I have the following problem to solve from this year's exam

The picture and the asc files are attached!
I have chosen random mosfets and BJTs, but for the problem I have the following transistors info

BJT's
hFE = 200
1/hoe = 200 kΩ
hie = 25 mV/Ib

MOSFETs
K= 0.5 mA/V²
Vth = 1 V

The Vo is at R4.

First question is to find the Q-point for the MOSFETs and for the BJTs. Teacher also says to ignore the base current of Q3 when finding the Q-point of M1 and M2.

This is the circuit:


I started with the following:

Q1

Vbe = 0.67 V
Vce = Vbe = 0.67 V

Vc = -Vcc + Vce = -15 V + 0.67 V = -14.33 V

I_R3 = (Vcc - Vc) / R3 = (15 V - (-14.33 V)) / 7.2 kΩ = 4.07 mA

I'm not sure if what I done is correct or not, but I think it is!

Now, I'm also not sure if the following is correct or not:

I_R3 = I_C1 + I_B1 + I_B2
I_B1 = I_B2 = I_C1 / hFE

--------------------------------------------------------------------
-Vcc + R3*I_R3 + Vce1 - Vcc = 0

R3*(I_C1 + I_B1 + I_B2) + Vce1 = 2*Vcc - Vce1

R3*(I_C1 + 2*(I_C1 / hFE) ) = 2*Vcc - Vce1

R3*I_C1*(1 + 2/hFE) = 2*Vcc - Vce1

I_C1 = (2*Vcc - VCe1) / ( R3*(1 + 2/hFE) )

I_C1 = (30 V - 0.67 V) / ( 7.2*(1 + 2/200) )

I_C1 = 4.03 mA
--------------------------------------------------------------------
I_B1 = I_B2 = 4.03 mA / 200 = 20.15 μA

So, I already have Q1 Q-point which is:

Q_point Q1 -- (0.67 V; 4.03 mA)

Now, I'm not sure if I already can evaluate Vce2, but I think I can't.

I tried to go now for the MOSFETs:

I_C2 = I_B2 * hFE = I_C1 = 4.03 mA

I_D1 = I_S1 = I_C2 / 2 = 2.02 mA

I_D1 = K* (Vgs - Vth)²
2.02 mA = 0.5 mA/V² * (Vgs - 1)²
Vgs1 = -1 V or Vgs1 = 3 V

Only the 2nd value is of interest because for the MOSFET to be on the active zone, Vgs must be greater than Vth, so

V_gs1 = 3 V

Now, I'm stuck here! I have a couple of questions:

In this kind of circuits, both MOSFETS V_GS must be equal or not? If so, I think I can keep going, if not, I'm really stuck! I can't figure out a way to find V_DS!!!

 

I'm going to assume that both V_GS are the same.

So, with I_D1 I can find the voltage drop at R1 and with that, I can find V_DS1.

I_D1 = 2.02 mA

V_R1 = I_D1 * R1 = 2.02 mA * 6.8 kΩ = 13.74 V

So,

V_DS1 = V_D1 - V_S1 = 13.74 V - (- 3V) = 16.74 V

If this is correct,

Q-point of M1 and M2 are supposed to be the same:

Q-point M1 = M2 = (16.74 V; 2.02 mA).

Forgot to mention that if Vgs1 = Vgs2, then the voltage at Q2 collector is -3 V, meaning that:

Vce2 = -3 V - (-15 V) = 12 V

So, also found Q2 Q-point:

Q2 Q-Point = (12 V; 4.03 mA)...

 

Q1 and Q2 form a current mirror, so the 4 mA flowing in R3 will equal Ic in Q2 and split evenly between M1 and M2. So the bias point of M2 will be Vcc+ -(R2*2mA).

 

Well, I think I can write one more equation to finish the Q-points!

-Vcc + R2*I_D2 + Vbe3 + R4*Ie3 - Vcc = 0

Ie3 = (2*Vcc - R2*I_D2 - Vbe3) / (R4)

Ie3 = (2*15 V - 6.8 kΩ * 2.02 mA - 0.67 V) / 4.7 = 3.32 mA

Ic3 = Ie3 / (hFE + 1 / hFE) = 3.32 mA / (201/200) = 3.30 mA

And with this I can find Vce3 as:

V_R4 = 4.7 kΩ * 3.32 mA = 15.6 V

Ve3 = -Vcc + V_R4 = -15 V + 15.6 V = 0.6 V

Vc3 = Vcc

Vce3 = Vc3 - Ve3 = 15 V - 0.6 V = 14.4 V

So, Q3 Q-point is:

Q3 Q-point = (14.4 V; 3.30 mA)

 

Q1 and Q2 form a current mirror, so the 4 mA flowing in R3 will equal Ic in Q2 and split evenly between M1 and M2. So the bias point of M2 will be Vcc+ -(R2*2mA).

Thanks for your reply, but I already passed that point! I have been a bit more pedantic than that! but, yeah, the error would be minimal!

Actually, if my assumptions are correct, probably I figure out all Q-points! Just wanted someone to check all my calcs! Thoroughly if possible!

 

So, to summarize:

Q1 Q-point = (0.67 V; 4.03 mA)

Q2 Q-point = (12 V; 4.03 mA)

M1 Q-point = M2 Q-point = (16.74 V; 2.02 mA)

Q3 Q-point = (14.4 V; 3.30 mA)

I would really like that someone could check these results!

Thanks!

 

M1 Q-point = M2 Q-point = (16.74 V; 2.02 mA)

If the power supply voltage is only ±15V, then where does the 16.74V come from?

 

1. Q1/Q2 form a current mirror. Current going through R3 = (Vcc+-Vcc--Vbe) / R3;
2. Half of that current flows through R2: Ir2=Ir3/2;
3. Q3's emitter sits at Vcc-Ir2*R2-Vbe.
4. from that, you can calcualte Q3's Ie.

 

Hum, I need to go back and check where I gone wrong. Indeed, it's weird the voltage above the power supply...

 

1. Q1/Q2 form a current mirror. Current going through R3 = (Vcc+-Vcc--Vbe) / R3;
2. Half of that current flows through R2: Ir2=Ir3/2;
3. Q3's emitter sits at Vcc-Ir2*R2-Vbe.
4. from that, you can calcualte Q3's Ie.

Why you say that I_R2 is half of I_R3?

 

Maybe I went wrong when I forgot to say that :

V_R1 = 16.74 V

but the voltage at M2 (or M1) is:
Vcc - V_R1= 15 V - 16.74 V = -1.74 V

So V_DS2 = -1.74 V - (-3 V) = 1.26 V

So maybe the correct M1/M2 Q-point = (1.26 V; 2.02 mA)

 

morning Psy,
On your asc sim circuit you have a VERY important wire missing from Q3 , Collector to Base.!

Eric

 

Hi there!

it's correct that I(r3) = ((Vcc+ - Vcc-) - Vbe )/R3, so it's about 4mA. Then Q1 and Q2 form a current mirror, hence, under the assumptions that Q1 and Q2 hfe's are very high, and that there is no early effect, you can assume that Ic(Q2) is 4mA too.
Then you know that, to calculate the bias point, you short circuit Vsig, therefore Vg(M2) and Vg(M1) are 0. Assuming Ib(Q3) = 0, then, the current equally splits to the sources of M2 and M1.
Assuming they are in their saturation regime, Id= k * (Vgs-Vt)^2 Therefore (Vgs-Vt)^2 = Id/K = 2 V^2, and as you correctly found, the only correct answer is Vgs = 3V.
Now, Vd(M1) and Vd(M2) is simply Vcc - V(R1) (or R2), therefore it is 15V-2mA*6.8kOhm = 1.4V. Since Vg = 0, and Vgs = 3V, it follows that Vs = -3V, and Vds = Vd-Vs= 4.4V.

Beside, considering that Vds - Vth < Vgs (4.4-1 < 3), you find that we correctly assumed that M1 and M2 were in the saturation regime (you MUST ALWAYS check your assumptions!).

Therefore you find also Vce(Q2) = Vs - Vcc- = -16.6V.

Vo is simply Vd(M2) - Vbe, i.e. about 0.7V. Therefore Ie(Q3) = I(R4) = (Vo - Vcc- )/R4 = 3.3mA.

cheers

 

morning Psy,
On your asc sim circuit you have a VERY important wire missing from Q3 , Collector to Base.!

Eric

You mean the wire from Q2 collector to Q2 Base? Or you actually mean Q3?

In my exam circuit I don't have that wire from Q3 collector to Q3 base.

The correct circuit, that matches my exam's, is the one in the picture. When I get home, I'll correct the .asc file.

Now, Vd(M1) and Vd(M2) is simply Vcc - V(R1) (or R2), therefore it is 15V-2mA*6.8kOhm = 1.4V. Since Vg = 0, and Vgs = 3V, it follows that Vs = -3V, and Vds = Vd-Vs= 4.4V.

I understand that your values are wrt to a voltage drop, hence, the positive values. But if Vgs = 3 V and Vg = 0 V, then Vs = - 3 V.

If Vd = Vcc - R1*Id1 = 15 V - 13.74 V = 1.26 V

Finally, Vds = Vd - Vs = 1.26 V - (- 3 V) = 4.26 V.

The difference is probably approximations I have done. I'm replying from my phone but when I get home I'll try to see where did I go wrong...

 

Yep, I'm using only approximate values

 

Yep, I'm using only approximate values

Nice, I also already spotted my mistake.

Back up there I wrongly assumed that Vd was the V_R1 drop... It's was there my mistake.

Later when I get home, I'll gather the most important values all in one post to be easier to check them. Next step will be to find Adm and Acm, aka Differential and Common gains...

Firstly I need to find the output impedance of the current mirror. Then I replace the current mirror by that impedance and finally I'll try to find the referred gains.

 

You mean the wire from Q2 collector to Q2 Base? Or you actually mean Q3?
In my exam circuit I don't have that wire from Q3 collector to Q3 base.
The correct circuit, that matches my exam's, is the one in the picture. When I get home, I'll correct the .asc file.

hi,
On your opening circuit its Q1 and on the LTS circuit its Q3.

E

 

Firstly I need to find the output impedance of the current mirror. Then I replace the current mirror by that impedance and finally

If my memory serves me well, you already have found the output impedance for the current mirror when we last talk about differential amplifier.
But know your task is easer because you do not have a RE resistor. And this is why the output impedance is 1/hoe = 200kΩ.

y I'll try to find the referred gains.

What gain ?? I never herd about "referred gain".

The voltage gain of a whole amplifier is defined as Av = Vo/Vsig. And we can find it quickly. M1 together with M2 form a differential amplifier with single-ended output. And the gain for a differential amplifier for single-ended output is 0.5*gm*RL. Where gm = 2*√(K*Id) = 2mS.
And RL = Rc||RinQ3 = 6.75kΩ where RinQ3 ≈ (hfe+1)*R4 = 945kΩ.
Q3 wok as a voltage follower with gain equal to one. So the overall gain is:
Av = Vo/Vsig = Rg1/(Rs + Rg1) * 0.5*gm*RL* (gm3*R4)/(1 + gm*R4) ≈ 0.5*gm*RL = 6.75 V/V.
And the simulation confirm this result (see the file).

 

I meant Common mode gain and Diff mode gain. Those were the gains I was referring to earlier...

The output impedance for the current mirror is the same as the other example I did before? I need to check it when I get home.

 

The output impedance for the current mirror is the same as the other example I did before?

In this case without emitter degeneration resistor Zo is simply equal to 1/hoe.