Please have a look on these diode circuits

[PainterGuy]

hello everyone,

i am in disagree with someone over the calculations in the circuits i will discuss with you.

circuit #1:

here is the diagram:-
http://img156.imageshack.us/img156/2734/diodecircuit.jpg

do you find it correct? please check it carefully. the circuit will only conduct between 2.3v-10v. frequency is 2hz and used diode is silicon so therefore voltage drop of 0.7v.

circuit #2:

here is diagram:-
http://img34.imageshack.us/img34/1634/diodecircuit2.jpg

do you find it correct?

5v = -0.7 + 2v -Vr
5v = 1.3 -Vr
-Vr = 5 - 1.3
Vr = -3.7

Vr is voltage drop across the resistor. i don't understand why answer is "-3.7"?

i am very grateful if you can tell me if these circuits are corrent... your approval means the other person i am having argument is wrong. many thanks.

cheers

 

[u_know_who]

Both of the calculation is wrong. At first circuit diode will conduct not from 2.3v. You set the n voltage lower than the p voltage. So the diode will conduct before 2.3v.

And second circuit the calculation is also wrong. You did wrong while applying KVL.

 

[PainterGuy]

Both of the calculation is wrong. At first circuit diode will conduct not from 2.3v. You set the n voltage lower than the p voltage. So the diode will conduct before 2.3v.

And second circuit the calculation is also wrong. You did wrong while applying KVL.

would someone please explain why the circuits are wrong? please

 

[Averagesupernova]

The AC driven circuit: An AC voltage source without a peak to peak spec is usually assumed to be RMS. Your graph does not reflect this.
-
The DC driven circuit: Voltage across the load resistor will be 6.3 volts. Do you dispute this? If so, why?
-
I will explain why, but I want your take on it first.

 

[vk6kro]

In each case, the result should be more obvious if you move the 3 V or 2 V battery to the other side of the diode. The current flowing will not be changed by doing this.

You can then see that, in the first diagram, the battery produces +3 volts on top of the AC waveform, so the diode will conduct sooner in the AC cycle than it would with just the AC.

In the second circuit, the two batteries just add their voltages to give 7 volts.

 

[PainterGuy]

The AC driven circuit: An AC voltage source without a peak to peak spec is usually assumed to be RMS. Your graph does not reflect this.
-
The DC driven circuit: Voltage across the load resistor will be 6.3 volts. Do you dispute this? If so, why?
-
I will explain why, but I want your take on it first.

many thanks everyone and Averagesupernova for all your help.

yes i dispute this 6.3 volts out of my limited knowledge!

i think of a diode as a 0.7v battery which only let the current flow in one direction.

1:- in the first AC circuit during +ve half cycle the diode will let the current pass. there would be drop of 0.7v (in equation it would be -0.7) across diode. then as current flows from -ve terminal toward +ve terminal therefore it would rise in voltage of 3v. the all the volts remaining will drop across Rl and as the current flows from more +ve toward less +ve therefore it -ve drop across Rl. Now according to KVL applied voltage equals voltage drops across the circuit. in this example so fat i have consider this an DC current in fact it's AC siglal because even if it was dc circuit current will not flow until it is above 2.3v as in the diagram i show current will start flowing when voltage is just above 2.3v.


2:- in the second diagram (that of dc powered). there is going to drop of 0.7v across diode. then there is rise of 2v across the battery (the battery will be recharges in this way). then the remaining volts will drop across the resistor. 5v = -0.7v + 2v - Rv => 5v = 1.3 - Rv => Rv = -3.7.

now i have explained how i reached my calculation. i hope you can know where gaps are in my understanding, where i am making mistake. please now lead me to correct way. many thanks.

cheers

 

[vk6kro]

Start at the bottom rail in the second diagram (marked with a ground symbol).

Relative to this, the top of the 5 V supply is +5 V.

Then there is a drop of 0.7 V across the diode to give 4.3 volts.

Then there is an increase of 2 volts across the 2 volt battery to give 6.3 volts which is the voltage across the resistor.

[PLAIN]http://dl.dropbox.com/u/4222062/cct.PNG
-------------------------------------------------------------------------
In the first diagram, consider the voltage from the AC supply when it is at -2.3 volts relative to the bottom rail.
There is a drop of 0.7 volts across the diode to give -3 volts. Then there is an increase of 3 volts from the 3 V battery to give 0 volts out.
When the AC supply gets to -2.2 volts then there will be a drop of 0.7 volts across the diode to give -2.9 volts. Then there will be an increase of 3 volts from the battery to give 0.1 volts across the resistor.

So, the diode starts to conduct at -2.3 volts from the AC supply.

[PLAIN]http://dl.dropbox.com/u/4222062/cct2.PNG

 

[PainterGuy]

In the first diagram, consider the voltage from the AC supply when it is at -2.3 volts relative to the bottom rail.
There is a drop of 0.7 volts across the diode to give -3 volts. Then there is an increase of 3 volts from the 3 V battery to give 0 volts out.
When the AC supply gets to -2.2 volts then there will be a drop of 0.7 volts across the diode to give -2.9 volts. Then there will be an increase of 3 volts from the battery to give 0.1 volts across the resistor.

So, the diode starts to conduct at -2.3 volts from the AC supply.

[PLAIN]http://dl.dropbox.com/u/4222062/cct2.PNG[/QUOTE]

i am very much grateful to you, vk6kro.

sorry, i don't understand this AC problem. the AC voltage in its +ve half cycle starts from 0v peaks to 10v then again to 0v, and then in second -ve half cycle starts from 0v peaks to -10v then again to 0v. one cycle completes here. time take taken is 0.5s.

i think diode only let's the current during the +ve half cycle. because during this +ve cycle, AC supply can be considered to be equivalent to variable DC power supply which can supply voltage from 0v to 10v. please check the attached diagram.

to this point my limited knowledge only confirms this that from 0v-0.7 the diode will not let current pass trough it. why i think so? think of diode as 0.7v battery with its +ve terminal toward +ve terminal of battery (for sake of simplicity i have assumed during +ve half cycle AC behaving as variable DC supply). so the DC power supply should push with greater pressure than 0.7V to get across the diode. i hope now you can see where the problem lies in my understand. now guide me to right solution please. many thanks for the help.

cheers

 

[vk6kro]

The diode doesn't know if it is in a circuit with negative voltages. The only thing that matters is if the anode is 0.7 volts more positive than the cathode..


So, if you just remove the resistor for a moment, the cathode of the diode will be at MINUS 3 volts because of the 3 V battery. So if the anode is at a greater voltage than (MINUS 3 PLUS 0.7 Volts), then the diode will conduct.
MINUS 3 Volts PLUS 0.7 Volts is MINUS 2.3 volts.

So, both ends of the diode are negative, but it will still conduct because the anode is less negative than the cathode.

It may make your head hurt for a while, but draw a diagram and see how it works.

 

[PainterGuy]

In the first diagram, consider the voltage from the AC supply when it is at -2.3 volts relative to the bottom rail.
There is a drop of 0.7 volts across the diode to give -3 volts. Then there is an increase of 3 volts from the 3 V battery to give 0 volts out.
When the AC supply gets to -2.2 volts then there will be a drop of 0.7 volts across the diode to give -2.9 volts. Then there will be an increase of 3 volts from the battery to give 0.1 volts across the resistor.

So, the diode starts to conduct at -2.3 volts from the AC supply.

[PLAIN]http://dl.dropbox.com/u/4222062/cct2.PNG[/QUOTE]

hi vk6kro,

many thanks for this help. the important thing learned so far:- the potential difference across should be +0.7V or more and the anode should be more positive compared to the cathode.

okay, as you say (though it will take some time to convince my owl self), the diode will start conducting at -2.3V (because -2.3 is 0.7 degree more positive than -3). the diode will not conduct at -2.4, -2.5, -2.6, ..., -10 (because in case some voltages such as -2.4 the potential difference would not be +0.7V (for example, -2.4 is 0.6 degree positive than -3) and in some voltages anode of the diode will be more negative comapred to cathode. so the diode will conduct through all positive half cycle of AC supply and it will also conduct during negative half cycle from 0V to -2.3. do you think it is correct? please tell me.

another question: the voltage drop in case of silicon diode is 0.7V. in solving problems do we take 0.7V as the voltage when diode starts conducting or as the voltage when diode is on brink of conducting?

i'm very much grateful to you for teaching me this. hope all my confusion settles down soon.

cheers

 

[vk6kro]

Yes, I think you have that correct.

A diode is not really like a battery. It is more like a very stubborn resistor which changes resistance to always have the same voltage across it, even if the current through it changes widely.

There is an area of diode operation where you have to be careful, though.

If the possible current is a few microamps, due to there being large resistors in circuit (say more than 500 K), then you might use 0.5 volt for the diode drop (assuming silicon diodes).
If there is more than 1 mA flowing, then you can use 0.6 volts or 0.7 volts.
Power diodes will drop 1 volt or more when they conduct currents of 1 amp or so.

Often a question will tell you what voltage to use for the diode. Otherwise, you just say what voltage you intend to use.

Below 0.5 volts, a silicon diode acts like it was reverse biased. There may be some small current flowing, but it is effectively open circuit.

 

[PainterGuy]

Start at the bottom rail in the second diagram (marked with a ground symbol).

Relative to this, the top of the 5 V supply is +5 V.

Then there is a drop of 0.7 V across the diode to give 4.3 volts.

Then there is an increase of 2 volts across the 2 volt battery to give 6.3 volts which is the voltage across the resistor.

[PLAIN]http://dl.dropbox.com/u/4222062/cct.PNG

hello vk6kro,

in the past i used to make calculations using KVL (which says, applied voltage = sum of voltage drops) this way:

5V = -0.7 + 2V - V_r

i took 0.7V because current is flowing from high to low potential. then i took +2V because current flows from low to high potential. and while traversing the resistor it goes from high to low potential. i don't understand why my analysis is incorrect? would you help me on this please?

yes, your analogy of considering a diode a stubborn resistor is very good. i like it.

i don't think one can use KVL in case AC.

suppose voltage drop across silicon diode is 0.7V then does it mean for the current to flow the voltage should be "equal to or greater than 0.7V" or "greater than 0.7V". tell me please.

i'm very much grateful to you for helping in many problem. you are very kind.

cheers

 

[vk6kro]

5V = -0.7 + 2V - V_r

I don't know how you got this.

Can you see that the total EMF in this circuit is 5V + 2 V or 7 volts?
This 7 volts is dropped across the diode and the resistor

So, 5 V + 2 V = 7 V = Vr + Vd.
But Vd = 0.7 volts so the voltage across the resistor is 6.3 volts as we got earlier.

suppose voltage drop across silicon diode is 0.7V then does it mean for the current to flow the voltage should be "equal to or greater than 0.7V" or "greater than 0.7V". tell me please.

The diode will conduct if the voltage across it is greater than about 0.5 volts. As the current increases due to resistors in the circuit being reduced in size, the voltage across the diode gets greater, reaching a maximum of about 1 volt at large currents.

 

[PainterGuy]

5V = -0.7 + 2V - V_r

I don't know how you got this.

Can you see that the total EMF in this circuit is 5V + 2 V or 7 volts?
This 7 volts is dropped across the diode and the resistor

So, 5 V + 2 V = 7 V = Vr + Vd.
But Vd = 0.7 volts so the voltage across the resistor is 6.3 volts as we got earlier.

hi vk6kro,
thank you for your help again. I'm very much sorry, now it is revealed i was applying KVL in wrong way. i was including only that emf source where i started to traverse the circuit. in previous post i started at 5V so i only include it on left hand side. sorry for the confusion. this was wrong. now it is clear to me KVL is:
sum of all EMFs = sum of all IR terms

many thanks for teaching this stuff. you are really nice and kind.

cheers