- #1

cjm181

- 69

- 1

## Homework Statement

The circuit shown below is part of the interface of a relay output module.*I*b is 1 mA and

*V*CC 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

*R*1

• value of

*R*1

• voltage across the relay coil

• voltage across

*R*2

• value of

*R*2

• collector of current

*I*c.

Assumptions:

Logic '1' = V

Logic '0' = V

Transistor forward current gain

*h*fe =

LED current = 10 mA

LED voltage drop at 10 mA = V

Base/emitter voltage = V

Collector emitter voltage when transistor is on = 1 V

## Homework Equations

## The Attempt at a Solution

So, to complete the list off assumptions:

*Logic '1'*= This is usually 5V.

*Logic '0'*= This is usually 0V

*Transistor forward current gain hfe*=

Gain = (50+10)/1 = 60

*LED current = 10 mA*

LED voltage drop at 10 mA = V

2.4V

LED voltage drop at 10 mA = V

2.4V

*Base/emitter voltage = V*

Assuming transistor is silicone. voltage drop of this transistor between the base and emitter is 0.6V

*Collector emitter voltage when transistor is on = 1 V*

So, now to find the answers:

*• voltage across R1*

V(R1) = 5-0.6 = 4.4V

*• value of R1*

R=V/I = 4.4/1mA = 4.4kOhms

•

*voltage across the relay coil*

8V. Transistor has 1v drop, supply is 9, each leg in parallel circuit is the same.

•

*voltage across R2*

If the LED has a drop of 2.4V, and the leg has 8V, then the voltage across R2 is:

8-2.4=5.6V

*• value of R2*

R2=V/I = 5.6/10mA = 560 Ohms

R2=V/I = 5.6/10mA = 560 Ohms

•

*collector of current Ic.*

Gain of 60, so:

60xbase current = 60x1mA = 60mA

Is this looking good?

Thanks