Help Understanding Circuit Requirements for Conducting Diode

Click For Summary
The discussion revolves around understanding the circuit requirements for a diode to conduct, specifically the voltage at point X, which must be at least 0.6V. The equation in question involves the expression (0.6 - (-10))/2.2x10^3, prompting confusion about why 10 is subtracted from 0.6 instead of the other way around. Participants emphasize the importance of applying Kirchhoff's Voltage Law (KVL) to clarify the current flow and polarities in the circuit. It is suggested that confirming one's intuition with KVL can help resolve such misunderstandings. The conversation concludes with an affirmation of the correct approach to analyzing the circuit.
zsolt2
Messages
19
Reaction score
2
Hi all. I need some help with the circuit attached.
For the diode to conduct, at point X the voltage must be 0.6V minimum. There is an equation to find the minimum required voltage input, Vin to give 0.6V at X. What I don't understand is the right hand side of the equation; why is it:
(0.6-(-10))/2.2x10^3
why subtract 10 out of 0.6 and not the other way around? I understand that ((Vin-0.6)/10^3) is the current developed in the 1k resistor and (0.6-(-10))/2.2x10^3 is the current in the 2.2k resistor. So the 0.6V is comming from the 1k towards the diode and the 2.2k resistor where it meets with -10V. Shouldn't the 0.6 be subtracted from the -10V then?
what I thought is that it supposed to be:
((Vin-0.6)/10^3) + (-10-0.6)/2.2x10^3
((Vin-0.6)/10^3) + -10.6/2.2x10^3
((Vin-0.6)/10^3) - 10.6/2.2x10^3
((Vin-0.6)/10^3) = 10.6/2.2x10^3

It will give the same result as the equation provided. Can you explain the circuit a bit and that where did I go wrong?
Thank you!
 

Attachments

  • diode.PNG
    diode.PNG
    23.6 KB · Views: 435
Engineering news on Phys.org
zsolt2 said:
What I don't understand is the right hand side of the equation; why is it:
(0.6-(-10))/2.2x10^3

Always go back to your basics
draw in currents and polarities then write KVL

KVLdemo .jpg
 

Attachments

  • KVLdemo .jpg
    KVLdemo .jpg
    37.1 KB · Views: 512
zsolt2 said:
What I don't understand is the right hand side of the equation; why is it:
(0.6-(-10))/2.2x10^3

upload_2018-12-28_14-41-55.png


looks to me like an addition not a subtraction
and i think that's what you said
zsolt2 said:
Shouldn't the 0.6 be subtracted from the -10V then?

Your intuition was right
form the habit of confirming your intuition with KVL

old jim
 

Attachments

  • upload_2018-12-28_14-41-55.png
    upload_2018-12-28_14-41-55.png
    1.2 KB · Views: 465
Thank you Jim!
 
  • Like
Likes jim hardy
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

Similar threads

  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
3
Views
1K
Replies
10
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
93
Views
8K
  • · Replies 9 ·
Replies
9
Views
8K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 9 ·
Replies
9
Views
2K