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cjm181
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Homework Statement
The circuit shown below 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
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
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
• collector of current Ic.
Gain of 60, so:
60xbase current = 60x1mA = 60mA
Is this looking good?
Thanks