# DC Biasing

#### juansta

Joined Mar 25, 2005
7

Can anyone point me in the right direction in finding the DC bias point for the attached circuit... Vds, Id, Vbe, Ib... and the rest...

The MOSFET is a n-type working in the depletion mode (Vt=-2)...

Any help would be greatly appreciated...

oh, and by the way, V1 is drawn incorrectly... the negative side should be at ground!

#### hgmjr

Joined Jan 28, 2005
9,027
Originally posted by juansta@Mar 25 2005, 08:41 AM

Can anyone point me in the right direction in finding the DC bias point for the attached circuit... Vds, Id, Vbe, Ib... and the rest...

The MOSFET is a n-type working in the depletion mode (Vt=-2)...

Any help would be greatly appreciated...

oh, and by the way, V1 is drawn incorrectly... the negative side should be at ground!
[post=6381]Quoted post[/post]​
You have posed an interesting DC circuit analysis challenge.

It would be helpful if you could supply the resistor values used in the circuit along with the MOSFET part number.

Also you have accidently assigned two resistors the same reference designation of R4.

#### juansta

Joined Mar 25, 2005
7
hehehe... interesting or not... Its driving me mad!!!

ok,
R2=10e6
R4(Left)=4.7e3
R4(Right)=1e3
R6=2.2e3

There is no model number for the mosfet, only the following params:

mue (SP?)=6.7e-2
Cox=4.3e-4
Vt=-2
Lambda (SP?)=0.03
W=35e-6
L=1e-6

For the BJT:
Beta=250
Va=75
Vbe=0.7

Now please understand that this is actually a uni assignment, and I do want to understand how this actually works... So it would be pretty pointless just posting an answer... Maybe a bit of working our or something?

I have gotten most of the way through, but I cant seperate Ib from Id to solve for both...

#### hgmjr

Joined Jan 28, 2005
9,027
I start by observing that due to the circuit configuration, the bipolar junction transistor is setting the Vgs to be -0.7 volts. That is because the base-emiiter voltage of the BJT is forward biased. Note that the input impedance looking into the gate of the MOPSFET is high enough to make any voltage drop across the resistor connectiing the emitter of the BJT to the gate of the MOSFET insignificant.

Armed with that fact and the mu = 0.067, I multiply 1/mu by 0.7 volts to get the resistance drain-to-source. That yields a value of 11.2 ohms. I then compute the voltage Vs by the formula Vs = (+V*4.7K)/(4.7K + 11.2 + 2.2K). +V is the power supply voltage. That would put Vg = Ve = (Vs - 0.7) at about 7.5 volts for a power supply voltage of +12V.

I think that's right.

The Rds of 11.2 ohms seems to be too low. It seems like it should be between 1k and 2K. With a value of 2K the quiescent DC operating point would be close to halfway between the power supply and ground which would be the ideal place for it in order to give the input signal plenty of signal headroom as it passes through this circuit.

Also, juansta can you provide the power supply voltage that is being use to power this circuit? As you can see I used +12 volts but that is probably not correct.

#### juansta

Joined Mar 25, 2005
7
The PSU is actually 10V DC.

The resistor values havent been chosen by me, thats just the way the asignment was written...

Why dont I get these values when I run it throughPSpice?!?!

That sorts out most of the MOSFET stuff... what about the BJT?

#### juansta

Joined Mar 25, 2005
7
Just having another look at your equation for the Vs voltage... its not a simple voltage divider, as you have current Ib exiting at Source node...

#### juansta

Joined Mar 25, 2005
7
I got it...
I'll post up a method tomorow when I get a chance...
I feel like kicking myself in the head for not seeing this earlier!!!

#### hgmjr

Joined Jan 28, 2005
9,027
Originally posted by juansta@Mar 27 2005, 04:53 AM
I got it...
I'll post up a method tomorow when I get a chance...
I feel like kicking myself in the head for not seeing this earlier!!!
[post=6426]Quoted post[/post]​
I look forward to your posting of the solution.

#### juansta

Joined Mar 25, 2005
7
Well here it is...

Some of the numbers dont completely match up, but close enough is good enough...
We are using the simple model of a BJT in hand calculations, and Im sure Spice uses something a bit more complex than this, so I'll just put it down to that...

Its just basically KVL loops...

I think its pretty self explanatory, but let me know if you want more info...

Now to draw the small signal equivalent!
I should be right for that, its all good fun...

Thanks for the pointers though, that really helped me look at the problem differently.

oh, and by the way, feel free to point out errors or anything that you might not agree with...

#### hgmjr

Joined Jan 28, 2005
9,027
Originally posted by juansta@Mar 27 2005, 12:23 PM
Well here it is...

Some of the numbers dont completely match up, but close enough is good enough...
We are using the simple model of a BJT in hand calculations, and Im sure Spice uses something a bit more complex than this, so I'll just put it down to that...

Its just basically KVL loops...

I think its pretty self explanatory, but let me know if you want more info...

Now to draw the small signal equivalent!
I should be right for that, its all good fun...

Thanks for the pointers though, that really helped me look at the problem differently.

oh, and by the way, feel free to point out errors or anything that you might not agree with...
[post=6444]Quoted post[/post]​
Congratulations!!!!. You solved it.

I looked at the diagram and all of the node voltages and branch currents look good.

After seeing your results, I see where my equation for Vs was in error. I missed the point that the mue = 6.7e-2 was in units of millimhos rather than mhos as I had assumed. Also I had assumed the power supply voltage at 12 volts when it turned out to be 10 Volts. That would have made my equation Vs = (4.7*10)/(2.2K+11.2K+4.7K). Solving for Vs would then have yielded 2.596 Volts. A value that is in close agreement with the value you obtained in the simulation of 2.568 Volts.

I don't know if you are familiar with Millman's Theorem but I recommend you take a close look at the technique since it helps simplify those nasty loop equation you have to deal with when using KVL. There is a pretty good introduction to the Theorem in the tutorials located on this website.

Way to go!

#### juansta

Joined Mar 25, 2005
7
hmmm...
Never heard of that theorem yet... Might have to read up on it as it sounds like a time saver...

Thanks for that...

#### mozikluv

Joined Jan 22, 2004
1,435
Originally posted by juansta@Mar 27 2005, 09:15 PM
hmmm...
Never heard of that theorem yet... Might have to read up on it as it sounds like a time saver...

Thanks for that...
[post=6466]Quoted post[/post]​
hi,