Schmitt trigger, what does the voltage divider at the base of the second transistor do?

Question

The following Schmitt trigger has a low threshold of about 3V and a hight threshold of about 7V:

^{simulate this circuit – Schematic created using CircuitLab}

As far as I understand Q1 is setting the low threshold and Q2 is setting the high threshold through the resistors RC1,RE and RC2,RE respectively (the transistors work as a voltage divider according to this). However I don't undersand what does the R1, R2 voltage divider do at the base of Q2, I tried to remove it and connect the base of Q2 directly to the collector of RC1 and didn't notice much effect on the low threshold, the high threshold on the other hand jumped to 8.6V then I tried to change RC2 value but it didn't change the high threshold significantly.

I thought that the voltage divider could be setting the base voltage of Q2 and that would also set the voltage of Q2 emitter (as in emitter follower) but shouldn't RC1 be setting the base voltage to be equal to \$V_{BE} + \frac{R_E}{R_{C2}+R_E} Vcc\$? What's going on in this circuit and how do you calculate the resistors values without using a simulator?

Answer 1

for better / finer control of gain. with the values used, it doesn't have much use here.

more common is to have a divider from Q2's emitter to Re.

Answer 2

Place Vin at zero volts. Current can only flow in the right half of the diffpair Q2. Voltage on base of Q2 will be R2/(Rc1 + R1 + R2) or about +8 volts. If less than 8 volts, its because base current causes extra voltage drops across Rc1 and R1.

Now raise Vin in a ramp (triangle waveform from function generator or SPICE PWL. Take your time, so the transistors' parasitics do not obscure the precise switching.

As Q1 base rises, it starts to steal some of the current from Q2 and let that stolen current flow thru Rc1. The voltage at Rc1 drops, the base of Q2 drops, current drops thru Q2 and has to flow thru Q1; the voltage at Rc1 drops even more, all within 10nS to 100nS with positive feedback around the loop we just discussed.

Looking at Rc1 versus Re, I bet the Q1 goes into saturation (less than 200mV across it) and Rc1 drops from near +9v to near 3 volts. The source resistance will determine how low the emitter of Q1 can drop. Here Rsource is zero, huge base currents are possible, thus emitter cannot drop after the positive-feedback action. That puts Q1 heavily into saturation.

What is the going-low threshold? Depends on when Q1 starts to exit the saturation. And that is about 3 volts.

Don't be afraid to use a simulator. You'll learn lots about the various modes of operation of bipolars.

Answer 3

Just change RC1 to a 20K pot and change the resistance till you achieve the desired low threshold.

The voltage divider on the base of Q2 determines what magnitude the analog signal must be to trigger the circuit. By changing RC1 you change how much voltage V1 must be to trigger it. Once the circuit triggers at the desired magnitude at V1, just remove RC1 and measure the resistance. Replace RC1 with that size resistor.