Problems with calculating circuit current with diode in parallel with resistor

[trespaser5]

I have a diode with a resistor in parallel(R2), but also diode resistor circuit has a resistor in series with it(R1). I have uploaded a drawing as well but I think it takes a bit of time for them to be authorised.
The values of the circuit are
Vs = 20v
R1 = 3kΩ
R2 = 470Ω
Diode = Has a Volt drop of 2v

I've found the formula Vo-Vd/Rl = I for a diode in series in a resistor but am not sure how this relates if the diode also has resistor in parallel with it. I'm guessing from what I've read that the diode does not have a resistance as such but only 'consumes' 2v, so can I use 18v along with the resistance in the remainder of the circuit to calculate It ? Or am I heading down the wrong path entirely. This is my second night on this question so could really do with some help. Thankyou

 

[MrAl]

Hi,

Can you upload a jpg or gif file instead of .doc?

 

[crutschow]

The diode drop ( it doesn't "consume" the volts, it exhibits a voltage drop) will also appear across the 470 ohm resistor so that means the drop across the 3k ohm resistor is indeed 18V.

 

[Ratchit]

trespaser5,

The values of the circuit are
Vs = 20v
R1 = 3kΩ
R2 = 470Ω
Diode = Has a Volt drop of 2v

Your schemat specifies a "bidirectional" diode. What's that? If a diode was bidirectional, it would not be a diode, would it? Do you mean something like two zener diodes connected back to back? Your schemat does not show that. Also you are showing the diode reverse biased.

Anyway, lets assume the diode is connected for conduction. The circuit without the diode will have a voltage of 2.71 volts across R470, so the diode will conduct and hold the voltage at 2.0 volts. That leaves 18 volts for the 3k resistor carrying 6 ma. The current will divide between the R470 resistor and the diode. 2.0/470 is 4.26 ma, so the diode will carry the remainder 6-4.26=1.74 ma. 2.0 volts/1.74 ma gives an internal diode resistance of 1k146 at that current.

Ratch

 

[meowth08]

Hi

I also want lo learn from this thread. Please post a clearer schematic. There are different ways to do it. You can do it in eagle and upload the schematic in pdf or use the simulators then upload an image. Please clarify the bidirectional diode. I don't know if there's such a thing. Well, I know of the zener diode. When forward biased, it acts like a normal diode. But normally, it is often in reverse biased mode for regulation purposes.

meowth8

 

[MrAl]

Hi,

Also, if the schematic is posted as a gif or jpg more people will be able to view it. Not all .doc files are viewable by everyone so that's a poor choice.

 

[WTP Pepper]

Many ways to solve this.
1. Kirchoff law as shown by Ratch above.
2. Thevenin equivalent circuits.
3. Cheating using LT Spice (but you need to know the basics to check the result).

Look on the diode in forward conduction mode as a voltage source pushing back and solve.
I assume I have interpreted the circuit correct as I get the same results, but cannot read you .DOC file posted.

I also assume the diode is an LED with such a high Vf at low currents.View attachment 64068

 

[MrAl]

Hi,


From the drawing provided by WTP Pepper the voltage at the junction of R2 and R1 with the diode open circuited is:
Vo=V1*R1/(R1+R2)

With the diode voltage drop being Vd, the diode conducts if the following condition is met:
Vo>=Vd

If that condition is not met then the voltage at the junction of R2 and R1 is Vo, but if that condition is met then the voltage at the junction is equal to the diode voltage drop Vd.

Since when V1 is 20 volts that condition is met, the voltage across the diode is equal to Vd which here is 2 volts.
With that voltage being 2 volts, the current through R2 is:
iR2=(V1-Vd)/R2
where here is:
iR2=(20-2)/R2

The current through R1 is:
iR1=Vd/R1

and so the current through the diode is:
iD=iR2-iR1

That should allow you to calculate whatever you need for this circuit.

 

[trespaser5]

Thank you all ever so much, sorry about posting the .doc file. Now I understand that the voltage drop is equal to the volts across the diode I managed to get the answer. Thanks for all your time