# Diode current textbook problem

#### starLED

##### Member

Calculate current Id through diode D and output voltage Uiz,
for following values of input voltage Uul:
a) Uul = 6V
b) U
ul = -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?

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##### Member
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

Regards, Dana.

#### starLED

##### Member
The input V is an arbitrary V source in this problem.
Does input V adds to E?
Meaning, total voltage is Uul+E=6+4.5=10.5 V?

##### Member
KVL states Uul = Vd + Id x R + (-E) = 0.

Regards, Dana.

#### Visitor

##### Well-Known Member
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.

#### starLED

##### Member
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.

##### Member
When the ideal diode conducts its V = 0, so Uul is impressed
across the R and E, which is = Uiz.

Regards, Dana.

#### starLED

##### Member
When the ideal diode conducts its V = 0, so Uul is impressed
across the R and E, which is = Uiz.
Ideal diode has 0 voltage drop, I get that, but still I don't get Uiz=6V.
Can you explain bit more?

#### Pommie

##### Well-Known Member
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.

#### starLED

##### Member
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.
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?

#### Pommie

##### Well-Known Member
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.

##### Member
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.

#### starLED

##### Member
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.
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.

#### Pommie

##### Well-Known Member
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.

#### starLED

##### Member
If I understand correctly, these are the voltages in circuit:

##### Member
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.

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#### starLED

##### Member
I understand why is 1.05 mA, I don't understand why Uiz is 6 V and not 10.5 V.

##### Member
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.