How to solve this transistor problem?

  • Thread starter lazyaditya
  • Start date
  • #1
176
7

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




Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Attachments

  • 15042012369.jpg
    15042012369.jpg
    24.3 KB · Views: 444

Answers and Replies

  • #2
368
12
Imagine that Q2 didn't exist, and the base resistor was hooked up directly to Q1. What would your first step be in that case?
 
  • #3
176
7
The approach i used was that i considered I2=Ie1 since its series current.And used kvl equation around Q2 [ .7 + .7 -Ib2(100 kohm) -I2(10 kohm) -Ie1 (100 ohm)]

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 ?
 
  • #4
176
7
The thing is my book doesn't has end results so if anyone could help me with this ?
 
  • #5
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,073
The approach i used was that i considered I2=Ie1 since its series current.
I2 drains to earth and is lost, it doesn't go through Q1.
And used kvl equation around Q2 [ .7 + .7 -Ib2(100 kohm) -I2(10 kohm) -Ie1 (100 ohm)]
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
 
  • #6
176
7
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 ?
 
  • #7
368
12
You should be able to use KCL to write a relationship between I1, I2, and Ib2. After that, you should be able to use the fact that you know the voltage across the 82k + 10k combo to get a second equation, at which point you should have two equations in two unknowns.
 
  • #8
176
7
You should be able to use KCL to write a relationship between I1, I2, and Ib2. After that, you should be able to use the fact that you know the voltage across the 82k + 10k combo to get a second equation, at which point you should have two equations in two unknowns.

Ya i can use the kcl but how to use the kvl to my advantage ?
 
  • #9
368
12
Remember that the Vcc at the top is technically connected to ground via a voltage source, so that gives you another loop in your schematic. Using KVL on that gives you an equation in terms of I1 and I2. Your existing KVL gives you an equation in terms of I2 and Ib2. Using KCL, you can describe I1 in terms of I2 and Ib2, so you can eliminate that variable from the equations. What you're left with is two equations in two unknowns.
 
  • #10
176
7
Remember that the Vcc at the top is technically connected to ground via a voltage source, so that gives you another loop in your schematic. Using KVL on that gives you an equation in terms of I1 and I2. Your existing KVL gives you an equation in terms of I2 and Ib2. Using KCL, you can describe I1 in terms of I2 and Ib2, so you can eliminate that variable from the equations. What you're left with is two equations in two unknowns.

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 ?
 

Attachments

  • 17042012372.jpg
    17042012372.jpg
    19.1 KB · Views: 349
  • #11
368
12
That looks pretty close. Check your signs in the second KVL equation, though--I think some of them may be backwards. Also, since Ie1 is going in the other direction from the base currents, you'll need another minus sign in the bottom equation as well.
 
  • #12
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,073
0.7 + 0.7 + Ie1(100 ohm) + I2(10 kohm) - Ib2(100 kohm) = 0
Did you not like my corrections to this?

I calculate IE1 = 10.14mA. but have not checked my working. See whether you disagree.
 
  • #13
368
12
If you're going to do it like that, the 0.7's need to be negative, don't they?
 
  • #14
176
7
That looks pretty close. Check your signs in the second KVL equation, though--I think some of them may be backwards. Also, since Ie1 is going in the other direction from the base currents, you'll need another minus sign in the bottom equation as well.

Did you not like my corrections to this?

I calculate IE1 = 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 ?
 
  • #15
176
7
The current in the emitter lead shows the actual current right ? But the question showed the current direction of Ie1 opposite to the that "Direction shown in transistor 1"
 
  • #16
368
12
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?)
 
Last edited:
  • #17
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,073
The method over here is correct then ?
Over where? I can't see where you are pointing! :tongue:
And i am always confused with the sign convention !
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.
the voltage drop is from emitter to base right ?
For an NPN the base is more positive than the emitter. Picture that emitter arrow as the arrow in an ordinary diode.
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 ?
The voltages across the 100kΩ and 100Ω will have the same sign.

EDIT: I just noticed that you have the arrow for IE1 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!
 
  • #18
176
7
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 IE1 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
368
12
I posted a response that details how to do this, but it got put on the first page right before it rolled over to page 2. Did you see it?
 
  • #20
176
7
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
:)
 

Related Threads on How to solve this transistor problem?

  • Last Post
Replies
1
Views
8K
Replies
2
Views
5K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
1K
Replies
0
Views
2K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
4K
Top