The Student Room Group

A2 Electronics Options Questions...

ok, i have a few questions that i am completely puzzled by:

im having to explain these questions, but can give links to the actual paper if my explanations suck

1) A zener diode is used in a potential divider to produce a stabilised output voltage of 3.9V, with a 6V input and a 10 ohm resistor. The zener diode must have aminimum current of 10mA through it to function correctly.
The output is over the zener diode.
Calculate the maximum current that can be provided to the output.


I used the traditional potential divider formula Vin+Vout x R2/(R1+R2)

I used the values of 10mA and 3.9V to get the zener diode with a resistance of 17.1ohms, then used I=V/R to get the current flowing through it

They have done it a different way in the answer and my answer is about 10% off theirs

have i done this right??

thanks
Reply 1
Consider the 10 ohm resistor - 2.1 v across it. Use ohms law to work out the current through it which turns out to be 0.21 A. Now, after this the current splits down 2 paths, one through the diode and one to the output. The minimum current through the diode is 0.01 A. Therefore by Kirchoff's First Law, the current to the output must be 0.2 A.
Reply 2
malolis
Consider the 10 ohm resistor - 2.1 v across it. Use ohms law to work out the current through it which turns out to be 0.21 A. Now, after this the current splits down 2 paths, one through the diode and one to the output. The minimum current through the diode is 0.01 A. Therefore by Kirchoff's First Law, the current to the output must be 0.2 A.


yeah thats what the mark scheme shows, i didnt think of that, i just dived in with the potential divider equation but i still dont see why my answer is wrong (0.223A)

anyway thanks for your help, dont seem to be many people on here who do electronics, wish i actually had a choice :frown:
Reply 3
bigdaveracer
yeah thats what the mark scheme shows, i didnt think of that, i just dived in with the potential divider equation but i still dont see why my answer is wrong (0.223A)

anyway thanks for your help, dont seem to be many people on here who do electronics, wish i actually had a choice :frown:


If you post all the steps of your calculation I may be able to show you where you went wrong.
Reply 4
using the potential divider formula:

3.9=6R/(R+10)
0.65=R/(R+10)
0.65R+6=R
R=17.1ohms

I=V/R
=3.9/17.1=0.223A
Reply 5
bigdaveracer
using the potential divider formula:

3.9=6R/(R+10)
0.65=R/(R+10)
0.65R+6=R
R=17.1ohms

I=V/R
=3.9/17.1=0.223A


Ok firstly, the third line should be 0.65R+6.5=R giving a R of 18. something.

However that mistake isn't what's wrong with your method. You've basically calculated the resistance of the diode and then substituted that back in to work out the current through the diode as if there was no output. The question is to work out the current to the output, not the diode (which you already know to be 0.01 A) You must remember that the current splits after going through the resistor.
Reply 6
malolis
Ok firstly, the third line should be 0.65R+6.5=R giving a R of 18. something.

However that mistake isn't what's wrong with your method. You've basically calculated the resistance of the diode and then substituted that back in to work out the current through the diode as if there was no output. The question is to work out the current to the output, not the diode (which you already know to be 0.01 A) You must remember that the current splits after going through the resistor.

yeah this seems to be where i went wrong, kirchoff and all that

cheers for the help :smile: have some rep
Reply 7
bigdaveracer
yeah this seems to be where i went wrong, kirchoff and all that

cheers for the help :smile: have some rep


No problem :smile:

Latest