KCC's Quizzes AQQ249 about a double-stage amplifier

Vout = 8*Vsrc

1) Zin=Vsrc/Is

     R=120k -> Zin=24k 

     R=30k -> Zin= -7.5k 

2) R=70k

3) considering -12V<Vout<+12V  for both amplifiers, it has to be -1.5V<Vsrc<+1.5V

4) Amp 1 -8V, amp 2 saturated at 12V 

Many thanks Cosimo for your prompt reply!

And big applause! All your 4 answers are correct and you are among the very first ones (there is only one before you and he answered via email...

Since gain=8 and since Is=Vsrc/10 - 7*Vsrc/R, you get:

1. Zin=Vsrc/Is= 24kohm (when R=120kohm) and Zin=-7.5kohm (when R=30kohm) 

2. Is= Vsrc/10 - 7*Vsrc/R =0--> R=70kohm

3. Since Vsupply=+/-12V, the range of linearity is +/-1.5V (dictated by second stage)

4. Output of first stage is 8V (gain=4), output of second stage (gain=2) is saturated at 12V

1. The amplifier amplifies *8. With this, the input impedance is 10k to GND and, R to 8*VIN. The latter makes up for a negative resistance of:r=-1/7R. Thus, input resistance is 1/(1/10K-1/7/R)= 24K for R=120k, and -7.5k for R=30k. 

2. For this, R=70k

3. 12V/8=+-1,5V (at most, with R2R out)

4. It goes into saturation at 12V output. First Op Amp has -8V out.

Bravo Gaetano; all your answers are correct; congratulations! obviously you know what are analog circuits!

Thanks Bernhard for your prompt reply! And yes you are correct for all the 4 questions; congrtaulations!

Happy (Western) New Year 2024 to all!

After a quite long "lethargy", our brains have enough time to be fully recharged!

Let's then publish the solutions of the last quiz (AQQ249 about a double stage amplifier:

Questions :
1. What’s the input impedance seen from the source VSRC for R= 120kΩ and for R=30kΩ (when
opamps are in linear mode)
Impedance seen from the source VSRC is simply VSRC/IS
One can observe : Vout = (‐4) * (‐2) * VSRC = 8 VSRC
IS = IR + I10k with IR = (VSRC ‐ Vout )/R = ‐7 VSRC / R and I10k = VSRC/10k
Therefore Is = ‐7 VSRC/R + VSRC/10k = VSRC (‐7/R + 1/10k)
The impedance = R*10k / (R‐70k)
For R = 120k, impedance = 24k and for R= 30k, impedance = ‐7.5k (thus negative !)

2. Find value of R to cancel the current (IS=0) from VSRC (we, here, suppose VSRC is such that opamps
don’t get saturated)
From the above formular : Is = VSRC (‐7/R + 1/10k) = 0 if R = 70k

3. What’s the values range for VSRC to still let the 2 opamps in linear mode?
Opamps output should stay inside the power rails. Thus the second opamp output, with
the cumulative gain of 8 must stay within ‐12V and +12V : ‐12V < 8VSRC < +12V
Therefore ‐1.5V < VSRC < +1.5V

4. What’s the values at the output of the 2 opamps when VSRC= +2 volts?
Opamp 1 output = ‐4 * 2 = ‐8 volt and Opamp2 output = ‐8V * (‐2) =”16V”, thus
saturation at +12V

Big applause for our 4 first winners:

Barry KULP (answer from email) : retired from Maxim

Ozan TASCI (answer from email): FAE at ADI Turkey

Cosimo CARRIERO, Key Account Manager at ADI Italy

Gaetano PIARINO, Key Account Manager at ADI Italy

Now, be prepared for our first quiz in 2024!

1- 1.32K for R = 120K and = 3K for R=30K

2- = -70/R

3- -1.5V to 1.5V

4- 12V

1, 1.32K when R 120K and 3K when R= 30K

2- R = -70/VSRC

3- -1.5 to 1.5V

4- 12V

Thanks for your reply mmsya836 ! Answers 1, 3 and 4 are correct; congratulations!

Answer 2 looks strange: is -70/R an impedance unit? If so, is it negative? An indication is the right answer is a positive value in kOhm...