# Solving Circuits

#### hgmjr

Joined Jan 28, 2005
9,027
Originally posted by davidand@Jun 8 2005, 08:43 PM
-8.6V/15 Ω =0.57mA

10V-(0.57)(60 Ω )=9.49V
[post=8339]Quoted post[/post]​
Aaaahhhh! I see where you are off the track.

The value -8.6V is the correct voltage at the emitter but it is not the voltage that is being "dropped" across the 15 ohm emitter resistor.

The voltage you need to be using is the voltage that you would measure across the 15 ohm resistor if you used a voltmeter to measure the voltage. Recall that the current flowing in the resistor by Ohm's Law is V/R where V is the voltage across the resistor R.

hgmjr

#### celect

Joined May 31, 2005
18
Originally posted by hgmjr@Jun 8 2005, 10:17 PM
Aaaahhhh!  I see where you are off the track.

The value -8.6V is the correct voltage at the emitter but it is not the voltage that is being "dropped" across the 15 ohm emitter resistor.

The voltage you need to be using is the voltage that you would measure across the 15 ohm resistor if you used a voltmeter to measure the voltage.  Recall that the current flowing in the resistor by Ohm's Law is V/R where V is the voltage across the resistor R.

hgmjr
[post=8343]Quoted post[/post]​

#### davidand

Joined Jun 2, 2005
43
Originally posted by hgmjr@Jun 7 2005, 12:14 PM

You need to take a second look at your Vc and Ve calculations.

HINT: The voltage at the emitter will be more negative than the voltage at the base. Keep in mind that you are dealing with an NPN silicon transistor.

Once you have Ve then you will have most of what you need to calculate Vc. Since you have no value for the beta of the transistor we will be forced to assume a value of say 100. That is unless you know the actual value posed by the problem. Or we can assume that Ie is equal to Ic and proceed on that basis.

Hang in there.
hgmjr
[post=8289]Quoted post[/post]​
use: Voltage at collector minus (current at emitter)(resistance at collector)

-10V/15 Ω =-0.66
-10 - (0.66)(60 Ω ) = -10 -(0.0396) = -10.0369b

#### hgmjr

Joined Jan 28, 2005
9,027
Originally posted by davidand@Jun 8 2005, 11:01 PM
use: Voltage at collector minus (current at emitter)(resistance at collector)

-10V/15 Ω =-0.66
-10 - (0.66)(60 Ω ) = -10 -(0.0396) = -10.0369b
[post=8348]Quoted post[/post]​
I can see that calculating the emitter current is an area where you are a bit uncertain on how to proceed. Don't worry, I'm sure together we can clear this up for you.

You have successfully determined Vb and Ve so we are well passed the halfway point. Once you get a handle on the emitter current it will be a sprint to the finish line, since you already have the formula correct for calculating Vc.

Ve as you have already correctly determined is -8.6V. This means that on one end of the 15 ohm resistor you have -8.6V. You also know the voltage at the other end of the resistor is -10V. Knowing these two values allows you to determine the voltage drop across the 15 ohm resistor. It is this voltage drop that you should use to compute the current that is flowing through the resistor.

HINT: The current flowing out of the emitter is equal to the current flowing through the 15 ohm resistor.

hgmjr

#### davidand

Joined Jun 2, 2005
43
Originally posted by hgmjr@Jun 9 2005, 07:48 AM
I can see that calculating the emitter current is an area where you are a bit uncertain on how to proceed. Don't worry, I'm sure together we can clear this up for you.

You have successfully determined Vb and Ve so we are well passed the halfway point. Once you get a handle on the emitter current it will be a sprint to the finish line, since you already have the formula correct for calculating Vc.

Ve as you have already correctly determined is -8.6V. This means that on one end of the 15 ohm resistor you have -8.6V. You also know the voltage at the other end of the resistor is -10V. Knowing these two values allows you to determine the voltage drop across the 15 ohm resistor. It is this voltage drop that you should use to compute the current that is flowing through the resistor.

HINT: The current flowing out of the emitter is equal to the current flowing through the 15 ohm resistor.

hgmjr
[post=8353]Quoted post[/post]​

Ok
so I have -10V - -8.6V = -1.4V dropped across the 15 Ω resistor.

so I use this -1.4V/15 Ω =0.0933

10V - (0.0933 x 60 Ω )= 10V - 5.598 =4.4V

V
c = 4.4V

#### hgmjr

Joined Jan 28, 2005
9,027
Originally posted by davidand@Jun 9 2005, 11:20 AM
Ok
so I have -10V - -8.6V = -1.4V dropped across the 15 Ω resistor.

so I use this -1.4V/15 Ω =0.0933

10V - (0.0933 x 60 Ω )= 10V - 5.598 =4.4V

V
c = 4.4V
[post=8357]Quoted post[/post]​
Congratulations, You've done it!!!!

#### hgmjr

Joined Jan 28, 2005
9,027
Originally posted by JoeJester@Jun 9 2005, 01:08 PM
Vb will be one of two numbers, depending on where you place the ground. It could be -7.9 Volts if the ground was placed at the +10 side of the 10 volt supply. Since you didn't define the ground, it could be placed anywhere. It's more common to see the ground placed at the -10 side of the 10 volt supply when using NPN transistors.

I can't think of too many applications where you would use the positive supply side as the ground when using NPN transistors.
[post=8359]Quoted post[/post]​
That's a good observation JoeJester.

All of my computations have been based on the use of a dual supply. A positive 10V and -10V for a total of 20V. I realize that the diagram does not have a -10V annotation. The drawing did not contain the usual ground symbol so that lead me to a possible mis-interpretation. I will go back and review the postings to see where I got off on the wrong track myself.

davidand,

Would you please verify whether your problem involves a dual supply or a single 10V supply? If it is not a dual supply then I need to revisit my calculations. Regardless, the approach I took was correct and applies to either case. Only the calculated values will change depending on the power supply configuratoin used.

My sincere apologies davidand if I have mislead you by my misunderstanding of your schematic. Once I get your clarification I will run through the calculations again if needed.

hgmjr

#### davidand

Joined Jun 2, 2005
43
Originally posted by hgmjr@Jun 9 2005, 02:39 PM
That's a good observation JoeJester.

All of my computations have been based on the use of a dual supply. A positive 10V and -10V for a total of 20V. I realize that the diagram does not have a -10V annotation. The drawing did not contain the usual ground symbol so that lead me to a possible mis-interpretation. I will go back and review the postings to see where I got off on the wrong track myself.

davidand,

Would you please verify whether your problem involves a dual supply or a single 10V supply? If it is not a dual supply then I need to revisit my calculations. Regardless, the approach I took was correct and applies to either case. Only the calculated values will change depending on the power supply configuratoin used.

My sincere apologies davidand if I have mislead you by my misunderstanding of your schematic. Once I get your clarification I will run through the calculations again if needed.

I understand it to be a dual supply.
I will continue with more problems until I get better at solving them.
hgmjr
[post=8360]Quoted post[/post]​

#### davidand

Joined Jun 2, 2005
43
Originally posted by JoeJester@Jun 9 2005, 02:08 PM
Vb will be one of two numbers, depending on where you place the ground. It could be -7.9 Volts if the ground was placed at the +10 side of the 10 volt supply. Since you didn't define the ground, it could be placed anywhere. It's more common to see the ground placed at the -10 side of the 10 volt supply when using NPN transistors.

I can't think of too many applications where you would use the positive supply side as the ground when using NPN transistors.

I remember that book ...

Since you solved it while I was typing, here are two simulations ... using your circuit. One with the ground at the +10 side of the supply and one with it at the -10 side of the supply.

Good work on solving the problem.

On edit ... since I didn't consider the dual supply ... I added the simulation with a dual supply.
[post=8359]Quoted post[/post]​
Thanks for your input, did you do the simulation with personal software or
with tools from this site?
looks neat.

#### hgmjr

Joined Jan 28, 2005
9,027
I believe you are ready to tackle the last problem HW#1.

HW#2 was good practice in the skills that will assist you in the solution of HW#1.

hgmjr

#### hgmjr

Joined Jan 28, 2005
9,027
Does HW#1 give a specific value for the input labeled Vin?

hgmjr

#### mozikluv

Joined Jan 22, 2004
1,435
hi hgmjr,

your assumption on dual rail supply of the amp is appropiate for demo. it's a good thing you didn't use a -5.0v

#### davidand

Joined Jun 2, 2005
43
Originally posted by hgmjr@Jun 9 2005, 09:29 PM
Does HW#1 give a specific value for the input labeled Vin?

hgmjr
[post=8370]Quoted post[/post]​
No V in has no label it does ask to find the DC voltages with respect to ground.

#### hgmjr

Joined Jan 28, 2005
9,027
Originally posted by davidand@Jun 9 2005, 10:13 PM
No V in has no label it does ask to find the DC voltages with respect to ground.
[post=8376]Quoted post[/post]​
So be it.

hgmjr

#### davidand

Joined Jun 2, 2005
43
Originally posted by hgmjr@Jun 9 2005, 11:26 PM
So be it.

hgmjr
[post=8377]Quoted post[/post]​
V
E = -5V

V
C1 = 10V

V
C2 = 10V

#### hgmjr

Joined Jan 28, 2005
9,027
Originally posted by davidand@Jun 10 2005, 11:50 AM
V
E = -5V

V
C1 = 10V

V
C2 = 10V
[post=8382]Quoted post[/post]​
Pardon me if I seem a little gun shy after the previous confusion over the power supply on HW#2.

I took a second careful look at HW#1 and it is definitely showing a dual supply. A positive 20V and a negative 20V.

It is interesting that your values for VC1 and VC2 are correct assuming that the voltage Vin equals 0V. I say interesting because your value for Ve is not correct according to my calculations.

HINT: The voltage at the base of the current source transistor is set by the zener. Pay close attention to where the anode of the zener diode is referenced. Also don't forget to account for the base-emitter voltage when calculating the voltage at the emitter of the current source transistor.

hgmjr

#### davidand

Joined Jun 2, 2005
43
Originally posted by hgmjr@Jun 10 2005, 01:59 PM
Pardon me if I seem a little gun shy after the previous confusion over the power supply on HW#2.

I took a second careful look at HW#1 and it is definitely showing a dual supply. A positive 20V and a negative 20V.

It is interesting that your values for VC1 and VC2 are correct assuming that the voltage Vin equals 0V. I say interesting because your value for Ve is not correct according to my calculations.

HINT: The voltage at the base of the current source transistor is set by the zener. Pay close attention to where the anode of the zener diode is referenced. Also don't forget to account for the base-emitter voltage when calculating the voltage at the emitter of the current source transistor.

hgmjr
[post=8383]Quoted post[/post]​
Emitter voltage = voltage at base - voltage base emitter
voltage at base = -5V - 0.7 = - 4.3

#### hgmjr

Joined Jan 28, 2005
9,027
Originally posted by davidand@Jun 10 2005, 06:12 PM
Emitter voltage = voltage at base - voltage base emitter
voltage at base = -5V - 0.7 = - 4.3
[post=8387]Quoted post[/post]​
The voltage you have computed for the base needs another look.

HINT: Note that the anode of the zener diode is referenced to -20V.

Emitter voltage = voltage at base - voltage base emitter
This equation is correct for all of the NPN transistors being used in HW#1 including the constant current source.

hgmjr

#### celect

Joined May 31, 2005
18
Originally posted by hgmjr@Jun 10 2005, 07:55 PM
The voltage you have computed for the base needs another look.

HINT: Note that the anode of the zener diode is referenced to -20V.
This equation is correct for all of the NPN transistors being used in HW#1 including the constant current source.

hgmjr
[post=8388]Quoted post[/post]​

#### davidand

Joined Jun 2, 2005
43
Originally posted by hgmjr@Jun 10 2005, 07:55 PM
The voltage you have computed for the base needs another look.

HINT: Note that the anode of the zener diode is referenced to -20V.
This equation is correct for all of the NPN transistors being used in HW#1 including the constant current source.

hgmjr
[post=8388]Quoted post[/post]​
What is the 5V zener diode doing is it acting as a open circuit toward the -20v supply.