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A 10-bit binary number can take on 210 = 1024 values. Therefore, if you want something to settle "within 10 bits," you're saying that it needs to settle to within 1 part in 1024: $\frac{1}{1024} = 0.0009766$, or 0.1% in round numbers.

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Yes, your understanding is correct. In reality, a MOSFET is a four-terminal device. The body is not necessarily connected to the source. For the planar MOSFETs used in VLSI design the source and drain are physically the same kind of structure. So, the source of the NMOS transistor is the terminal with the lower voltage, out of the two terminals that could ...

2

If you take only the output from the final amplifier, then the transfer function is: At some value of input, the output stage will saturate and not increase any further. If we then add the other outputs, we will get (not to scale): As each amplifier saturates, then the slope reduces until the next amplifier in the chain saturates until all 4 amplifiers ...

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This is called a differential amplifier simulate this circuit – Schematic created using CircuitLab A constant current source draws a constant combined current through the emitters of transistors Q1 and Q2. If the voltage at V_in1 is equal to the reference voltage REF, and assuming that the transistors are identical, then the currents through Q1 and ...

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I want to understand why, for the REF output of INA333, there is a pull-up of 0.05V? The minimum voltage that the INA333 can produce when the negative power rail is attached to 0 volts is typically 50 mV so, if the REF pin were tied directly to ground, the output would not change for inputs that might otherwise have caused a 50 mV rise in the output. By ...

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Look carefully, M1 and M2 are simply made up of different number of the SAME transistor, i.e (W/L)N. M1 contains 1 unit and M2 contains K units. This is how you achieve the best matching. Trying to match Vth of different sized transistors is a hopeless venture even on the same chip! Heck, Vth matching of same sized transistors in different locations is ...

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It appears that the published specifications for the sm-s4315r are poor. Sending a 1.5 ms control pulse tells the servo to stop. It appears that normal operation assumes a continuous pulse train of about 50 pulses per second. Pulses shorter than 1.5 ms result in continuous rotation in the clockwise direction. Pulses longer than 1.5 ms result in ...

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Your circuit is Ok from the point of view of the input signal polarity since it adds an offset to the input in order to avoid its saturation when $V_\mathrm{in}$ goes below zero: note however, from the picture you show, it seems that you need a voltage gain of at least  A_v=\frac{V_\mathrm{out}}{V_\mathrm{in}}\approx\frac{1.5\mathrm{V}}{50\cdot 10^{-3}\...

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Use an opamp circuit, non-inverting config, with a gain of 30. I. E. R2/R1 = 29 Connect R1 to gnd through a large cap. Make a resistive divider across your Vsupply to get midpoint of 1.5v Connect your signal through sufficiently large cap to non-inverting pin. Connect resistive divider centre point also to inverting terminal through large resistor (say ...

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The only way to control the current is to control the voltage. If you don't want to do this digitally, and your FET can handle some dissipation, put in a current sense resistor and differentially measure the current through the resistor and the voltage across the load. An analog multiplier will take these readings and output power. Compare your power ...

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Everything you create in the IC layout contributes to a parasitic in some form. Each metal wire encounters inductance since current will create a magnetic field, and likewise metals near other metal will create stray capacitances. Pre-layout simulation cannot account for routing in this way (but it does account for the known capacitances of your FETs and ...

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Every conducting structure on an integrated circuit has capacitance, resistance, and inductance. You usually didn't ask for a capacitor or a resistor or an inductor, but they are present inherently. To perform an accurate simulation you must include the effects of these parasitic elements. The process of parasitic extraction is the estimation of these ...

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