Hey friends, welcome to the YouTube channel

ALL ABOUT ELECTRONICS. Now, in the previous videos of the diode,

we understood that what is a diode and we had also seen the V-I characteristics of the

diode. Now suppose, if this diode is connected in

a particular circuit then, first of all, we need to find that whether this diode is conducting

or non-conducting. And if it is conducting then what is voltage

and current in that particular circuit. So, in this video, let’s learn that how to

solve the diode circuits and how to find the voltage and current in circuits which contains

these diodes. Now, there are two ways by which we can solve

this diode circuits. The first is a graphical method in which using

the load line we can find the voltage and current in the circuit. And the second method is the diode approximation. So, first let’s see how we can solve this

circuit using the graphical method. Now, like I said, in this graphical method

using the load line we can find the voltage and current across the diode. But first of all. let’s understand that what is load line. So, here we have a very simple circuit in

which some DC voltage is applied to this circuit and some current limiting resistor R is connected

in series with this diode. Noe, here let’s assume that the applied voltage

is sufficient that it will forward bias this diode. And let’s assume that the voltage across this

diode is equal to Vd and the current that is flowing through this circuit is equal to

Id. So, now if we apply the KVL, then we can write

this voltage V is equal to Vd+ ID*R. So, if you see this equation, we have two

variables. The voltage across this diode Vd and the current

that is flowing through this diode. That is equal to Id. So, from this expression, we can find the

line on this Id Vs Vd curve and we can get the different possible values of Vd and Id. So, for example, if we put Vd is equal to

zero, in that case, we will get the extreme value of this Id. That is equal to V divided by R.

So, we will get one point on this Y- axis. That is equal to V/R.

And similarly, if we put Id is equal to zero, in that case, we will get one point on this

horizontal axis. And that will be equal to Vd is equal to V

volt. So, we will get one more point on this X-axis. And by connecting these two points, we will

get the load line for the given circuit. So, this intersection point will give us the

operating voltage and current for the diode. So, in this way, using this load line we can

find the operating voltage and current for the diode. Although this graphical method gives us very

accurate values of voltage and current, it very time-consuming process. So, there is another way by which we can not

only quickly solve the circuits but we can even troubleshoot the circuits. And this method is diode approximation method. Now, in the introductory video of this diode,

we had seen the V-I characteristic of the ideal diode. So, if the diode is forward biased in that

case it will simply act as a closed switch. On the other end, in the case of reverse bias,

it will act as an open switch. Thereafter we had seen the second approximation

in which we had assumed that the diode will start conducting only when the applied voltage

crosses some threshold voltage. And if the applied voltage is less than this

threshold voltage, in that case, it will act as an open switch. Thereafter we had also seen the third approximation. Where we have also included some series resistance

of this diode. Now, in most of the circuits, the current

limiting resistor which is connected in series with the diode used to be much larger than

the diode resistance. So, in most of the cases, we can neglect this

diode resistance. So, in our analysis, we will use the second

approximation. Where we will assume that whenever the diode

is forward biased, then the voltage across the diose will be equal to 0.7 V and whenever

it is reversed biased then simply it will act as an open switch. So, using this approximation let’s solve some

diode circuits. And let’s start with a very basic example. So, in this circuit, the 5V of voltage is

connected to this diode. And here we also have 1Kilo-Ohm of a current

limiting resistor. So, let’s find out the voltage and current

for the given circuit. Now, first of all, we need to identify that

whether this diode is conducting or non-conducting. And that can be found by finding the Thevenin’s

equivalent voltage across this diode. Now here, if the applied voltage across this

anode and cathode of the diode is more than 0.7 V, in that case, we can say that the diode

is conducting. And here, we are assuming that the diode that

is connected is silicon diode. In case of a germanium diode, the threshold

voltage used to be around 0.3V. So, in the case of a germanium diode, the

diode will start conducting whenever the applied voltage across the diode is more than 0.3

V. Now, here as the diode is conducting let’s

represent it by its equivalent circuit. And here let’s assume that the current that

is flowing through the circuit is equal to Id. So, this current Id will be equal to 5V minus

0.7 Volt divided by 1Kilo-Ohm. That is equal to 4.3 mA. So, this is the current which will flow through

this particular circuit. And the voltage across the diode will be equal

to 0.7 V. Now, in this circuit, if we assume that the

diode that is connected is silicon diode, and the applied voltage is 0.5 V, in that

case, it will not cross the barrier of this silicon diode. So, the circuit will act as an open circuit. And hence the current that is flowing through

the circuit will be equal to 0. But instead of a silicon diode, suppose we

have germanium diode, in that case, this diode will conduct. So, as you can see, in the case of a germanium

diode, this circuit will conduct. While in the case of silicon diode this circuit

will act as an open circuit. So, while solving the circuits we also need

to see the type of diode which is connected in the circuit. Alright so now once again let’s assume that

the diode that is connected in the circuit is silicon diode. But now the direction of the diode has been

reversed. Now, in this case, if you see, if you find

the Thevenin’s equivalent voltage then now across this anode and cathode the voltage

will be equal to -5V. Because now the anode is on the right-hand

side and the cathode is on the left-hand side. So, the equivalent voltage across this diode

will be equal to -5V. And hence, this diode is reversed biased. So, there will not be any flow of current

through this circuit. and simply we can say that this diode current

Id will be equal to zero. So, as you can see, using this diode approximation

we can easily analyze and solve the circuit problems. Now using this approximation the values of

the voltage and current that we have got are not accurate values but they are quite close

to the actual values. So, for analyzing or troubleshooting the circuits

we can use this diode approximation. Alright, now let’s see a few more examples

based on this diode circuits. So, in this example, two silicon diodes are

connected in series and we have asked to find the voltage and current across each diode. As well as we have asked to find the output

voltage across this 1 Kilo-Ohm resistor. Now, for a moment let’s assume that these

two diodes are conducting and as they are silicon diodes, so in case of the forward

bias the voltage across each diode will be equal to 0.7 V. So, turn ONthese two diodes the voltage across

the two terminals of the diode pair should be more than 1.4V.

And here as the applied voltage is 5V so these two diodes will turn ON. So, let’s replace these two diodes with their

equivalent circuit. And if you see the equivalent circuit then

it will look like this. Now, here let’s assume that current that is

flowing through the circuit is equal to I. So, the current I can be given as 5 – 0.7-0.7. And that is divided by 1 Kilo-ohm resistor. So the value of this current I will be equal

to 3.6 mA. Now, the output voltage Vout will be the drop

across this 1 Kilo-Ohm resistor. So, the output voltage Vout will be equal

to 3.6 mA, multiplied by 1Kilo-ohm resistor. So, the output voltage Vout across this 1

Kilo-ohm resistor will be equal to 3.6 V. So, in this way in case of the series connection

of the diode the threshold voltage to turn on the diode will increase. So, instead of two silicon diodes if we three

silicon diode which are connected in series, the threshold voltage to turn on this diode

will increase to 2.1 V. Similarly, try to find the current and output

voltage in the given circuit. Now, this circuit is similar to the previous

circuit. But here the one silicon diode is replaced

by this germanium diode. So, try to find the answer and do let me know

your answer in the comment section. Alright so now let’s move to the next example. So, in this example, we have been asked to

find the value of these voltages V1, V2, and Vout. Now, if you see this circuit, the anode of

this diode is connected to the 10V through this 4 Kilo-ohm resistor. While the cathode is connected to -5V through

this 3Kilo-ohm resistor. So, intuitively you can say that this diode

will turn on. So, here if we apply the KVL then we can write

10 minus V1, minus 0.7 V, that is drop across this silicon diode, minus V2 that is a drop

across this 3 Kilo-ohm resistor, minus (-5V). and that will be equal to 0. So, if we simplify it then we will get 14.3V

that is equal to V1+ V2. Now, here we have assumed that the current

which is flowing through the circuit is equal to I. So, this V1 will be equal to 4 kilo-ohm times

this current I. And this V2 will be equal to 3 kilo-ohm times

this current I. So, if we simplify it then we can say that

this current I will be equal to 14.3V divided by 7 Kilo-ohm. That is roughly equal to 2.04 mA. So, this is the current which will flow through

this circuit. From this, we can say that this voltage V1

will be equal to 4 kilo-ohms multiplied by this current I. That is equal to 8.16 V.

And similarly, V2 will be equal to 2.04 mA, multiplied by 3 kilo-ohms. That is equal to 6.12 V. So, these are the values of voltage V1 and

V2. Now, here the output voltage Vout will be

equal to V2- 5V. That is equal to 6.12V – 5V. And that is equal to 1.12 V. So, this is the actual value of the output

voltage. Alright so now let’s move to the next example. Now, here these two diodes are connected in

the parallel connection. And we have been asked to find the value of

the output voltage as well as the current I that is flowing in the given circuit. And we have been also asked to find the value

of the diode currents Id1 and Id2. Now, if you see this circuit, here the two

terminals of the diode are connected to the ground potential. And the anode of these two diodes is connected

to the 10V through this 1 kilo-ohm resistor. It means that these two diodes will become

forward biased. It means that the output voltage which we

will get across these two diodes will be equal to 0.7 V.

It means at this node the voltage is equal to 0.7 V. So, from this, we can easily find the current

that is flowing in the circuit. So, the current I will be equal to 10V – 0.7

V, divided by 1 kilo-ohm resistor. And that is equal to 9.3 mA. So, this is the total current which will flow

in the given circuit. Now, here we are assuming that these two diodes

are identical. So, this current I will get equally divided

between two diodes. It means that the current Id1 and Id2 will

be equal to 4.65 mA. Now, in this example, as the two diodes which

are connected in parallel are identical, it is easy to solve this particular problem. But if two diodes which are connected in parallel

are made up of different materials then we need to see the threshold voltage of each

diode. So, based on this we will see the next example. So, in this example, this silicon and the

germanium diodes are connected in the parallel connection. Now, here for a moment let’s assume that the

two diodes are conducting simultaneously. And first of all, let’s assume that this silicon

diode is conducting. So, if this silicon diode is conducting then

there will be a voltage drop of 0.7V across this diode. So, at this node, the voltage will be equal

to 12- 0.7, that is equal to 11.3 V Similarly, let’s assume that this germanium

diode is also conducting. So, if this diode is conducting then the voltage

drop across this diode should be equal to 0.3 V.

And the voltage drop across this silicon diode should be equal to 0.7 V. Now, as these diodes are connected in parallel

connection, the voltage drop in the parallel connection should remain constant. But here the one diode has a voltage drop

of 0.7V, while the other diode has a voltage drop of 0.3 V.

It means that only one diode should conduct and one diode should remain off. So, let’s find out which diode will conduct

and which diode will remain off. Now, in this circuit whenever the 12 V voltage

is applied to the diodes, in reality, this voltage will take some time to reach from

0 to 12 V. Now, during the course of this transient,

if you see first of all this germanium diode will turn on. Once this diode is turned on, it will try

to maintain voltage across the two terminals. So, because of that, the voltage across the

silicon diode will remain 0.3 V And hence this silicon diode will not be able

to turn on. So, if you see actually the voltage across

the two terminal will remain 0.3 V. So, the output voltage Vout, in reality, will

be equal to 12- 0.3 V. That is equal to 11.7 V. So, in this way, whenever the two different

types of diodes are connected in parallel then we need to see the threshold voltage

or the cut- in voltage for each diode. So, the diode which has lower cut-in voltage

will turn on first. Alright so based on this let’s see the last

example. Noe, in this example also we have been asked

to find the output voltage. Now here, to find the output voltage let’s

assume that both diodes are conducting simultaneously. And first, let’s talk about the first diode. So, if the first diode is conducting then

the voltage at this node should be equal to 2V minus 0.7 V.

That is equal to 1.3 V. Now, the same voltage will also appear at

the cathode of this second diode. And the anode of the second diode is connected

to the 4 V. It means that the second diode will also become

forward biased. And if this diode is turned on then there

will be a 0.7 V of the voltage drop across this diode. So, if w consider this second diode as a conducting

diode then the voltage at this node should be equal to 4V minus 0.7 V.

That is equal to 3.3 V. But here if both diodes are conducting simultaneously

then you can see that there is a contradiction at this node. It means that both diodes can not conduct

simultaneously. Now, suppose if we assume that this second

diode is conducting only, in that case, the voltage at this node will be equal to 3.3

V. And at this node also the voltage should be

equal to 3.3 V. So, if this is the case, then the first diode

will turn off. And there will not be any contradiction of

the voltage. So, it means that assumption that we have

made is correct. That means the first diode will remain off

and the second diode will remain on. And the actual voltage at this node will be

equal to 4V minus 0.7V. That is equal to 3.3 V. So, this is the output voltage of the given

circuit. So, I hope in this video, you understood how

to solve the problems based on the diode. And we will solve some examples in the upcoming

quiz. So, you can check the community tab section

of the channel for the more information. So, if you have any question or suggestion,

do let me know in the comment section below. If you like this video, hit the like button

and subscribe to the channel for more such videos.

The timestamps for the different topics covered in the video:

1:02 Graphical Method (Using the Load Line)

3:19 Diode Approximations

4:29 How to Solve a circuit problem using diode approximation

7:46 Example 1 ( Series connection of Diode)

9:54 Example 2

12:11 Example 3 (Parallel Connection of Diode)

13:41 Example 4 (Parallel Connection of Diode with different diodes (Si and Ge))

16:11 Example 5 (Parallel connection of diode with different voltages)

For more Solved Questions on the Diode Circuits, check out this playlist:

https://www.youtube.com/watch?v=f0xMLVpu0zo&list=PLH9R5x7JVXCEVJBjahN-wXv7r1l9dZbjv

Tq

Worthful video

This really helped. Thanks!

thank you,well and clear!!!!!!!!!!!!!!

I think on 9:35, you meant 2.9 volts not 2.1 volts, am I right?

I was so confused with this diodes stuff but you saved my llife. Thank you, you are the best

@ 17:32 did you just assume 4V circuit to be on to solve? or is it the case that if both path are powered, only the path with 4V will turn on.

Sir me ap ko istarh ak pic send kardonge mera kal both zarore assigmnt hai plz

Sir plz send me your whats app no mera kal boht important asigmnt hai mere face book ud nahe hai me kise karo

I didn't understood the last one .how u told it is 4 volt

The voltage across the 1k ohm resistor is 4 V and the current for the circuit is 4 mA for the example with one silicon diode and one germanium diode.

Nice explanation sir…

Sir , thanks for informative video , sir I have questions , that's I have two different DC source ,now I want to add these source by using diode ,so how will connect two different sources . ? Please answer me .

Kathode

You are life saver!!!!👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍

thanks alot

In last question why are we assuming that 2nd diode is conducting only

Tamam… Your explanations are always Crystal Clear…

Needs English subtitles.

My niggggggaaaa

Great explanation g.

4mA ,4v

Excellent video. Very clear and helpful.

17:39 why not 1.3v

4mA , 4V

TUBOL HAHAHAHA

very clear explaination..thank you…

At 12:09 how did we get vout=v2-5??please help

I had a exercise with a circuit with a Vo, but to resolve it we should put a second output voltage Vo2, how to know when to put a second output voltage?

Mera life save kar dhiya

Thanks bhai

I understood it was helpful

Awesome, thanks !!!! Guys subscribe it …..

Does it matter which diode you turn off in the last example?

10:58 I thought The Current " I " was going to split at the node between the 3k resistor and the V-out (KCL) causing the 4k & the 3k resistor currents to be of different values ? Am I Wrong?

What a hot accent hahahah

Nice sir which is this app you use

in 10:38 , why you wrote -(-5) and not -5 in calculating the voltage ??

thankYou for the video. 🙂

its very helpful.

in example 4 why we don't take second one as turn off

4mA and V0 is 4v

Quite helpful for 12th boards

Hell clear now. Thank you sir!

Good teaching, but i have some question. Why in example 2, Vo = v2-5

Any one have the answer of my question?

in example 2

it is a open circuit So How do u apply KVL Rule

Plz Someone do reply

Sooo nice video sir…sooo nice.it is physics with beautyy.thanks for your good explanation.

Thnk uhh muchhhhh😊

For example 2, can you explain how Vo=(Id)(5k)+10 ?

I call the diode the scientific YouTube Play Button. Why? Because diodes and the YouTube Play Button has one thing in common, they can only go in one direction.

13:00 WHY 0.7V across both?

Thank you

thank you sir

In example 1, wouldn't the polarity of the two diodes with respect to each other affect if current will flow?

from the first diode: + to –

then the second diode is also + to –

but wouldn't the second diode be reversed biased because of the first diode?

thanq so muchh superb vedio

I'm struggling to understand diodes, and I'm a beginner, so my question might seem obvious. At 17:12 you said there was a contradiction. Why exactly is there a contradiction?

nice sir

last answer is 2.3v

Thank you !

4mA

Why , in example 2 for the value of output voltage,we subtract V2 -5 to get 1.12 volts???

Last example is not clear

What if the value or type of diode is not given… Then how can we solve

9:35 someone can explain this example in case i only have back curents of two diferent diodes please?

Very bad teaching. You’re just talking through and not actually teaching. But thanks anyways

in the example of the si and ge , is the threshold voltage .7+.3 = 1 v,that will be required to turn on ?

4mA and 4volts

in the last example how about the -1V source and 1k resistor ? do they have any effect on the output ?

i didnot understand last problem why the first diode did not conducts ? please who knows reply fast

Curious why Vo is calculated different in example 2 and example 5? Why not care about the voltage drop across the 1k ohm resistor?

How to explain question no 5

Dusre ko bebkuf batante ho kya

Thanks man ! Love from Pakistan 🇵🇰

In example 5 if both the voltages are equal then what will happen?

In 12:02, how is Vo=V2-5?

Nice explanation…Thanks Sir.

💓💓💓💓💓💓💓💓

Sir,in example 5….why resistance is not considered?

In last problem why 2nd diode is in on state?

What happens when both diodes are off in last example?

how did u calculate vo in example 2

awesome explanation

I=4mA and V= 4A

At 12:04, why is the output voltage only V2-5V? What happened to 10V , V1, and the 0.7V of the diode?

Thank you brother. Nice illustration and presentation.

Why In 2 example , u didn't consider V1?

THANK YOU!

सर आप पढ़ा अच्छा रहे लेकिन हिंदी में भी पढ़ाया करिए l

4MA answer

Very well done!!!

This is the most perfect video ever. Clear and concise with perfect explanations. Thank you!

So helpful thank you Sir and keep going🖒

Nice video, helps me a lot

output voltage 4 and current is 4

For every course I cannot understand, I can rest assured that there's and indian guy with a dedicated youtube channel for it somewhere.

Perfect video. So clear and direct.

Thank you indian guy, you're saving a student for doing this tutorial video

Worst lecture ever seen

12:06 I don't get it ..why is v0=v2-5? why did we ignore the other resister ?

Again you proved better than my professor.

Thanks have a test on diodes your way made everything more clear.

For item 9:38 the value of I= 4mA and Vo= 4V

Sir , what is full load and no load thing ??