Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Diode states.

Status
Not open for further replies.

SeanHatch

New Member
Today in class we learned about diodes. Our prof explained to us that the only manual way to find the correct states of the diodes is guess and check. Each time, we ended with one correct circuit; that is Kirchoff's laws coincided with the diode constraints.

Now I'm doing this circuit on my own, and I'm finding that there are 2 combinations which fulfill Kirchoff's laws. Is this possible? or must I be making a mistake somewhere?

The circuit is simple really. DC sources resistors and diodes.
 
What is a diode state?
If a diode is forward biased then it conducts, and if it is reverse biased then it doesn't conduct. That's all.
 
Try getting some sleep and then start from scratch again without looking at your old work. Assume it's forward biased and do the calculations and check to see if the voltage across the diode actually is correct. If it's not forward biased then assume it's reverse bias and then solve the circuit again and check to see if the voltage across it would reverse bias it.

if you get positive checks both times, you did something wrong. Are you sure you understand what it means for a diode to be forward biased? The voltage from the anode to cathode must be greater than 0.7V (I am assuming that is the number you are using).
 
Last edited:
I'm pretty sure I understand the 'ideal' diode fairly well. We are using 0V as the forward biasing point because we are speaking in 'ideal' diodes.
 
Attach the schematic to your reply here. You don't need to host it somewhere else.
 
Here it is.
 

Attachments

  • Untitled-1.jpg
    Untitled-1.jpg
    771.5 KB · Views: 138
The bottom "state" drawing is correct. The diode in the middle blocks the 5mA at the top from returning to the 5V battery.
 
You can't just look at the currents you calculated and go "oh! it's flowing forward through the diode so the diode is conducting!". Remember that you >>>ASSUMED<<< the diode was conducting so you shorted it and based your V=IR calculations based on that modified circuit. Of course the currents will satisfy V=IR since that's what you used to calculate them! Of course the current will flow through the diode since you pretty much just drew the diode back into the picture while still ignoring it in your math!

What you must do is use your calculated currents and go back to your circuit and calculate the voltage drop across each and every diode. If there is a postive voltage drop across the diode, and it is high enough to exceed the forward voltage drop of the diode, then it conducts and you are right- if not, you must change your assumption and redo your calculations. The currents tell you nothing about whether the diode is actually forward biased or not- it's the voltage across the diode that you need to know.

If you calculate the voltage at D1's anode, you see that it is zero since it starts (from the 5V source) at 5V and then there is a 5V drop across the resistor because of the currents you calculated with your assumption. The result is 5V-5V = 0V at D1's anode. And then from seeing that the 10V is connected directly to D1's cathode, you see that there is
5V-10V = -5V across the diode. -5V is less than the forward voltage drop of the diode (assumed to be an ideal 0V in your case). Therefore D1 is not forward biased and not conducting (not what you assumed). So you do the calculation again with D1 assumed to be non-conducting. You are supposed to do this for D2 as well to check.

2 diodes = 4 possible combinations = 3 changes to get your assumptions wrong! = 3 times maximum to redo your calculations!

That's the key to these problems (it's also the only thing that makes these problems different from source-resistor problems). Check the diode voltages with the currents you calculated from your assumption. It's not the current that matters, it's the voltage.
 
Last edited:
Ok, one more thing. If the node where the two diodes meet is indeed 10V, why isn't the current moving from there to the battery?; how does it still flow from the 5V battery to the 10V node?

Man, you totally need to be much more careful with analysis with non linear elements!
 
It doesn't. You don't need to current to have a voltage difference.

If you look at your correct schematic, notice the unconnected resistor? What is the voltage at the open end of the resistor? It's 5V! That's because one end is connected to the 5V source, but the resistor is not part of a closed circuit and no current flows in it therefore no voltage drop across it!

Funky huh? You are probably going to need know that when you do Thevin and Norton's thereom.
 
Last edited:
yes but if you have a difference current will flow, right? (unless there's a diode haha ok lame) ?

so basically the current i ASSUMED was incorrect? I think I'll just be safe if I check that both the voltage AND current is correct for each of my assumed states; I think that's what I've not been doing.
 
Last edited:
No no. The check itself IS the calculation of the voltage across the diodes. It's not just being safe, it is necessary to finish the problem. You must do this to see if your assumption is right. Your current calculations will ALWAYS agree with whatever you assumed because you removed the diodes when you make the assumptions. After you stick the diodes back into your circuit, you must check that the voltage conditions produced by your assumed currents still produce the necessary voltage bias for conduction (or blocking) of the diode.

A voltage difference will try to drive current to flow, but if the resistance is is infinite like in an open circuit no current will flow, even if there is a voltage difference. You don't need current flow for a voltage difference.
 
Last edited:
true true. but in this case, there is a 10 volt node with a 1k resistor followed by a 5V node from the battery. I found that the current flows from the 5V to the 10V node. that doesn't make sense to me.

I appreciate your help alot by the way. I know you're busy and probably upset with me so you need not reply anymore haha. thanks again.

Actually I think your last edit to your last post helped a whole ton. Thanks alot.

/END.
 
This is what I am saying- the reason you found that it flowed that way was because you replaced the diode with a short-circuit when you assumed it was conducting. It's a different circuit than the original one you started with. You did not take into account the diode in this new circuit.

After calculating the assumed currents, you stick the diode back into the circuit. So now the circuit is different from the one you used in your calculations. Now you must check to see if those currents still produce the right voltage around the diode for it to actually conduct (and become a short-circuit, ideally).

EDIT: Replied before you edited.
 
Last edited:
diode circuit

diode
The circuit can be solved by the LUCKY choice of case_2
NO further examination would be necessary since the circuit can have only 1 state.

HOWEVER, the states of the diodes was ASSUMED:
If cases_1, case_3, case_4 are assumed before case_2,
then these cases must be shown to be false assumptions

A circuit with multiple battery sources and 100 or 10,000 diode,resistor, would
require a computer program to solve the many equations involved.!
-----------------------------------------------------------------
refer to your ATTACHMENT

eq
a1 [1k 0][i1] = [ 5-Vd1+Vd2] // where i1,i2 are CW
b1 [0 1k][i2] [10-Vd2] // currents in loop A,B

a2 i1*1k + i2*0 = [ 5-Vd1+Vd2]
b2 i1*0 + i2*1k = [10-Vd2]

c1 Id1 = i1
d1 Id2 = -i1 +i2
-----------------------------------------------------------------

case_1
assume diode_1 conduct;
diode_2 cutoff so equation d1 above becomes
eq
1-d1 Id2 =0

1-a2 i1 =5v/1k =5ma
1-d2 gives Id2 =0 = -i1 +i2

1-d3 i2 =i1 =5ma !

1-d3 gives diode_2 conducts, BUT this contradicts assumption that diode_2 cutoff
therefore< diode_2 cutoff is a false assumption
and equation 1-d3 is not true
-----------------------------------------------------------------------
case_2
assume diode_1,diode_2 both conduct:
: : : :
: : : :---> NO contradiction with assumption
therefore diode_1,_2 both conduct
-----------------------------------------------------------------
case_3
assume diode_2 conducts,diode_1 cutoff
: : : :
: : : :----> contradicts this assumption
-----------------------------------------------------------------
case_4
assume diode_1,diode_2 both are both cutoff
: : : :
: : : :----> contradicts this assumption
-----------------------------------------------------------------
hawk2eye
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top