Finding R, I & V in an PLC output interface

1. Jan 27, 2017

cjm181

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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

2. Jan 28, 2017

CWatters

Looks ok. I think I might drive the base a bit harder. I don't think 4k4 is a standard size so I'd used the next E24 value down which I think is 4k3 or 3k9.

In real life you would check the resistance of the relay coil and perhaps put a resistor in series with the coil. You wouldn't normally rely on the base current and the transistor gain to set the relay coil current. Transistor gains are usually very variable. What if the shop sold you a very good one and the gain was actually 120?

3. Jan 28, 2017

rude man

Putting a resistor in series with the coil diminishes that current. "In real life" the coil resistance is designed so that a saturated transistor voltage acrosss the coil (i.e. ~9V here) provides at least the minimum required pull-in current. The transistor is therefore operated in the saturated mode.

4. Jan 28, 2017

cjm181

Is it not 8V across the coil? 9V supply, 1V drop across the transistor?

I am just asking as I have answered 8V across the coil and across the leg containing R2 and the LED.

Thanks

5. Jan 28, 2017

rude man

Why 1V across the transistor? You're right in that there will be some drop, but if you look at typical transistor characteristic graphs you will find that the drop is more like 0.1V, maybe 0.2V. The main thing is to ensure sufficient base drive to ensure saturation, that typically is 0.1ic.

6. Jan 28, 2017

cjm181

Does the question not say it's 1v?

"Collector emitter voltage when the transistor is on = 1v"

Thanks

7. Jan 29, 2017

rude man

It was listed under "assumptions". It's not realistic. Besides, the list had almost no numbers in it (Vce was one of 2/7) so I didn't pay too much attention to it ....