# Interview ques? which was asked. Need some help in solving

Status
Not open for further replies.

#### gmailbond

##### Junior Member level 1 1.One inverter is connected to 20 inverters. Wp/Wn of all 20 inverters is 2/1. Scale the driver inverter appropriately.

2.What are the critical issues during synchronising between 2 clock domains?

3.Two capacitors in parallel with a switch in between. Initially switch is open and C1 is charged to 10V. After that the switch is closed. Find the voltage on C2.

4.Design a logic to count the no. of 1's in an 8 bit register?

#### research_vlsi Re: Interview ques? which was asked. Need some help in solvi

Solution for 4 th Ques

declare count=4'b0; data is ur 8 bit input

for(i=1;i<=7;i=i+1)
if(data==1)
count=count + 1;
else
count=count;

Count is ur output.

#### LECS

##### Newbie level 5 Perhaps 4 is in a list of standard interview questions somewhere, because I was asked that at an interview for a PC software post. My answer was that it depended on project priorities such as small code, fast code, or easy-to-maintain maintain code. I presented solutions based around bit-shifting, a lookup table and bit-by-bit checking.

#### shanmugaveld for 4th Ques
count : std_logic_vector(3 downto 0);

count <= reg(0) + reg(1) + reg(2) +
reg(4) + reg(5) + reg(6) +
reg(3) + reg(7) + reg(8);

#### arjun1110 Re: Interview ques? which was asked. Need some help in solvi

For the third question:

The voltage on C2 will be five volts. As soon as switch is closed the charge stored on C1 will start discharge till C2 is equal charged.

Charge on parallel plate capacitor
Q=CV -----------> 1

V=Q/C ----------->2

When the switch is closed, charge on each capacitor

Q1=Q2=Q/2 ---->3

Voltage on capacitor C2 is

V2=Q2/C
=Q/2C (substituting value of Q from equation number 3)
=V/2 (Replacing Q/C from equation # 2 )
=10/2
=5 Volts

Regards

#### gharuda

##### Member level 3 Re: Interview ques? which was asked. Need some help in solvi

1) the size of the inverter could depended upon the speed. ie rise time and fall time of the circuit

so the 20 inverters are driven by one inverter. in this case all of the inverter are of same size the delay from the input to the output is

T phl=T plh =(Rn +Rp)(C out+20Cin) ,so from Tphl and Tplh size of the inveter can be design

2.What are the critical issues during synchronising between 2 clock domains?

1 operating frequency: when two domains are of different frequency.
2 phase : frequency been same but their phase differ.

#### eecs4ever

##### Full Member level 3 1.

the total load size is 40/20, so our driver should be roughly 1/3 to 1/4 the size. here i think i'd go with a 14/7 inverter.

3. arjun110 is correct only if C1 == C2.
Q= CV , Q here is 10 * C1, C=C1+C2, so
Vout = 10*C1/(C1+C2)

#### jorgito

##### Full Member level 4 Hello,

In the proposed solution to question 3, it is valid only if capacitors are the same value, condition I do not see stated in the question.

Regards!

#### arjun1110 Re: Interview ques? which was asked. Need some help in solvi

for question three.

Irrespective value of C2, voltage on C2 will be 5volts.

Reason:
Assume two water tanks t1 and t2 of different capacity with the diameter of the two tanks remains same, only height varies and both tanks are connected at the bottom through tap.

Now assume tank t1 is filled with ten liters of water and the tap is made to open, so what amount of water will be discharged into t2? It will be five liter only even though it is has higher capacity to store.

Any other answer with reason will be always welcome.

#### jorgito

##### Full Member level 4 Hello arjun1110,

Beware of analogies, specially if you dont use the correct one. The electrical capacitance is better modeled as tank's diameter than its height.

In your derivation, you state that Q1=Q2, when you must say V1=V2, being the capacitors in parallel.

If you want convince yourself, you can carge a 1uF capacitor and discharge it over a 1000uF.
If you prefer a more mathematical way, proceed calculating the initial stored energy in the small capacitor and the sum of energies on both after discharging.
With the given values, E1= (1 e-6 F * 10V**2)/2 = 50uJoule
After discharging, and assuming as you say that both capacitors have 5 V across them, the energy stored in the big capacitor will be:
E2 = (1 e-3 F * 5V**2)2 = 12.5 mJoule
Even desregarding the energy still stored in the small capacitor, the final energy in the system is orders of magnitude greater than the initial energy.

Regards

#### arjun1110 Re: Interview ques? which was asked. Need some help in solvi

Hi gmailbond,

>>3.Two capacitors in parallel with a switch in between. Initially switch is open and C1 is charged to 10V. After that the switch is closed. Find the voltage on C2.

When the switch is closed, will the supply voltage is still connected across C1 and C2 ?

Or the circuit is only two capacitor connected in parallel and capacitor C1 is initial charged to 10v before closing the switch and suuply is removed?

Hi jorgito,

If he says yes to first one then voltage on C2 will be 10V.

Or in case 2 what will be voltage on C2?

Regards,

#### gmailbond

##### Junior Member level 1 Re: Interview ques? which was asked. Need some help in solvi

Hi arjun,
The question was as it is with no other info given ,u need to assume the values for the capacitor,when we asked the interviewer.

So like wat is the voltage of C2 with the assumed conditions
1.C1=C2
2.C1≠C2

Status
Not open for further replies.