The circuit shown in FIGURE 3 is part of the interface of a relay output module.

Question

The circuit shown in FIGURE 3 is part of the interface of a relay output module. Ib is 1 mA and VCC is 9 V. The relay requires a minimum of 50 mA to energise. Complete the values of the assumptions listed below in order to calculate:

• voltage across R1
• value of R1
• voltage across the relay coil
• voltage across R2
• value of R2
• collector of current Ic.

Assumptions:
Logic '1' = V
Logic '0' = V
Transistor forward current gain hfe =
LED current = 10 mA
LED voltage drop at 10 mA = V
Base/emitter voltage = V
Collector emitter voltage when transistor is on = 1 V

Explanation / Answer

Logic 1 = High State = 9v = Vcc = Supply Voltage

Logic 0 = Low State = 0v

Current gain = hfe = 0.754

LED Current = 10mA

Voltage drop = I × R = 0.98

Vce = 1v so base emitter Voltage is 9V as equal to the Supply Voltage. So Ic = 10 mA

Voltage across Vr1 and Vr2 is respectively 5V and 2V.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.