Electrical engineer interview questions

  1. What types of CMOS memories have you designed? What were their size? Speed? Configuration Process technology?
  2. What work have you done on full chip Clock and Power distribution? What process technology and budgets were used?
  3. What types of I/O have you designed? What were their size? Speed? Configuration? Voltage requirements? Process technology? What package was used and how did you model the package/system? What parasitic effects were considered?
  4. What types of high speed CMOS circuits have you designed?
  5. What transistor level design tools are you proficient with? What types of designs were they used on?
  6. What products have you designed which have entered high volume production?
  7. What was your role in the silicon evaluation/product ramp? What tools did you use?
  8. If not into production, how far did you follow the design and why did not you see it into production?
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7 Comments on Electrical engineer interview questions

  1. Edric Sizemore
    Posted 2/19/2007 at 3:14 pm | Permalink

    We have a portable device that operates with 12 volts of direct current. We currently use 8 AA 1.5 volt
    batteries to generate the 12 volts. However, the
    8 AA batteries add too much weight to the device.
    Is it possible to substitute the 8 AA batteries with
    one 9 volt and two AA to create the 12 volts?

    Or, can we utilize two 9 volt batteries with a converter to create 12 volts of direct current? Our
    objective is to reduce the amount of batteries required, thus reducing the weight. Can you please assist me.

  2. Vince
    Posted 5/1/2007 at 8:16 am | Permalink

    Use two 9 V battery and build a simple resistive voltage divider. The setup is 18 V from the two 9 V batteries connected in series and in series with 2 resistor R1 and R2. The desired output voltage 12 V will be taken between the 2 series resistors R1 and R2. The formula is Vout=[R2/(R1+R2)]xVin. Or you can first get the ratio [R2/(R1+R2)] that when multiplied with 18 V will result 12 V. Then the resistance value can be computed.

  3. Skip
    Posted 8/2/2007 at 5:50 pm | Permalink

    While Vince’s suggestion above would work to make 12V there is a series flaw. If you have a load on the 12V (which of course you do) then you will draw current through R1. Thinking about that, you are going to drop your voltage due to power dissipation in the resistor. Besides, the resistors are always consuming power in this situation so it is not such a good idea in the first place.

    The 9V + 2x 1.5V battery idea works but is still kinda clunky.

    Is there more information on the design, ie. Current requirements, battery life expectation, etc. If so, then we can look into a more efficient design.

  4. Posted 4/8/2008 at 5:00 am | Permalink

    We have a ABB Motor whose spec is
    Model - M2 QA 160L 4A
    (Ref-ABB / low voltage motors / Cat. Asia region / 400 V 50 Hz EN 05-2006)

    1) Power 15 Kw
    2) Voltage - 400 V, 50 Hz
    3) power factor - 0.86
    4) Current - Is - 27.97 amps
    5.Rated RPM : 1452

    My Question is : How will the RPM vary when the Voltage Comes down to 380 V or it goes up to 415 Volts.

    Pls give me a curve/Charactersitics or a Forumale by which related Supply voltage & RPM.

  5. abhishek.k
    Posted 10/20/2008 at 1:17 pm | Permalink

    what is power factor?what is their purpose in loads..

  6. Edwin
    Posted 11/6/2008 at 8:06 am | Permalink

    Use two of the 9V batteries in series to get 18V. Now connect a 78L12 voltage regulator to get 12V supply. This voltage is for electronic devices which accept low current and is very much stable, otherwise use a voltage devider network.

  7. Edwin
    Posted 11/6/2008 at 8:38 am | Permalink

    Power Factor can be defined in various ways but either way points to one thing:
    1. It is the ratio of the power-producing current in a circuit to the total current in that circuit or
    2. It is the ratio of the working power(in kW)to the apparent power (in kVA)
    3. It is the factor by which apparent power must be multiplied to get the working power.

    Improved power factor may give economic and system advantages or even both. It results in
    1. Power-billing savings
    2 Increased power system capacity i.e vars are generated by the loads and fed back into the system. In this case very few quantities of vars will be consumed from the supply utility and hence current drawn from the system capacity falls.
    3. Reduction of power system losses
    4. Additional load can be added to a fully loaded transformeror distribution system by simply improving the power factor.

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