Analog IC Interview Questions

[Puppet1]

Any possible suggestions and/or solutions would be helpful!

**broken link removed**

 

[borodenkov]

Which question are you intersted in? All of them are very basic problems, although for some of the last ones I do not understand what they ask to find

 

[icpilipala]

basic information

 

[cetc1525]

a test to review the concept.

 

[Puppet1]

Questions 5,7 -- any ideas on answer for these ?

all questions require you to solve the output voltage/current at different points.

10 is for control theory.

 

[cmosbjt]

Question? I can see only graphs.

 

[Puppet1]

there is only a graph in question 10.

the others are all circuits.

what is 5?

 

[segabird]

what are the problems??

 

[Puppet1]

questions are here

**broken link removed**

 

[minxin_zhou]

Basic questions about mixer principle and amplifier.
1) mixer
2) fs should be larger than frenquency times 2
3) closed loop amplifier
4) comparator
5)Degeneration or source follower Bipolar amplifier
6)2 staged amplifier
7)Reference?
8) Current Reference
9) differentical input single output Bipolar amplifier
10) Bode diagram

 

[yyliang]

Where is the problem?

 

[duron999]

what is the question 1 answer ???

 

[eda4you]

sin (2 pi 1kHz) * sin (2 pi 10 kHz) = 1/2 (cos(2 pi 9kHz) - (cos(2 pi 11 kHz))


two sin signals with half the amplitude at 9 and 11 kHz.

Question 5 also solved! Will draw a transfer chart.

 

[Puppet1]

7 would be nice too!

have no idea what that is!!

 

[layes2]

2)the sample frequency is not enough
I think

 

[eda4you]

Question 5:

The gain for β>> is as follow:

A=-R1/R2=-3


The transfer characteristic:

for Vin 0V - 0.7V: Ve≈ 0V and Vc ≈ 10V.

From 0.7 V, a current Ic will flow. Since the gain is equall -3, with increasing Vin (from 0.7 on) the Vc will decrease with a gain of -3. E.g. for Vin = 1.7 V (=1+0.7), Vcc ≈ 7V (= 10 - 3*1).

But note: When Vin ≥ Vc, the basis-collector diode isn' in reverse state. It is in forward mode, resulting that the Transistor does not amplify anymore! It will act like a modified switch.
Therfore we have to calculate the voltage where Vin = Vc!

Vin=Vc

Vin = 10V - R1 * Ic

We know that Ic ≈ (Vin - Vbe)/R2, then

Vin = 10V - R1 * (Vin - Vbe)/R2 = 10V - 3*Vin + 2.1V

Solving for Vin yield to:

Vin ≈ 3V!

From vin 3V to Vcc=10V the Vc will have a gain fo +1, since it "follow" the Basis voltage vin.

Ve has a gain of +1 since it follow vin too.

The following transfer chart has been simulated with orcad.

 

[eda4you]

Question 7:

A kind of current source with positive temp. coeff.

Analysis: ( I supposed that the upper left transistor has an Area of A=1):
Ic=Ie since β>>

transistor referring:
Q1 Q3

Q2 Q4


vbe1 + vbe4 + Ic4*R = vbe3 + vbe2

since vbe=Vt*ln(Ic/Is)

Vt*ln(10µ/Is) + Vt*ln(Ic4/(4Is)) + Ic4*R = Vt ln(Ic4/Is) + Vt ln(10µ/Is)
since 4 times A!!!! for Q4
solving for Ic4 (which is equall Iout!)

Ic4 = Vt*ln4/R ≈ 36mV / 36kΩ = 1µA (correted to 1µA THANKS Jordan76!!)

And Vout min (that means Vout should be > than Vout min to work as above described): Vout min = Vbe1 + Vbe4 + 36mV ≈ 1.4V

Hope this is right! Maybe someone can be tell me the pros of this circuit to common ones!!!

 

[SkyHigh]

Answers to Quesntions in **broken link removed**

1) If you multiply two pure signals (Cosine or Sine), you get two sum and difference frequencies. Therefore 10KHz * 1KHz, you get 10KHz - 1 KHz = 9KHz, and the other 10KHz + 1KHz = 11KHz. This can is shown in the Table of Complex Identities of any two pure signals, Sine and Sine, Sine and Cosine, Cosine and Cosine. Basically Sine and Cosine are the same, just difference by a 90° phase.

2) ADC suffers from aliasing. Nyquist Sampling Theorem states that the Sampling Frequency should be at least 2 times the maximum frequency at the input of the ADC, i.e. Fsample >= 2Fin or Fsample should be at least 12KHz. For proper reconstruction of the original signal, DAC should also employ Fsample >=2Fin.

3) This is a DC Inverting Amplifier. Using Kirchoff's Current Law, Sum of currents at a point V+ is zero. Vin/R1 = -Vout/R2. Thus the DC Voltage Gain Vout/Vin is -R2/R1 = -3

4) This is a Window Detector, somewhat like a Comparator, such that V+ is (Vout-Vin)*1K/(1K +3K) = (Vout-Vin)/4.

5) This is just a BJT configured with Vin at Base. If load is tapped at Ve, this is described as an Emitter-Follower. The small-signal voltage gain is unity or 1. The phase difference is zero. Ve is one diode drop from Vin. If the load is tapped at Vc, this is the common-emitter collector output stage. The small-signal voltage gain is -R1/R2, thus there is a 180° phase difference between Vout and Vin. Vc is Vcc-Ve-Vce. In both cases, there is no DC gain. However there is a quiescent current of Ve/R2.

6) A simple 2-State CMOS Op-Amp Amplifier. A single-output (between P2 and N3) differential amplifier with a stable current source (by means of a current mirror using P1 and P2) and a biased pull-down resistor (by means of on resistance of a N3). P3 and N4 is a CMOS inverter to serve as the Pull-Pull output stage. This stage is also commonly known as a static driver and it is usually modelled to drive a capacitive load. Since the output of the differential amp is 180°, the inverter compensates this by inverting or adding another 180° to make it 0° or 360°.

7) This is the Back-to-Back Bistable. Older Bistables are made this way as memory device as such a BJT-based Flip-Flop.

8) This is the Wilson's Current Mirror using MOSFETs. Formerly used in BJTs. This kind of Current Mirror increases Linearity by reducing the current programming error of a single current mirror by having another mirror below it.

9) This is the Single-Output Differential Amplifier using BJTs with a simple current mirror as stable current source. However the output experiences a 180° phase difference.

10) Fp is commonly-known as the Maximum Linear Frequency. If Ao is 1, Fp is the Unity-Gain Bandwidth. Fc is the Maximum Operating Frequency or Maximum Bandwidth. This kind of plot is sometimes known as the Bode Plot for analysing Frequency Response of a circuit, a system or a device.

 

[eda4you]

High Skyhigh!
your answer to question 5 is partly wrong! You wrote there is no dc gain!? Of course there is !


Look at my pdf figure (simulation) !!!

 

[SkyHigh]

To simplify my answer, I am refering that there is no DC voltage gain.

It depends which point you are driving. There is a Small-Signal Transconductance Gain, gm and it's a function of AC current gain, hfe. There is however DC current gain, or HFE, which is fixed with the BJT.