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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?

enter image description here

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  • \$\begingroup\$ 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. \$\endgroup\$ – Circuit fantasist Jan 23 at 16:33
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Where does the number 11 come from?

enter image description here

Figure 1. The current path when the left transistor is on.

If we ignore the voltage drop across the left transistor when hard on then we have a potential divider given by the two resistors.

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

Why does one need to add the emitter voltage to the BE-voltage?

Since Q1's emitter is at 0.8 V we'll need to raise the base by 0.6 V to turn it on. Thus any voltage above 1.4 V will keep Q1 turned on and steal the bias for Q2. Only as you pull Q1's base below 1.4 V will it start to turn off and allow Q2 to turn on.

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