The Circuit Pro
Quiz - 22 January 2021
Correct answer: A
The power dissipated by each branch is simply 5V*I: in the branch a) the current will be higher than in branch b) because of the lower forward voltage of the Schottky diode (about 0.3V), that implies an higher voltage drop on the resistor, hence an higher current and higher power dissipation w.r.t. the branch b).
Effective calculations:
Ia = (5 - 0.3)/1k = 4.7mA => Pa = 5*4.7m = 23.5mW
Ib = (5 - 0.7)/1k = 4.3mA => Pb = 5*4.3m = 21.5mW
What people answered
Correct answer: B
If an nMOS is used as high-side switch, the gate voltage must be able to go higher than the supply voltage of the MOSFET in order to switch it on correctly (Vgs > Vt while Vs = Vcc). It's clear that the maximum gate voltage in this circuit is Vcc, so it would work only if we replace the nMOS with a pMOS.
What people answered
Correct answer: A
The relationship between output and input voltage for a buck converter (in CCM) is: Vo/Vin = D, where D is the duty-cycle of the signal Vc. So the output voltage clearly increases if D increases.
Quiz - 16 June 2020
Correct answer: C
"Op" is the select signal of the output MUX: with Op = 0 the selected signal is the output of the AND gate. Since Ainv = Bin = 1, the inputs A and B get negated before reaching the AND gate (the MUXs select the path with the NOT) and we know that, from the De Morgan's theorem, A*B = (A+B), that is a NOR operation.
Correct answer: A
How to obtain 2's complement:
- represent the absolute value of the number with 4-bits: -3 --> 3 (absolute value) --> 0011;
- invert all the bits: 0011 --> 1100;
- sum 1: 1100 + 1 = 1101 (2's complement of -3);
Correct answer: A
Every MIPS instruction is 32-bits long so it takes 32-bits to be stored in the instruction memory.
Quiz - 20 May 2020
Correct answer: A
A Dickson multiplier is a DC-DC voltage converter. In particular, this type has a 5x mutliplication factor, that means Vout = 5Vin.
To know more, check out the related post.
Correct answer: B
EEPROM indeed, stands for Electrically Erasable Programmable Read Only Memory. Not to be confused with an EPROM that must be erased with UV radiation, or with a PROM that can't be erased.
Correct answer: C
Please, refer to the image below. Note that the PROM is programmed, i.e. some fuses are broken, disconnecting some MOSFETs from GND and therefore making them inactive.
The vertical lines (readout bits), are all pulled-up by the PMOS transistors at the top, so if there is no pull-down action on a line by an NMOS, this line will be HIGH (read as '1'). An NMOS can pull-down the line only if the gate is HIGH (the decoder has selected its row) and it's connected to GND (fuse is not broken).
Since the address is "01", the active output is Y1: only the NMOS connected to this row can pull-down the line. Now notice that on this row the only NMOS still connected to GND is the second one, so only the second column (bit) will be LOW. The readout is therefore "1011".
