Download A LED Exercise

A LED Exercise
An LED which has the characteristics shown in this graph
is to be used in the circuit below in which both VS and R
can be varied. For this LED the switch on voltage (VD) is
1.7 volt which produces a current of 10 mA at which point
the LED will just glow dimly. Let us say that the diode
operates brightly at 40 mA, but will fail if the current
exceeds90 mA for too long.
For best value in doing these questions try them before
you read ahead. There is much more value in trying them
for yourself than just reading the solution. If you have
trouble, just read the next little bit for a clue.
1. Initially the power supply is set at Vs = 6V. What value
is required for the resistor so that the LED operates at 40
mA?
That was a nice simple question requiring only that we use the graph to determine that
with 40 mA through the diode there is a 2.0 V drop across it, leaving a 4 V drop across the
resistor which needs to have a current of 40 mA flowing through it. Thus R = V/I = 4/40 =
0.1 k = 100 
2. In a similar way, find the minimum value of the resistor that could be used without
damaging the LED. (Answer at end.)
3. Now try this variation: The supply voltage is set at 10 V and there is a 200  resistor in place. What current
will flow through the resistor now?
You will have found this difficult to answer unless you realised that it is closely related to the last question – if
you didn’t, look at the two situations again and see if you can find the answer to this one before you read on.
Hopefully you saw that the answer to this question is still 40 mA because in effect we doubled the voltage
across the resistor by adding another 4 V to the supply giving us 8 V across it, but we also doubled the resistor
so that the current will still be 8/0.2 = 40 mA, giving the required 2 V drop across the LED. But what if it had
not just ‘so happened’ that the answer was easy to see? Try another straightforward question first:
4. If a current of 20 mA is flowing and the resistor is 200 , what is the supply voltage?
That wasn’t too hard because the 20 mA told us there was a VD of 1.85 V (from the graph) and a VR of IR = 20
 0.2 = 4 V. Another reasonably simple question:
5. If we find a voltage across the 200  resistor of 2.0 V what is the supply voltage and will the LED be
glowing?
P.T.O
That wasn’t hard because the 2.0 V told us there was a current of 2/0.2 mA through the circuit and then it is just
a case of using the graph.....
6. If there is a voltage drop of 4.6 V across the resistor (R is not 200  this time) and 2.1 V across the LED,
what is the supply voltage and the resistance of R?
7. Let’s try something a little harder! If the supply voltage is 10 V and R is 400 , what will be the current in
the LED? First, why is this a harder problem?
Because we don’t know either VR or VD we can’t easily find the current! Neither can we set up a simultaneous
equation because we don’t know the function that describes the LED characteristic. So we must use guesswork!
8. Can you say approximately what the voltage across R is? And so what would be an approximate current?
9. With any moderate current through the LED the voltage across it is around 2 V and so VR is about 8 V which
would imply a current of 20 mA. Now how can we check to see if this is the right current? Must the correct
answer be larger or smaller than this?
10. As we can see from the graph, if 20 mA was flowing, there would be about 1.85 V across the LED. This
however would mean there was 8.15 V across R, which in turn would mean what current was flowing?
11. So what is the right answer?
12. Now find the current if VS = 5 V and R is 40 
13. Now try VS = 20 V and R = 360 , and then VS = 3 V with R = 15 
14. What are the relative merits of each of the two schemes in the last question? For example, what would
happen if the supply voltage rose by 1 V in each case?
ANSWERS
2. Minimum value of R: VD = 2.2 V (from graph) so VR = 6 – 2.2 = 3.8 V and R = 3.8/90 = 42 
4. Supply voltage = total voltage (Vs = VD + VR) = 1.85 + 4.0 = 5.85 V
5. Current is 10 mA and so VD = 1.7 from graph. LED doesn’t glow. Supply voltage = Vs = VD + VR = 1.7 + 2.0 = 3.7 V
6. VS = VD + VR = 6.7 V. VD = 2.1 V means I = 65 mA and so R = 4.6/0.065 = 70.8 
10. 8.15/400 = 20.4 mA. This will be a pretty good guess as it leads to the correct voltage across the LED.
12. VR  3 V  I  75 mA  VD = 2.15 V  VR = 2.85 and I = 2.85/40 = 71 mA, which would  VD = 2.1 V so a pretty good guess would be about 72 mA.
KB April 2004