hardware engineer interview questions shared by candidates
What are the effects of a mismatched circuit?5 Answers
unwanted heat and intermodulation products
IIP2 & DC Offsets
Mismatched circuits can produce excess heat, DC offsets, intermodulation products, and other unwanted complications.
IN the digital domain, zero crossings and ISI , as well as DC offset can be compromised.Loss of received signal strength with distortion, Overheating and damage to TX drivers
Worse is not working.
How do you represent a real-world quantity in a digital system?1 Answer
What he really wanted me to do was draw the schematic of an ADC input circuit. I don't know why he didn't just ask that in the first place. Since it was a broad question, I spent over 5 minutes explaining binary representation, word size (8/16/32bit), endian-ness, and finally when I mentioned ADCs, he latched onto that and tried to get me to dig deeper. So I drew a block diagram of an ADC, focusing mostly on the interface to the microcontroller, and the channel select multiplexer. He grew increasingly frustrated and tried to get me to draw more detailed sub-circuits for each of the blocks in my diagram. Finally, I got to the actual ADC front end, and I drew a classic textbook R2R ladder circuit, and he seemed satisfied with that. By that point we had spent over 10 minutes in an interview where he was already complaining that he had a lot of ground to cover in limited time. I was so frazzled by the question that I did not even think of mentioning a successive approximation ADC as an alternative to the R2R, even though it's a much more common implementation in my experience.
If you have a 10W light bulb and a 100W light bulb, and you connect them together, which one will be brighter?9 Answers
If you connect them in parallel, obviously the 100W bulb will be brighter. But if you connect them in series, the 10W bulb will almost certainly be brighter because it limits the current going to the 100W bulb to a fraction of its normal operating current. This assumes by "brightness" you really mean total radiated power in visible + IR wavelengths, and that both bulbs have "brightness" linear with power.
I do not agree with answer for "in-series". Must have been the next to last question you had on the last day. What incadescent bulb will current limit? Assuming incandescent bulbs, the bulbs would act as a voltage divider and the 100W bulb would still give off more light (lumens in the visible spectrum) than the 10W bullb. Use an analysis whereby use a smaller resistor for 100W and a 10x higher resistance for 10W bulb. Each bulb will be alittle dimmer than in parallel, but 100W still brighter. Fluorescent bulbs would be a whole new story. Would they even work at lower voltages?
@Don you are correct up until your conclusion that "Each bulb will be a little dimmer than in parallel, but 100W still brighter." The 100W bulb is 10x dimmer if you assume resistance does not depend on current. Reality is even worse since R increases with current. Since the 10W bulb has 10x higher resistance than 100W bulb, it will absorb 10x as much power in the voltage divider: 100W bulb = 1A at 100V, or 100 ohms 10W bulb = 0.1A at 100V, or 1000 ohms 100W bulb + 10W bulb in series driven off 110V line is 1100 ohms carrying 0.1A of current. Power delivered to 10W bulb is P = I^2 x R = (0.1A)^2 x (1000 ohm) = 10W. Power delivered to 100W bulb is P = (0.1A)^2 x (100 ohm) = 1W. So the 100W bulb only receives 1W delivered power because the 10W bulb absorbs all the power in the divider. The 10W bulb will be brighter.
Case 1: Bulbs B1, B2 are connected in series. Same current flows through B1 and B2. P = I^2 / R. And given P1 = 10 and P2 = 100. So, R2 = 10*R1 =>B1 will be brighter. Case 2: Bulbs B1, B2 are connected in parallel. Same voltage across B1 and B2. P = V*I. So, I2 = 10*I1 =>B2 will be brighter.
1. When connected in series: In a series connection, current flowing across each element is same. So when 40W bulb and 60W bulb are connected in series, same current will flow through them. To find which bulb will glow brighter we need to find the power dissipation across each of them. From the relation P=(I*I) R since current is same we can say that power dissipation will be higher for the bulb with higher resistance i.e. 40W bulb. Hence 40W bulb will glow brighter in series connection. 2. When connected in parallel: In a parallel connection, voltage across each element is same. So when 40W bulb and 60W bulb are connected in parallel, voltage across them will be same (100 V in the given case). To find which bulb will glow brighter we need to find the power dissipation across each of them. From the relation P=(V*V)/R since voltage is same we can say that power dissipation will be higher for the bulb with lower resistance i.e. 60W bulb. Hence 60W bulb will glow brighter in parallel connection.
Above parallel connection are good, but when connected in series 10W bulb cannot handle 100W current and filament will fail, then no bulb will be ON.
The 100W bulb will always be brighter as long as it's working on its nominal voltage. End of story.
I have attended the same question, 40W & 100W bulb with 110V supply , all are them connected in serial Answer : Who is having more resistance, he will have more power drop also he is the brighter one. Here 100W will have less resistance and 40W bulb has more resistance, So 40W will be glow brighter than 100W bulb R1 = 110 x 110 / 100 = 2.2 ohm R2 = 110 x 110 / 40 = 302.5 ohm Answer is 40W bulb will be brighter than 100W bulb
Describe a function to check if an integer is a power of 2.6 Answers
For an integer n: If n is less than 1, return false. If the bitwise & of n and n-1 is 0, return true. Otherwise, return false.
Write the number in binary and count the number of ones in that.If the number os ones is only 1 then it the number is indeed a power of 2
first check if no is 1 then return false else write the number in binary and then check number of ones in that.if only one 1 is there then its a power of 2
I think the main idea is to use recursion function, for the integer which is larger than 0, if it is 1 return true, else return function(n-1)
See if the sum of all bits is 1. If that's the case then the number is a power of 2.
function int power2(int i); if (i%2==1) return 0; else if (i/2==1) return 1; else return power2(i/2); endfunction
If you have a 600 digit number with only 0's and 1's, and exactly 300 1's, can the number be a square?6 Answers
I am trying to undersatnd why you say the answer is No. for a 6 bit binary number, of which 3 bits are exactly 1s. 25 and 49 are possible such numbers that happens to be a square... Why can't this possibility extend to a 600 bit binary number?
The answer should be YES: @gustion: your example with 49 (7 pwr2) is correct but with 25 (5 pwr2) is incorrect. 7 in binary is 111 (3 1-bits is half of 6 bits), but 5 in binary is 101 (2 1-bits does not equal half of 6 bits). In general, any binary number with n-bits, half of which are 0's and half are 1's is a square of a binary number with half the number of bits, all being 1-value bits. In addition, the number's magnitude will be n/2-1 1-bits followed by n/2 0-bits followed by the last 1-bit. For example, lets say we have a 16-bit number. The number which will have 8 1-bit and 8 0-bit binary digits and also be a square is: 1111 1110 0000 0001 (7 1-bits followed by 8 0-bits followed by 1-bit) and this number is a square of 1111 1111 Binary number with 600 bits and 300 1-bits will have a magnitude of 299 1-bits followed by 300 0-bits followed by 1-bit, and it will be a square of a 300-bit number with the magnitude of 300 1-bits.
Well on a simple note, 9's binary is 1001, a 4 digit binary with two 1's and two 0's, and is a perfect square. The same analogy should also be true for any such number.
can someone explain why this pattern works?
Because (2^n - 1)^2 = 2^(2n) - 2 * (2^n) + 1 = 2^(2n) - 2^(n + 1) + 2^0. Such that bit n + 1 to 2n -1 will be 1, bit n to 1 will be 0, and bit 0 will be 1, which makes it n bits of 1 & n bits of 0. You can test 15^2 = 225 to understand it.
10 Gbytes of 32-bit numbers on a magnetic tape, all there from 0 to 10G in random order. You have 64 32 bit words of memory available: design an algorithm to check that each number from 0 to 10G occurs once and only once on the tape, with minimum passes of the tape by a read head connected to your algorithm.3 Answers
Good luck with that. You'd better come up with an answer in <10 minutes.
a 32 bit number can have unique values from 0 to (2^32)-1. Meaning it can have max value of upto (4 Giga -1). If there are more than 4 Giga locations, then there is repetition which is obvious, even before one starts the question. Usually, questions of repetition can be solved by "XOR"ing all the values. If the value is not zero then some number is repeated. One should solve around that angle.
I think the employer was asking about the binary search tree algorithm. This is how I would have answered it. 1. First create a binary tree for all the data(which can be 4 gb only) 2. Every time a new data is to be inserted into the memory, we should traverse the tree to search if the element is present: if present, return "element already stored and so we cannot insert it again" else, insert the element into its appropriate place. Since, its a binary tree we can search an element in (log n) time and with minimum number of passes.
Q: What does a capacitor look like at high frequency?4 Answers
A: A capacitor looks like a capacitor, inductor, and resistor in series at high frequency.
Its an short circuit at very high frequencies. Look at the impedance for a capacitor and you'll notice that the frequency is inversely related to it. (also note that capacitor impedance is purely reactive). Therefore, as frequency increases the impedance approaches 0, which acts like a short. (an inductor is the opposite)
Describe a circuit that implements the following truth table using only NAND gates. A B OUT 0 0 1 0 1 1 1 0 0 1 1 17 Answers
If you increase the width of a PCB trace, does it decrease or increase the trace impedance?3 Answers
As you increase trace width, impedance is lowered.
The question is a little tricky (not really tho). The impedance of a straight line is described by Z=R+jX, where R is the resistance and X is the reactance. As the width of the trace is increased, R decreases. If we assume that there is no inductance, then the impedance will go down, as X=1/(jwC), and C=W*L*C0. However, in a very inductive space (like trace is curly and goes around itself a lot), then X = jwL, thus increasing the complex part of the impedance. I would go with the decreasing impedance, as on the PCBs resistance is the dominant term.
If you increase the width of the pcb trace then it will help decreasing the resistance of the PCB trace. I give you a simple analogy if for example we consider the inductance as the negligible amount then, ( R=c (L/A)) where c is resistivity . And we are increasing the Area of cross section that is A. Then that is inversely proportional to the resistance and hence it decreases the impedance.
List all the possible solutions to make a hole in any metal. 2 minutes are given.3 Answers
Drill Laser High pressure water jet Friction drill Punch
Drill Punch Laser Water Jet Plasma Cut Gas Torch Grind Cut with Knife Cut with oscillating saw
also wire cut, using chemical like acid, arc welding
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