Hello guys. i am wondering about the voltage across this diode... can you give me a hint if i am right or not :?
the voltage across a diode is usually .7V if we short the 8V power source it has .7 V if we short 5V source it has 8V because it acts as a reverse biased. if i substract them the voltage is 7.3V . The other thing that i am thinking is 8V - 5V = 3V across the diode ... can you please help me ?

 

The diode has a small reverse leakage current. That current flows through the 10 ohm resistor (pushed by 3 V). So the voltage across the diode is 3 V minus the IR drop across the resistor.

ak

 

The diode has a small reverse leakage current. That current flows through the 10 ohm resistor (pushed by 3 V). So the voltage across the diode is 3 V minus the IR drop across the resistor.

ak

can you help me the IR drop ? what exactly is that and does it changes the result ? Sorry ... a bit new in electronics :/

 

@A.Zanev
Current will flow from high voltage to low voltage. Current can only flow in the direction of the diode arrow (and is blocked for the reverse direction). Therefore, no current can flow when the batteries and diode are connected in this fashion.
For question
a) The voltage is 8 - 5 = 3 volts.
b) the diode is reverse biased.

 

and as a reverse biased diode has almost infinite resistance the total resistance is infiit so the current through the ressistor is extremely small and the IR drop across the resistor is close to 0.0

thank you GopherT for the answer!

 

In my experience if a homework question like this doesn't explicitly state "real-world" device models, things like reverse leakage current aren't meant to be considered.

 

In my experience if a homework question like this doesn't explicitly state "real-world" device models, things like reverse leakage current aren't meant to be considered.

That's why it is called "book learning."

 

Hello guys. i am wondering about the voltage across this diode... can you give me a hint if i am right or not :?
the voltage across a diode is usually .7V if we short the 8V power source it has .7 V if we short 5V source it has 8V because it acts as a reverse biased. if i substract them the voltage is 7.3V . The other thing that i am thinking is 8V - 5V = 3V across the diode ... can you please help me ? View attachment 80380

It sounds like you are trying to use superposition in some way here (shorting each supply in turn and combining the results). Superposition only works with linear circuits and diodes are nonlinear components, so superposition doesn't apply.

It's not enough to say that there is 3V across the diode because it doesn't indicate what the polarity is (now, combined with the answer as to whether it is forward-biased or reverse-biased, the polarity is indicated).

 

can you help me the IR drop ? what exactly is that and does it changes the result ? Sorry ... a bit new in electronics :/

The fact that you aren't given any information about the reverse leakage current implies that it can be neglected for this problem. It can be neglected in almost all "real world" applications as well (but not ALL of them). For a diode capable of carrying a few amps of forward current, the reverse leakage current is commonly a few dozen microamps. So let's call it 100uA. Through a 10Ω resistor you are looking at 1mV of drop across the resistor. It's negligible.

 

It can be neglected in almost all "real world" applications as well (but not ALL of them). For a diode capable of carrying a few amps of forward current, the reverse leakage current is commonly a few dozen microamps. So let's call it 100uA. Through a 10Ω resistor you are looking at 1mV of drop across the resistor. It's negligible.

You have to get a lot better than you are now to chase microamps through a circuit, but you will get there. I've been down in the nanoamp range a couple of times in 40 years.