Explain the operation of the circuit shown in Figure S1-1. What would happen in this circuit if switch S1 were closed?

Question

Accepted Answer

Diode [latex]D_1 ext { and } D_2 [/latex] together with the transformer form a full-wave rectifier. Therefore, a voltage oriented positive-to-negative as shown will be applied to the SCR and the control circuit on each half cycle.

(1) Initially, the SCR is an open circuit, since [latex]ν_1

(2) Voltage [latex]ν_1[/latex] is applied to the control circuit, charging capacitor [latex]C_1[/latex] with time constant [latex]R C_1[/latex].

(3) When [latex]ν_{ C }>V_{ BO }[/latex] for the DIAC, it conducts, supplying a gate current to the SCR.

(4) The gate current in the SCR lowers its breakover voltage, and the SCR fires. When the SCR fires, current flows through the SCR and the load.

(5) The current flow continues until [latex]i_D[/latex] falls below [latex]I_H[/latex] for the SCR (at the end of the half cycle). The process starts over in the next half cycle.

If switch [latex]S_1[/latex] is shut, the charging time constant is increased, and the DIAC fires later in each half cycle. Therefore, less power is supplied to the load.

Question S1.3: Explain the operation of the circuit shown in Figure S1-1. W...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions