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Load Line Analysis - Diode

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GilbertSam

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Hi All,

Please help on how to get the diode current and voltage across the load resistance?

As reference to the graph, one is for load another is for diode but how to get the solution as above question?

Urgently seek all your kind help.

Thanks & Best Regards,
Gilbert
 

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You can't get the solution unless you have the V-I curve of the particular diode you are using.
 
since your souce is 5V the 1st graph is wrong for voltage drop across the diode Vs current through the resistaor, is has to come from 5-5 not 4-4.

the intersection gives you the diode/ load current and diode voltage, the voltage across the load =Vs-Vd.
 
Hi mbarazeen and All,

I have checked with my tutor and this is confirmed that the Question nothing wrong.

Please help....

Thanks & Best Regards,
Gilbert
 
Hi Ron,

The IV curve is provided there. Can I say the VD is 3V?

hi,
Look at this image, the 'blue' line is the load line for the 1K without the diode in circuit and the 'red' line is the load line with the diode in circuit.

You should now be able to see the voltage drop across the diode and resistor.:)

untitled.JPG
 
Hi Eric,

I know that when ID = 0A, VD = VBB = 5V while when VD = 0V, ID = VBB/R = 5mA

That's why we have the load line 5V and 5mA. I have go though internet and reference material, they just mentioned 5V-5A load line. The other load line whereby 4V-4mA is weird for me. I really can't figure out why it exist here??

If what you say that 4V-4mA is for the load line for diode, that's mean the diode current should be 4mA while the voltage across load resistance is VR = ID x R = 4mA ( 1k ) = 4V

Let say I would like to get he power dissipated in the diode is that mean P = ID VD = 4mA ( 4V ) = 16mW

Please correct me if I'm wrong,

Really need your kind help on this,

Thanks & Best Regards,
Gilbert
 
i also have no idea about that 4V-4mA load line, i thought its your solution,
the charecteristic of the diode and the load line of 5V is enough for the solution. the intersection gives you the current and diode voltage for that particular resistance.
 
Hi Eric,

I know that when ID = 0A, VD = VBB = 5V while when VD = 0V, ID = VBB/R = 5mA

That's why we have the load line 5V and 5mA. I have go though internet and reference material, they just mentioned 5V-5A load line. The other load line whereby 4V-4mA is weird for me. I really can't figure out why it exist here??

If what you say that 4V-4mA is for the load line for diode, that's mean the diode current should be 4mA while the voltage across load resistance is VR = ID x R = 4mA ( 1k ) = 4V

Let say I would like to get he power dissipated in the diode is that mean P = ID VD = 4mA ( 4V ) = 16mW

Please correct me if I'm wrong,

Really need your kind help on this,

Thanks & Best Regards,
Gilbert
You really need to find out the reason for the 4V - 5V discrepancy.
 
Well, if you can't find out why the graph shows a 4V supply, while the schematic shows a 5V supply, my best guess is that you are supposed to draw the 5V/1k load line and graphically solve for the diode voltage and current. Below is how I would do it.
I get about 3.7mA and 1.3V. It could be 3.8mA and 1.2V. In any case, Vd+Id*1k=5V.
 

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Hi Ron,

It makes sense to do this but we need the same scale of graph to draw this.

Btw, anyone still got other idea?

Please advice,

Thanks ....
Gilbert
 
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