As shown in the figure, the self-locking circuit. When simulating on Multisim, it is found that if the resistance R2 is selected too large (for example, more than 10K), the LED will be on when the simulation starts; if R2 is selected appropriately (for example, below 1K), it will work normally , The light is off when the power is on, the light is on when the switch is pressed, and the light remains on when the button is released. Now I want to know how this R2 affects the circuit.
Sinclair 发表于 September 17, 2020
The power-on conduction is the result of positive feedback by the leakage current of the two transistors. The function of R2 is to connect the leakage current of Q2 to the ground, so that the base of Q1 is lower than the turn-on voltage and cannot be turned on, but only leakage current.
The size of R2 can be appropriately selected according to the size of the leakage current. If the leakage current is large, the resistance must be small, so that Q1 cannot be turned on, thereby preventing the effect of positive feedback.
Chauncy 发表于 September 17, 2020
I have seen this circuit before. The reason why the value of R2 is too large, which causes the LED to turn on when powered on, is because there is a leakage current in Q2, which generates a relatively large voltage drop on R2, thus turning on Q1.
Agnieszka 发表于 September 17, 2020
Determined by the characteristics of the triode element.
