right) collector ◽Rb - resistor between the base of Q2 and Ground ◽Ra - resistor between Q1 collector and Q2 base ◽Vb - node voltage at the base of Q2 ◽Ix - curent that would flow to Ra and Rb in series with Vcc. (Assume R1 is zero) 🔶Notes: 1.Ic1<Ic2 2.Assume Vce1 & Vce2 = O at saturation 3. Ix or bleeding current must be small🌠 🔶Design process: 1.Decide UTP ⚪Choose UTP = 5V 2. Decide Ic2 ⚪Choose 5mA 3. Solve for Re ⚪Re = (UTP/Ic2) = 5V/5mA = 1kohm 4. Solve for R2 ⚪R2 = [(Vcc - Vce2 - UTP)/(Ic2)] = [(10V - 0V -5V)/(5mA)] = 1kohm 5. Decide Ic1 ⚪Choose 2mA 6. Decide LTP ⚪ Choose 2V 7. Solve for R1 ⚪R1 = [(Vcc - Vce1 - LTP)/(Ic1)] = [(10V - 0V -2V)/(2mA)] = 4kohm 8.Calculate Vb ⚪Vb = UTP + 0.7 = 5.7V 9. Decide Ix ⚪ Ix = 500uA 10. Solve and Adjust values of Ra and Rb ⚪ eq1: Ra + Rb = Vcc/Ix = 10V/500uA = 20kohm ⚪eq2: [Rb/(Ra + Rb)](Vcc) = Vb : [Rb/( 20kohm)] = 5.7V/10V : Rb = 11.4Kohm > Ra = 20kohm - 11.4kohm = 8.6kohm ⚪ the previous calculations were made with the assumption that R1 is zero.Subract🌟 R1 to Ra for a new value of Ra ⚪Ra = 8.6kohm - 4kohm = 4.6kohm 11. Adjust to standard values 🌠I don't know how small it would be but perhaps at uA range. 🌟Not sure of this method, it looks stupid but works so its fine. ⚫Min Output = UTP (I'm not sure w/ this) ⚫Max Output = Vcc (Same w/ this) ⚫ I just extracted this method from a website i forgot. I hope i don't get sued for this. ⚫ These calculations only apply to this setup. ⚫I only accept constructive criticisms Haters gonna hate. ( ︶︿︶)_╭∩╮

published 9 years ago