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:<math>V_\mathrm{HT} = \frac{R_\mathrm{E}}{R_\mathrm{E} + R_\mathrm{C2}}{V_+}</math>. |
:<math>V_\mathrm{HT} = \frac{R_\mathrm{E}}{R_\mathrm{E} + R_\mathrm{C2}}{V_+}</math>. |
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The output voltage is low but well above ground. It is approximately equal to the high threshold and may not be low enough to be a logical zero for |
The output voltage is low but well above ground. It is approximately equal to the high threshold and may not be low enough to be a logical zero for next digital circuits. This may require additional shifting circuit following the trigger circuit. |
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'''Crossing up the high threshold.''' When the input voltage (Q1 base voltage) rises slightly above the voltage across the emitter resistor R<sub>E</sub> (the high threshold), Q1 begins conducting. Its collector voltage goes down and Q2 begins going cut-off, because the voltage divider now provides lower Q2 base voltage. The common emitter voltage follows this change and goes down thus making Q1 conduct more. The current begins steering from the right leg of the circuit to the left one. Although Q1 is more conducting, it passes less current through R<sub>E</sub> (since R<sub>C1</sub> > R<sub>C2</sub>); the emitter voltage continues dropping and the effective Q1 base-emitter voltage continuously increases. This avalanche-like process continues until Q1 becomes completely turned on (saturated) and Q2 turned off. The trigger is transitioned to the high state and the output (Q2 collector) voltage is close to V+. Now, the two resistors R<sub>C1</sub> and R<sub>E</sub> form a voltage divider that determines the low threshold. Its value is approximately |
'''Crossing up the high threshold.''' When the input voltage (Q1 base voltage) rises slightly above the voltage across the emitter resistor R<sub>E</sub> (the high threshold), Q1 begins conducting. Its collector voltage goes down and Q2 begins going cut-off, because the voltage divider now provides lower Q2 base voltage. The common emitter voltage follows this change and goes down thus making Q1 conduct more. The current begins steering from the right leg of the circuit to the left one. Although Q1 is more conducting, it passes less current through R<sub>E</sub> (since R<sub>C1</sub> > R<sub>C2</sub>); the emitter voltage continues dropping and the effective Q1 base-emitter voltage continuously increases. This avalanche-like process continues until Q1 becomes completely turned on (saturated) and Q2 turned off. The trigger is transitioned to the high state and the output (Q2 collector) voltage is close to V+. Now, the two resistors R<sub>C1</sub> and R<sub>E</sub> form a voltage divider that determines the low threshold. Its value is approximately |
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