# Schmitt trigger low threshold calculation

I have this classical Schmitt trigger circuit from an old Elektor magazine.

It describes the following, when U1 is lowered (from a high state):

If the right transistor cuts off entirely and the left transistor is completely turned on, then there is on the emitter resistor about 9V / 11 = 0.8 V. Then one should add the BE-voltage, that the left transistor needs to conduct. The low threshold is thus 1.4V.

Where does the number 11 come from? Why does one need to add the emitter voltage to the BE-voltage?

• It's a good idea to insert a base resistor between the potentiometer's wiper and Q1's base... Also, if you want to simplify the circuit more, you can omit the 100k-10k voltage divider connecting the Q2's base directly to Q1's collector. – Circuit fantasist Jan 23 at 16:33

$$V_{U2} = \frac {470}{4k7 + 470} {V_{CC} } = \frac 1 {11} V_{CC}$$