# Help - Amplifier with BJT and Mosfet

#### Alfa_ET

Joined Feb 26, 2011
17
Hi everybody,

I need help in this exercise of electronica. I'm not sure about the results I got. Please see my resolution and tell me where I am wrong.

Exercise:
http://img7.imageshack.us/img7/2826/digitalizar0001pi.jpg

Resolution part 1:
http://img259.imageshack.us/img259/2006/digitalizar0002d.jpg

Resolution part 2:
http://img832.imageshack.us/img832/4278/digitalizar0003v.jpg

Resolution part 3:
http://img12.imageshack.us/img12/3479/digitalizar0004wd.jpg

Resolution part 4:
http://img528.imageshack.us/img528/1580/digitalizar0005j.jpghttp://img522.imageshack.us/img522/3710/digitalizar0005im.jpg

Thank you all
greetings

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#### t_n_k

Joined Mar 6, 2009
5,447
Keep in mind for the PMOS case the source must be at a higher potential than the gate.

You have calculated for the PMOS

Vgs = 3.61V and Vg = 3.64V

Thus Vs2 should be 3.61V above Vg2.

Hence I would think Vs2=3.61 + 3.64=7.25V rather than 0.03V ....

#### Alfa_ET

Joined Feb 26, 2011
17
Keep in mind for the PMOS case the source must be at a higher potential than the gate.

You have calculated for the PMOS

Vgs = 3.61V and Vg = 3.64V

Thus Vs2 should be 3.61V above Vg2.

Hence I would think Vs2=3.61 + 3.64=7.25V rather than 0.03V ....
Yes you are right. And about Vdrain 2?

#### Alfa_ET

Joined Feb 26, 2011
17
Keep in mind for the PMOS case the source must be at a higher potential than the gate.

You have calculated for the PMOS

Vgs = 3.61V and Vg = 3.64V

Thus Vs2 should be 3.61V above Vg2.

Hence I would think Vs2=3.61 + 3.64=7.25V rather than 0.03V ....
the gain of first stage is -gm1*Rd. And about the gain of second stage?

#### t_n_k

Joined Mar 6, 2009
5,447
I would think the 'logical' thing to do would be to set VD2 [Vo] static bias voltage to 0V or mid-poiint between the ±10 rails. This is an unusual amplifier configuration I've not met before so I would have to look at the circuit more carefully to be convinced in my own mind.

In any case if VD2 is 0V and ID2 ≈ 7.27mA then RL would be 10V/7.27mA=1376Ω.

I also think your estimate for ID1 & RE might be off a little.

One can show for the NMOS

ID≈(gm)^2/(4*Kn)≈(4E-3)^2/(4*2.4E-3)=1.67mA

hence

RE≈0.815/1.67mA=488Ω

#### Alfa_ET

Joined Feb 26, 2011
17
I would think the 'logical' thing to do would be to set VD2 [Vo] static bias voltage to 0V or mid-poiint between the ±10 rails. This is an unusual amplifier configuration I've not met before so I would have to look at the circuit more carefully to be convinced in my own mind.

In any case if VD2 is 0V and ID2 ≈ 7.27mA then RL would be 10V/7.27mA=1376Ω.

I also think your estimate for ID1 & RE might be off a little.

One can show for the NMOS

ID≈(gm)^2/(4*Kn)≈(4E-3)^2/(4*2.4E-3)=1.67mA

hence

RE≈0.815/1.67mA=488Ω
Why set Vd2 to 0V? maximum excursion of the signal?

#### t_n_k

Joined Mar 6, 2009
5,447
That was my reasoning - yes. But as I noted earlier I'd have to take a closer look at the large signal conditions to confirm that supposition.

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#### t_n_k

Joined Mar 6, 2009
5,447
Did a simulation of the configuration out of interest.

I note that the onset of large signal output cut-off is fairly insensitive to the DC bias operating point of the PMOS drain.

The output THD is ~10% with an input of 400mV p-p.

Peak voltage gain is about 6x with phase inversion.

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• Alfa_ET

#### Alfa_ET

Joined Feb 26, 2011
17
Did a simulation of the configuration out of interest.

I note that the onset of large signal output cut-off is fairly insensitive to the DC bias operating point of the PMOS drain.

The output THD is ~10% with an input of 400mV p-p.

Peak voltage gain is about 6x with phase inversion.
Good work.
RL and RD had given the same value, except RE.
What simulation software do you use?
can you check the output resistance of the amplifier? I think the value is close to the value of RL.
the gain is gm1*RL but i don't know explain why.
thank you

#### t_n_k

Joined Mar 6, 2009
5,447
Yes I would think Rout≈RL. The PMOS effectively acts as a current source driver into the load, so looking back into the drain terminal one sees a high AC impedance. That leaves just RL in parallel with a high impedance.

As to the overall voltage gain I think it would be given by

$$A_v=-\frac{g_{m1}g_{m2}R_LR_D}{1+g_{m2}R_D}$$

which can be approximated by

$$A_v=-g_{m1}R_L$$

provided

$$g_{m2}R_D>>1$$

The latter is true in my simulation setup as the apparent gm2 value is much greater than the value of 5mA/V stated in the original problem definition. That's dictated by the PMOS type and its parameters.

Looking at circuit values in my simulation the gm1 value works out around 4.3mA/V and this equates well with the observed voltage gain of 6x. Since |Av|≈gm1*RL=4.3*1.4=6.0