Parallel and Series-Parallel Configuration of Diodes (Examples)
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- Опубликовано: 6 май 2016
- Analog Electronics: Parallel and Series-Parallel Configuration (Examples)
Topics:
1. Solved numerical problem on parallel diode configuration.
2. Solved numerical problem on series-parallel configuration.
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Hey guys,
in example 1, he used 4Id as a final result. The smaller parts were like this, 2Id * 2k R. he didn't write all steps, he just used the short formate. I hope you get why he used 4Id not 2Id :)
best wishes,
Thanks☺
I had never Thought Analog Electronics is so cool
was looking for diode biasing with two sources! thanks a lot :D
In example 2, why does he follow that specific path when going from 16v to 12v. Is he not considering the other silicone diode that is in parallel at all?
Plz explain sign convention when resistor is used in 1st sum when we are finding vo
In example 1, The equivalent resistor of the two parallell branches with 2k ohm each is a single branch and resistor with 1k ohm. To get V0, you can divide the voltage over that equivalent resistor and the one connecting to ground, giving you (9.3-0) * (2k ÷ (2k + 1k)) = 6.2v.
9.3v and not 10v because of the diodes of course. So now you have V0.
To calculate Id: 2*Id (total current) = (9.3-0)V ÷ 3k ohm = 3.1mA, which means 1*Id = 1.55 mA.
Don't know why he uses KVL and then mixes in currents in the expression, or where the 4id comes from... This is how I did it, hope it helps.
This is how almost every circuit professor would teach it. It's best to make use of altering circuits to path towards Impedance anyways
why 9.3 - 0?
Huh? Is 6.2 supposed to be a current? Sincr you have used V=IR?
Is there a current flow if the other diode is GaAs?
sir pls send problems on diode current eqn.how to substitute applied voltage in formulae of reverse saturation current io
Sir in problem2 instead of+16v if we apply -16v ...how to calculate Sir????
Sir one more resistor is there which is connected with parallel to another as we in parallel resistor we simply apply equivalent resistor so why not over here
What if we have one Ge diode and one Si diode instead of 2 Si diode in Quesyion 1?
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Thank you sir
2:23 how is this possible. Are we finding I first then Id right? I am not getting why you are taken 2Id and across output resistance
According to kcl at the node id+id( currents entering) this should be equal to the leaving currents
@@keerthim7149 then in 2nd Ques also along 4.7 k is I?? It should also be doubled
Sir please explain if two different diode connected in parallel. Then in what way current is distributed.
The current will split and distribute evenly between the two branches since they are equal in resistance. I believe when combining the diodes in parallel that the voltage does not add up instead 0.7v drops total across.
How did you get the 0.55? Please help
In example 1 why we use 2I and 4 I seperately?
Really cool..
Hello dear, why you used 4i_d instead of 2i_d? Ex 1. (10-0.7 -2*i_d -4i_d)
My question is if in the parallel section has different current, then what should I do...
You may get 2 equations for 2 unknowns then solve them you will get your required ans.
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if both diodes are different which is parallel to each other, which one will be used, the one with the higher value or the one with the lower?
Lower value. The diode requiring higher value will remain OFF ( open circuit).
What if the direction of current is no given, how will we know if it's a forward bias or reverse biased diode?
i think you'll based on the Voltage source and the direction of the diodes
In the first question, the branch currents were ID and after leaving the branch it got multiplied with 2*ID*2k resistor... But in the second question after leaving the branch the current did not get multiplied that is, it remained as ID*4.7k only and not 2*ID*4.7k... why?? Can anyone pls explain this for me?
Noticed that too
In first I'd is current through each diode making output current 2id whereas in 2nd current through each diode is not specified.i enters into 2 diodes combination so I should leave 2 diodes combination
Because it gets divided by 2 first for both of them and then added again when it comes out to make it the same value as before.
Sir you are very good at theory....but confuses at circuits😂...Anyways....Thanks for making such content ....you are doing great job
why is it 2Id in first one and not the same in second one second one also 2I shud come right? pls answer
From top to bottom:
1. we started with "I" as branch current
2. This "I" divided into 2 branches
3. The divided current from 2 branches merged into single branch
4. We again get "I" in the final single branch
so, we used 'I" for 4.7k resistor
please note that the current is not divided for V0 wire.
How did you consider 4Id in first but then ignored in the second question?
he considered I as total current in second question but 2Id as total current in first question.
what's your reference book?
Bro u have changed the terminals of the diodes in the 1dt circuit which u have drawn
Sir explain example 2.11 including blue light ?
V0 in the second example is 14.585~14.6v
What will happen if the two diodes are not same .... then the current through both arms will be different ... In that case how to calculate the I sub D???
The diode with the higher rating will then be converted into an open circuit. therefore no current will pass through it
vincent bryan Reyes , why bro
because current takes path of least resistance
then apply kvl inside the loop containing 2 diodes
@@Vince0208 nope, both diodes will act as resistors and we have seen examples for 2 resistors in parallel where the current passes through both of them.
Finding Vo for second problem is straight forward....
Vo=16 - 0.7 - 0.7= 14.6v.
why plugging in the value of l?
what if we have si diode in parallel with Ge diode and both are forward biased.. so we will have 0.3v across both or 0.7 ? as both r in paraller so they will have same voltage.
Ge diode will reach the ON state once the voltage is 0.3 V and it will maintain this voltage. So, the Si diode also gets the same 0.3 V and not 0.7 V, hence Si diode remains OFF ( open circuit).
and why is that so sir
Musaab Islam because the diode needs a minimum voltage which is 0.7V for silicon to be activated. So when it is provided with 0.3V then it's unable to activate itself with that amount of voltage. Thus, we convert it to an open circuit whereby no current flows. :)
Kaunaj Banerjee so will there be no current in the entire circuit
Ashish Rajput
When Ge and Si diodes are connected in parallel Ge diode gets forward biased at 0.3V itself where as Si diode needs to reach 0.7V. But even before Si diode reaches 0.7V, at 0.3V as Ge diode gets forward biased which makes it replaced by the ideal internal impedance of the diode in forward bias i.e rd=0 and hence maximum current flows through Germanium at 0.3V itself which makes the Si diode to never turn ON.
in example 1 r1 2k and r2 2k are parallel solving it we got get a 1k then add 2k from r2 cause it series then we can get a resistance equivalent of 3k..
why is the direction of id is same for both Si in the first problem?
very good question next question, please.
What happens when Vin < V(diode) and KVL gives a negative amp? Does that mean zero current is flowing?
If Vin is less then Vd then there will be no current flowing. As the input voltage cant get any higher then Vb, the potential barrier in diode will remain same, As far I know!
In 2nd ckt what about the equation 16v=0.7v+0.7v+Vo+12?
But this gives wrong ans
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I don’t understand how the resistor current is 2Id instead of Id like the other two???
Why 4ID sir?
Yaaa... Y!?
As V = IR
2k ohms × 2ID = 4ID
For problem 1. You took 2 Si diodes , but what if one of the diodes is GaAs ??? How to solve then ???
Since barrier potential of GaAs is greater than silicon's then the diode with higher barrier potential behaves as an open circuit, in this case GaAs will behave as an open circuit.
when current is flowing from one direction current at the resistance must be Id only instead of 2Id
how can u take 4Id?
2kohm(2Id )=4Id
In example 2 why in. The parallel diode the current is same..as..there voltage to be same..and current should be half
If u have basics every circuit looks easy
Can you explain why you have taken -0.7v and -2I0. Why you taken minus sign
interesting....................
WHY THE CURRENT IS 2Id in first problem and I in second problem can any one explain plz..
in Q1 Id is the current through one of the diodes only, not the total current. while in Q2, I is the total current and it will also be the sum of the currents through the two diodes connected in parallels. i.e. if we denote the currents through the diodes connected in parallels in Q2 as "Id" (since both diodes have similar characteristics, so same current will flow through both) then I =Id + Id =2Id.
What if the two resistors are not equal?
2. Exercise. 16V - 0.7V - 0.7V - 4.7I = 12V. In this case the I is half amp not 0.55mA.
(-4.7*1000), since the resistor is in K ohms
Csaba Molnar I wont be half.
why is 12 v negative ?
it similar to the source ! so is positive
Same I can't understand.
I cant understand that too
where did Vo come from.
Id1 + Id2= Id and since Id1 and Id2 is the same then we can say that Id= 2(Id1). So 2Id1 * 2kohm = 4Id ;)
What if the Id1 and Id2 arent the same?
you explain very nicely but make many mistakes while taking the units
In last example 16-0.7-0.7-4.7I= -12 hoga please chek it
wasnt the ans 0.4mA
How to come 4Id ? please explain...........
2Kohm x 2 x Id
Id is the current going through one of the branches that are in parallel while 2*Id is the current going through the last resistor
Sir how come 4Id *
V=2k*2ID
Current through the 2k resistor is 2id so,4Id
2nd confused
What if the other diode is a GaAs?
all the current will flow through 0.7 silicon diode as compare to 1.2
because 0.7 potential barrier is less that 1.2 hence 1.2 would be considered open.
Neso Academy what is your reference plz
I didn't understand where 4Id is coming from
I don’t understand
Resistance is 2k . In equation you just taken just 2.
Exactly, he did NOT explain this calculation well
The resistor is in kilo.ohm and the current is in mille.ampere
So, kilo will cancel out with mille.. Just like dividing 1000 over 1000..
I hope this was clear.
@@abbasaljanabi2667 Yes and...
I noted that in the case where you have micro amperes you should convert it to milli by dividing by 1000 then the other calculations go normally. In short we deal mostly with milli amperes for current, kilo ohms for resistance and volts for voltage.
why you know this is 0,7 V
wtf is 4id explain please
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