Diode current textbook problem

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starLED

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I am learning about diodes from a textbook, and I need help about this problem.

Calculate current Id through diode D and output voltage Uiz,
for following values of input voltage Uul:
a) Uul = 6V
b) Uul = -6V
Other given values are: E = 4.5 V and R = 10 kΩ.
Diode is ideal.

I am having trouble understanding what is input voltage Uul.
If you have E = 4.5 V, wouldn't Uul also be 4.5 V?


 
Last edited:
A diode is a non linear device, one would solve this with a load
line analysis.



The input V is an arbitrary V source in this problem.

Does the problem state Vdiode is some value ? If so then
its a standard KVL problem. But if problem wants you to
use diode equation as model from diode load line fastest
way to get answer.

Regards, Dana.
 
I think Dana, while correct, is far beyond the scope of the problem. It's stated "The diode is ideal", which removes all those pesky real-world properties. An ideal diode conducts perfectly without voltage drop when forward-biased and blocks completely when reverse-biased.
 
Solution for a) from the textbook is:
\[ U_{ul}-RI_{d}+E=0\\\\ I_{d}=\frac{U_{ul}+E}{R}\\\\ I_{d}=\frac{6+4.5}{10000}=\frac{10.5}{10000}=1.05\:mA\\\\ U_{iz}=U_{ul}=6\:V\]

I don't understand why is Uiz (output voltage) = 6 V.
 
When the ideal diode conducts its V = 0, so Uul is impressed
across the R and E, which is = Uiz.



Regards, Dana.
 
If the input voltage is 6V then the output must be 6V as there is zero across the diode. When the input voltage is -6V then the diode doesn't conduct so the output voltage is E (4.5V) as that's the only thing powering the output.

Mike.
 
Pommie said:
If the input voltage is 6V then the output must be 6V as there is zero across the diode. When the input voltage is -6V then the diode doesn't conduct so the output voltage is E (4.5V) as that's the only thing powering the output.

Mike.
Click to expand...
But isn't the total voltage 6+4.5=10.5?
Meaning, 10.5 V is input voltage accros the diode, so why isnt Uiz=10.5 V?
 
The two power sources aren't in series (wrong, see later in post) so don't add together. When the input is 6V then a current will flow through the resistor and charge the battery. Actually, just realised the battery is in "backwards" - negative up. However, all this means is when 6V is the input then 6+4.5V will be dropped across the resistor and the battery will be (slightly) discharged - the output will still be 6V. When the input is -6V then the output will all come from the battery so will be -4.5V.

Mike.
Edit, just noticed in the textbook answer they work out the current through the resistor (and diode) by using 10.5/10000.
 
Simply write KVL when diode conducts, with 0V across it because it is "ideal".

So Uul - Vdiode - Uiz = 0 or Uul - Uiz = 0 or Uul = Uiz


Regards, Dana.
 
danadak said:
Simply write KVL when diode conducts, with 0V across it because it is "ideal".

So Uul - Vdiode - Uiz = 0 or Uul - Uiz = 0 or Uul = Uiz


Regards, Dana.
Click to expand...
So where does Id = 1,05 mA comes from ?
I am confused with these 2 power sources Uul and E, I can't figure out what exactly is going on with voltage.
 
The text book answer calculates the current through the resistor and diode (ideal meter too). It uses ohms law, v=ir, or, switch it around, i=v/r. So, 10.5 volts across a 10k resistor is 1.05mA.

Mike.
 
Starting at Uul ground node Uul - Vdiode -Idiode x R +E = 0

But when Vdiode conducts it = 0

Then Uul -Idiode x R + E = 0 0r Idiode = (Uul + E) / R = (6 + 4.5) / 10K = 1.05 mA

Keep in mind Uiz is not an energy source, it is simple symbolic points where V is measured,
so there is no KVL loop associated with it and Uiz. its just measurement points in circuit.


Regards, Dana.
 
Last edited:
Uiz - Idiode x R + 4.5 = 0 when the ideal diode is conducting
Uiz = Idiode x R - 4.5 = 10.5 - 4.5 = 6


Regards, Dana.
 
starLED said:
I understand why is 1.05 mA, I don't understand why Uiz is 6 V and not 10.5 V.
Click to expand...
If you understand this then you understand why the voltage across the resistor is 10.5V. So just using the resistor battery leg of your circuit, the voltage across the resistor is 10.5V, the voltage across the battery is -4.5V (note minus). Add them together and you get 6V.

Mike.
 
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