If I'm being honest to calculate I(R) in (ii) I assumed ohm's law would come into play. I'm just not sure how only one switch being pressed will effect the values. I think I may be over thinking this question as it is only 2 marks.

 

hi jj,
This part


Look up the 'standard approx for Vfwd of a Diode' , thats the assumed voltage drop across a diode due to the current flowing thru it.

Once you know that approx Vfwd, subtract that from the Vsupply in order to get the effective voltage across 'R'
E

 

The original circuit has no lamps so don't talk about lamps.
JJB wrongly guessed that the 510 ohms resistor has 0.7V across it.
Why only 0.7V? From magic??

 

The original circuit has no lamps so don't talk about lamps.
JJB wrongly guessed that the 510 ohms resistor has 0.7V across it.
Why only 0.7V? From magic??

Is the forward voltage of a silicon diode not approx 0.7V? (I got this from the internet) please feel free to offer some information rather than just criticising the help that E gave me.

 

guru,

If I consider that the TS would benefit from me using Lamps as an analogy to explain the circuit I will do so, OK.

The 0.7Vfwd comes from the question paper, why don't you read before the post properly before you criticise the work of others.

I suggest you focus your thoughts on helping the TS with his problem.

 

hi jj,
That approx 0.7Vfwd is OK.
So assume one of the switches is Closed, what will be the current thru the 510R.?

E

 

Is the forward voltage of a silicon diode not approx 0.7V? (I got this from the internet) please feel free to offer some information rather than just criticising the help that E gave me.

Is the 510 ohms resistor a diode?? NO.
The resistor is fed +5V and is in series with a diode or two. Then it has a little more than 4V across it, not 0.7V.
With a little more than 4V across the resistor then its current is a little more than 4V/510 ohms= >7.8mA and the graph shows about 0.725V across one diode.

 

The resistor is fed +5V and is in series with a diode or two.

It is not in series with a diode or two, its in series with two diodes in parallel.

Your maths are also wrong, there is 4.3v across the 510R , so the current thru the 510R is 4.3v/510R= 8.4mA NOT 7.8mA.

 

I

The resistor is fed +5V and is in series with a diode or two.

It is not in series with a diode or two, its in series with two diodes in parallel.

Your maths are also wrong, there is 4.3v across the 510R , so the current thru the 510R is 4.3v/510R= 8.4mA NOT 7.8mA.

Thank you very much E, sorry it took me such a long time to get there. Out of interest - how did you end with exactly 4.3v?

 

hi jj,
Well you now you know that the nominal Vfwd for diode is 0.7v ,
So if you have a 5V supply its just a subtraction.

Its importnat to note that the Vfwd is a function of current thru the diode, higher current , higher Vfwd.
E

 

One question said to assume the "standard" forward voltage of a diode to estimate the current in the resistor and another question said to refer to the diode's graph of forward voltage vs current to see the exact forward voltage.

I said "about numbers" so that JJB could calculate the exact current but he wrongly guessed that the resistor has only 0.7V across it.

 

I said "about numbers" so that JJB could calculate the exact current but he wrongly guessed that the resistor has only 0.7V across it.

guru,
If the guy knew the answers to the query , he would not be here on AAC asking for help and hopefully some tuition , rather than Why only 0.7V? From magic?? sarcasm.

E