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## Homework Statement

The circuit has two transistors 'Q1' and 'Q2' as shown in the figure with 'Beta1'=100 and 'Beta2'=50. The circuit figure is shown below and Vbe1=Vbe2=0.7 volts.

Need to calculate I1,Ib1,I2,Ib2,Ie1,Ic1,Ic2

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- Thread starter lazyaditya
- Start date

- #1

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The circuit has two transistors 'Q1' and 'Q2' as shown in the figure with 'Beta1'=100 and 'Beta2'=50. The circuit figure is shown below and Vbe1=Vbe2=0.7 volts.

Need to calculate I1,Ib1,I2,Ib2,Ie1,Ic1,Ic2

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Then i converted the currents in the form of Ib2 by considering that [(Beta2 + 1)Ib2 = Ib1] and further using equation [(Beta1 + 1)Ib1 = Ie1]. So i converted whole equation in form of Ib2 and calculated Ib2 from there.

From Value of Ib2 I calculated Ib1 and Ic2 using Formula's [(Beta2)Ib2=Ic2] and [(Beta2 + 1)Ib2 = Ib1]

Now i have the value of Ib1,Ib2 and Ic2 so i calculated the Value of Ie1 using equation [(Beta1 + 1)Ib1 = Ie1] and also the value of Ic1 from [(Beta1)Ib1 = Ic1]

Now again using Kvl in the loop formed by 24 volt, Vcb2 ,10 kohm, 100 ohm and 1Kohm i calculated Vcb2.

By using KVL in top left loop (24V,Vcb2,82Kohm and 100Kohm i calculated I1)

I am confused that approach i used is it correct ? or what else should i do to solve this problem in minimum amount of time ?

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The thing is my book doesn't has end results so if anyone could help me with this ?

- #5

NascentOxygen

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I2 drains to earth and is lost, it doesn't go through Q1.The approach i used was that i considered I2=Ie1 since its series current.

KVL: the sign of each term is dependent on your assigned current directions.And used kvl equation around Q2 [ .7 + .7 -Ib2(100 kohm) -I2(10 kohm) -Ie1 (100 ohm)]

.7 + .7 + Ib2(100 kohm) -I2(10 kohm) + Ie1 (100 ohm) = 0

- #6

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I2 drains to earth and is lost, it doesn't go through Q1.

KVL: the sign of each term is dependent on your assigned current directions.

.7 + .7 + Ib2(100 kohm) -I2(10 kohm) + Ie1 (100 ohm) = 0

So on that basis I2 is not equal to Ie1

Thanks

So how should i calculate all the current values ?

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Ya i can use the kcl but how to use the kvl to my advantage ?

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Then two equations will be like this ?

Equation (1)

24 - I1(82 kohm) - I2(10 kohm) = 0

Equation (2)

0.7 + 0.7 + Ie1(100 ohm) + I2(10 kohm) - Ib2(100 kohm) = 0

Then using Kcl :- I1 = Ib2 + I2

and getting Ie2 = Ib2 [(Beta1 + 1)(Beta2 + 1)]

? Should i solve like this ?

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- #12

NascentOxygen

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Did you not like my corrections to this?0.7 + 0.7 + Ie1(100 ohm) + I2(10 kohm) - Ib2(100 kohm) = 0

I calculate I

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If you're going to do it like that, the 0.7's need to be negative, don't they?

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Did you not like my corrections to this?

I calculate I_{E1}= 10.14mA. but have not checked my working. See whether you disagree.

The method over here is correct then ?

And i am always confused with the sign convention ! the voltage drop is from emitter to base right ?

I am considering my current flow in the direction opposite to "I2" and in the direction of "Ib2" and taking Voltage rise = +ve and Voltage drops = -ve then won't 0.7 volt be positive, and voltages across 10 kohm and 100 ohm be +ve while voltage across 100 kohm be -ve ?

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Let's back up...the sign conventions here are turning what should be a relatively simple problem into a huge mess that I think is confusing all of us.

First, flip [itex]Ie_1[/itex] around. Now it goes the same direction as [itex]Ib_2[/itex], so your equations involving the Betas don't have any negative signs in them.

Second, do the KVL by going clockwise. That means that every stage along the way is a voltage drop, except for the 10k resistor (the voltage drop is from base to emitter in an NPN transistor). I always find it easier to sum voltage DROPS in KVL equations, because then you're following the arrow of the current, instead of going against it.

By doing it this way, I get the following equations:

[tex]

82k\Omega\cdot I_1 + 10k\Omega\cdot I_2 - 24V = 0\\

100k\Omega\cdot Ib_2 + 0.7 + 0.7 + 100\Omega\cdot Ie_1 - 10k\Omega \cdot I_2 = 0\\

I_1 - I_2 - Ib_2 = 0\\

Ie_1 = (\beta_1 + 1)(\beta_2 + 1)Ib_2

[/tex]

Mixing all of those together should give you two equations in the unknowns [itex]I_2[/itex] and [itex]Ib_2[/itex]. NascentOxygen, do you agree?

(P.S. Once you've solved it, don't forget to flip [itex]Ie_1[/itex] back around to agree with the original definition of the problem. This is a really bizarre convention that I think is responsible for most of the confusion here--are you sure you copied that down right?)

First, flip [itex]Ie_1[/itex] around. Now it goes the same direction as [itex]Ib_2[/itex], so your equations involving the Betas don't have any negative signs in them.

Second, do the KVL by going clockwise. That means that every stage along the way is a voltage drop, except for the 10k resistor (the voltage drop is from base to emitter in an NPN transistor). I always find it easier to sum voltage DROPS in KVL equations, because then you're following the arrow of the current, instead of going against it.

By doing it this way, I get the following equations:

[tex]

82k\Omega\cdot I_1 + 10k\Omega\cdot I_2 - 24V = 0\\

100k\Omega\cdot Ib_2 + 0.7 + 0.7 + 100\Omega\cdot Ie_1 - 10k\Omega \cdot I_2 = 0\\

I_1 - I_2 - Ib_2 = 0\\

Ie_1 = (\beta_1 + 1)(\beta_2 + 1)Ib_2

[/tex]

Mixing all of those together should give you two equations in the unknowns [itex]I_2[/itex] and [itex]Ib_2[/itex]. NascentOxygen, do you agree?

(P.S. Once you've solved it, don't forget to flip [itex]Ie_1[/itex] back around to agree with the original definition of the problem. This is a really bizarre convention that I think is responsible for most of the confusion here--are you sure you copied that down right?)

Last edited:

- #17

NascentOxygen

Staff Emeritus

Science Advisor

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Over where? I can't see where you are pointing! :tongue:The method over here is correct then ?

Then you should spend some time on this, to adopt a technique that overcomes this failing. No one is going to accept that as an excuse for getting circuit analysis wrong.And i am always confused with the sign convention !

For an NPN the base is more positive than the emitter. Picture that emitter arrow as the arrow in an ordinary diode.the voltage drop is from emitter to base right ?

The voltages across the 100kΩ and 100Ω will have the same sign.I am considering my current flow in the direction opposite to "I2" and in the direction of "Ib2" and taking Voltage rise = +ve and Voltage drops = -ve then won't 0.7 volt be positive, and voltages across 10 kohm and 100 ohm be +ve while voltage across 100 kohm be -ve ?

EDIT: I just noticed that you have the arrow for I

- #18

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Over where? I can't see where you are pointing! :tongue:

Then you should spend some time on this, to adopt a technique that overcomes this failing. No one is going to accept that as an excuse for getting circuit analysis wrong.

For an NPN the base is more positive than the emitter. Picture that emitter arrow as the arrow in an ordinary diode.

The voltages across the 100kΩ and 100Ω will have the same sign.

EDIT: I just noticed that you have the arrow for I_{E1}pointing in an illogical direction. This of course changes what I've previously written. I sharn't change now, or I'll be confused, too!

Ok thanks alot !

I will be working on sign conventions then !

- #19

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- #20

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Let's back up...the sign conventions here are turning what should be a relatively simple problem into a huge mess that I think is confusing all of us.

First, flip [itex]Ie_1[/itex] around. Now it goes the same direction as [itex]Ib_2[/itex], so your equations involving the Betas don't have any negative signs in them.

Second, do the KVL by going clockwise. That means that every stage along the way is a voltage drop, except for the 10k resistor (the voltage drop is from base to emitter in an NPN transistor). I always find it easier to sum voltage DROPS in KVL equations, because then you're following the arrow of the current, instead of going against it.

By doing it this way, I get the following equations:

[tex]

82k\Omega\cdot I_1 + 10k\Omega\cdot I_2 - 24V = 0\\

100k\Omega\cdot Ib_2 + 0.7 + 0.7 + 100\Omega\cdot Ie_1 - 10k\Omega \cdot I_2 = 0\\

I_1 - I_2 - Ib_2 = 0\\

Ie_1 = (\beta_1 + 1)(\beta_2 + 1)Ib_2

[/tex]

Mixing all of those together should give you two equations in the unknowns [itex]I_2[/itex] and [itex]Ib_2[/itex]. NascentOxygen, do you agree?

(P.S. Once you've solved it, don't forget to flip [itex]Ie_1[/itex] back around to agree with the original definition of the problem. This is a really bizarre convention that I think is responsible for most of the confusion here--are you sure you copied that down right?)

Thank you

:)

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