KVL and figuring out current in a circuit

[MikeMike]

Circuit in question

now, i left out the voltage, which is given. however we are supposed to use KVL to figure out current and voltage at certain spots (Vc Vb Ve Vce) which are all around the op amp which i forgot to label the emitter side ,but its the lower portion.

now, i THINK the equation to be used should be Vcc+R1*Ir1+R2*Ir2+R3*Ir3 = 0

and then slowly solve from there (r1=2.2, r2=100 r3=1) but im not positive, do i have to split it up into 2 different loops? with R2 being on its own loop? i am very confused.

and Beta is 100

 

[MikeMike]

anyone at alllll?

 

[TuxDave]

You forgot Vbe ~ 0.7

 

[Gibson486]

No....that whole network does not constitute one loop....You have teh right idea, but you are just not applyng what you know about BJT's....


also, which one is R1, R2, etc....

 

[MikeMike]

Originally posted by: TuxDave
You forgot Vbe ~ 0.7

yea, yes i did
the R1 and R2 is listed in the post, R1 is the 2.2k, R2 is 100K, R3 is 1k.

 

[Gibson486]

I'll give you a couple hints....

notice that VE is connected to ground.... What does that mean? This should easily give you VB.


To find VC, notice that it is connected to a resistor....use it to find the voltage

or you can do it the easier way and work backwards from finding VB.....

 

[MikeMike]

Originally posted by: Gibson486
I'll give you a couple hints....

notice that VE is connected to ground....


To find VC, notice that it is connected to a resistor....use it to find the voltage

can i use Thevinans Theorem for this? Vth = (R2/(R1+R2))(VCC)???

 

[Gibson486]

Originally posted by: MIKEMIKE

Originally posted by: Gibson486
I'll give you a couple hints....

notice that VE is connected to ground....


To find VC, notice that it is connected to a resistor....use it to find the voltage

can i use Thevinans Theorem for this? Vth = (R2/(R1+R2))(VCC)???

now you are just trying too hard......

 

[JohnCU]

if I wasn't drunk, I could figure this out. However, I just spent the evening figuring out the magnetic field around a coaxial cable.

 

[Gibson486]

Originally posted by: JohnCU
if I wasn't drunk, I could figure this out. However, I just spent the evening figuring out the magnetic field around a coaxial cable.

That course sucks, I got a D and i do not plan on retaking it.

 

[MikeMike]

Originally posted by: Gibson486

Originally posted by: MIKEMIKE

Originally posted by: Gibson486
I'll give you a couple hints....

notice that VE is connected to ground....


To find VC, notice that it is connected to a resistor....use it to find the voltage

can i use Thevinans Theorem for this? Vth = (R2/(R1+R2))(VCC)???

now you are just trying too hard......

i have a full page of hints and calculations and BJT relationships on the back of the hmwk that im trying to relate to the hmwk.

so if i try to hard, thats my bad.

so lets see, Ie=Ib+Ic can i get Ic from doing XXv/2200 (ohms) to get X.XmA and that would be Ic???

 

[JohnCU]

Originally posted by: Gibson486

Originally posted by: JohnCU
if I wasn't drunk, I could figure this out. However, I just spent the evening figuring out the magnetic field around a coaxial cable.

That coarse sucks, I got a D and i do not plan on retaking it.

Yeah, but it's kinda cool that I got the integration to work out to wear I learned why a coaxial cable doesn't cause interference, there's no magnetic field outside of it.

 

[Gibson486]

Originally posted by: MIKEMIKE

Originally posted by: Gibson486

Originally posted by: MIKEMIKE

Originally posted by: Gibson486
I'll give you a couple hints....

notice that VE is connected to ground....


To find VC, notice that it is connected to a resistor....use it to find the voltage

can i use Thevinans Theorem for this? Vth = (R2/(R1+R2))(VCC)???

now you are just trying too hard......

i have a full page of hints and calculations and BJT relationships on the back of the hmwk that im trying to relate to the hmwk.

so if i try to hard, thats my bad.

so lets see, Ie=Ib+Ic can i get Ic from doing XXv/2200 (ohms) to get X.XmA and that would be Ic???

VE = 0. plain and simple. VBE=VB-VE=.7, so VB = .7

(VC-VB)/100K [or is it VB-VC??? I always got that wrong] will give you the current [see, now we just used KCL/KVL], but we don't have it.....actually we do......we have the current equation you just typed.....

there, that is more than enough info to solve this.

 

[Gibson486]

Originally posted by: JohnCU

Originally posted by: Gibson486

Originally posted by: JohnCU
if I wasn't drunk, I could figure this out. However, I just spent the evening figuring out the magnetic field around a coaxial cable.

That coarse sucks, I got a D and i do not plan on retaking it.

Yeah, but it's kinda cool that I got the integration to work out to wear I learned why a coaxial cable doesn't cause interference, there's no magnetic field outside of it.

Wow, that's further than i got. I never got the integration to work...in any problem

 

[MikeMike]

Originally posted by: Gibson486

Originally posted by: MIKEMIKE

Originally posted by: Gibson486

Originally posted by: MIKEMIKE

Originally posted by: Gibson486
I'll give you a couple hints....

notice that VE is connected to ground....


To find VC, notice that it is connected to a resistor....use it to find the voltage

can i use Thevinans Theorem for this? Vth = (R2/(R1+R2))(VCC)???

now you are just trying too hard......

i have a full page of hints and calculations and BJT relationships on the back of the hmwk that im trying to relate to the hmwk.

so if i try to hard, thats my bad.

so lets see, Ie=Ib+Ic can i get Ic from doing XXv/2200 (ohms) to get X.XmA and that would be Ic???

VE = 0. plain and simple. VBE=VB-VE=.7, so VB = .7

(VC-VB)/100K [or is it VB-VC??? I always got that wrong] will give you the current [see, now we just used KCL/KVL], but we don't have it.....actually we do......

there, that is more than enough info to solve this.

now my only question to you is why must Ve be zero? is it because there is only one ground, and that happens to be at VE? would this change if i were to add a ground connection before VB?

 

[Gibson486]

Originally posted by: MIKEMIKE

Originally posted by: Gibson486

Originally posted by: MIKEMIKE

Originally posted by: Gibson486

Originally posted by: MIKEMIKE

Originally posted by: Gibson486
I'll give you a couple hints....

notice that VE is connected to ground....


To find VC, notice that it is connected to a resistor....use it to find the voltage

can i use Thevinans Theorem for this? Vth = (R2/(R1+R2))(VCC)???

now you are just trying too hard......

i have a full page of hints and calculations and BJT relationships on the back of the hmwk that im trying to relate to the hmwk.

so if i try to hard, thats my bad.

so lets see, Ie=Ib+Ic can i get Ic from doing XXv/2200 (ohms) to get X.XmA and that would be Ic???

VE = 0. plain and simple. VBE=VB-VE=.7, so VB = .7

(VC-VB)/100K [or is it VB-VC??? I always got that wrong] will give you the current [see, now we just used KCL/KVL], but we don't have it.....actually we do......

there, that is more than enough info to solve this.

now my only question to you is why must Ve be zero? is it because there is only one ground, and that happens to be at VE? would this change if i were to add a ground connection before VB?

you know, your right....my bad.... in order to ve to equal zero, there cannot be a resistor....i made that mistake in electronics and never learned from it.....

 

[MikeMike]

Originally posted by: Gibson486

Originally posted by: MIKEMIKE

Originally posted by: Gibson486

Originally posted by: MIKEMIKE

Originally posted by: Gibson486

Originally posted by: MIKEMIKE

Originally posted by: Gibson486
I'll give you a couple hints....

notice that VE is connected to ground....


To find VC, notice that it is connected to a resistor....use it to find the voltage

can i use Thevinans Theorem for this? Vth = (R2/(R1+R2))(VCC)???

now you are just trying too hard......

i have a full page of hints and calculations and BJT relationships on the back of the hmwk that im trying to relate to the hmwk.

so if i try to hard, thats my bad.

so lets see, Ie=Ib+Ic can i get Ic from doing XXv/2200 (ohms) to get X.XmA and that would be Ic???

VE = 0. plain and simple. VBE=VB-VE=.7, so VB = .7

(VC-VB)/100K [or is it VB-VC??? I always got that wrong] will give you the current [see, now we just used KCL/KVL], but we don't have it.....actually we do......

there, that is more than enough info to solve this.

now my only question to you is why must Ve be zero? is it because there is only one ground, and that happens to be at VE? would this change if i were to add a ground connection before VB?

you know, your right....my bad....

i am right about what???

 

[RaynorWolfcastle]

Originally posted by: Gibson486

Originally posted by: JohnCU

Originally posted by: Gibson486

Originally posted by: JohnCU
if I wasn't drunk, I could figure this out. However, I just spent the evening figuring out the magnetic field around a coaxial cable.

That coarse sucks, I got a D and i do not plan on retaking it.

Yeah, but it's kinda cool that I got the integration to work out to wear I learned why a coaxial cable doesn't cause interference, there's no magnetic field outside of it.

Wait till you start working with time varying E fields in waveguides, there's a party in the making. Try working out the field equations for the various guided modes inside a cylindrical dielectric waveguide. Now THAT is a good time Bessel functions FTW!

 

[Yossarian]

1.21 jigawatts

 

[Gibson486]

wait....are you given beta? or is beta assumed to be really big?

 

[MikeMike]

Originally posted by: Gibson486
wait....are you given beta? or is beta assumed to be really big?

beta is given at 100

i have to go to sleep, i have a 7:30

 

[Gibson486]

Originally posted by: MIKEMIKE

Originally posted by: Gibson486
wait....are you given beta? or is beta assumed to be really big?

beta is given at 100

i have to go to sleep, i have a 7:30

that changes everything....IC = IB....

 

[ramuman]

Originally posted by: Gibson486

Originally posted by: MIKEMIKE

Originally posted by: Gibson486
wait....are you given beta? or is beta assumed to be really big?

beta is given at 100

i have to go to sleep, i have a 7:30

that changes everything....IC = IB....

How is Ic=Ib?, that would be a very crappy BJT

It would help for the sake of reference if you identified the BJT as a NPN or PNP. I'll assume one and try to get you started.

Anyways Ie=Ib*(beta+1) and Ic=Ib*beta. Just pick from one of the many equations you have:

(Vs-Vc)/2200 = Ic
(Vc-Vb)/100k = Ib
(Ve-0)/1000 = Ie

You know Vb-Ve (or at least we assume it to be .7V). You can solve that system of equations to find any current because you know Vs, ground and Vbe.

 

[MikeMike]

Originally posted by: ramuman

Originally posted by: Gibson486

Originally posted by: MIKEMIKE

Originally posted by: Gibson486
wait....are you given beta? or is beta assumed to be really big?

beta is given at 100

i have to go to sleep, i have a 7:30

that changes everything....IC = IB....

How is Ic=Ib?, that would be a very crappy BJT

It would help for the sake of reference if you identified the BJT as a NPN or PNP. I'll assume one and try to get you started.

Anyways Ie=Ib*(beta+1) and Ic=Ib*beta. Just pick from one of the many equations you have:

(Vs-Vc)/2200 = Ic
(Vc-Vb)/100k = Ib
(Ve-0)/1000 = Ie

You know Vb-Ve (or at least we assume it to be .7V). You can solve that system of equations to find any current because you know Vs, ground and Vbe.

yea, its Ie ~ Ic

 

[ramuman]

Originally posted by: MIKEMIKE

...

yea, its Ie ~ Ic

Exactly, thats safe to assume to within 1% since beta is 100. Were you able to solve this?