How to Solve the Diode Circuits (Explained with Examples)


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.

100 thoughts on “How to Solve the Diode Circuits (Explained with Examples)

  1. 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

  2. @ 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.

  3. 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.

  4. 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 .

  5. You are life saver!!!!👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍

  6. 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?

  7. 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?

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

    Any one have the answer of my question?

  9. 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.

  10. 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?

  11. 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?

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

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

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

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

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

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

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