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Well, one of the tricks used in CMOS design to simplify logic is to selectively invert entire signals. The overall logic function remains the same, but it is possible to simplify the stages and remove extra inverters. Sometimes it requires adding a couple of inverters to the inputs and outputs, but in general this technique can significantly simplify the ...


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Logic gates get faster when the supply voltage increases. What you call the "time constant" of the charging path, isn't constant: it depends on supply voltage. If you think of it as an \$RC\$ circuit, the capacitance \$C\$ remains roughly constant, but \$R\$ decreases with supply voltage (remember that with increased \$V_{GS}\$ on a transistor, its ...


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Boolean operators AND, OR, and NOT are the basic building blocks: any Boolean expression can be expressed with these operations. But that's not the only possible orthogonal set of Boolean operators. NAND by itself is sufficient to express all Boolean logic expressions, because it's possible to build each of the basic AND, OR, and NOT equivalents out of ...


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Quick answer - a FET is not a bipolar transistor. Vth relates to the minimum voltage present between Gate and Source: Vgs(threshold) if you like. The voltage present at the source will depend on the resistance between source and drain for that particular Vgs. For an N FET, the diagram you have will not work. If the Source voltage approached the Drain ...


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The illustration is correct. VGS must be above Vth for the NMOS to be ON, otherwise it will be OFF. Therefore in your circuit the maximum possible voltage at the source is VDD-Vth, otherwise the NMOS would be OFF. Note that in your circuit VGS=VDD-(VDD-Vth)=Vth, so it is ON, but the voltage at the source cannot increase further because it would turn OFF the ...


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Complex CMOS gates yield better transistor counts and fewer gate delays in this sort of factorization. F = (!a && !c) || (a && d) F = !( !(!a && !c) && !(a && d)) F = !( (a && c) && !(a && d) ) That's a OAI21 and a NAND2 -- 6 FETs in the OAI and 4 FETs in the NAND2. 10 Total. ...


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Meet the power plant for the Nutball UAV. It contains, among other things, a servo, a killer brushless fan motor, and a 2.4 GHz spread spectrum receiver. Without having the ability to perform a physical experiment, we have to perform a thought experiment. The battery recommended for the pod has no controls, save the speed controller for the prop, which is ...


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As long as it's under the logic low threshold for the end-device, what's the problem? And yes, the low-side Rds_on resistance is most likely causing the discrepancy. The guaranteed "low" for a digital system (for the 1.8-5V systems i've worked on) is usually 0.5V-0.7V or less. So if your 50MHz signal goes from 0.2V to 3.1V for example, that will still ...


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My suspicion is that the plane's physical wiring was momentarily changed: likely a transient disconnect of a wire connected to the motor. Nominally, energy is delivered to the motor through the ESC from the battery via MOSFETs configured in "half-bridge" configurations. The rate at which magnetic fields in the motor grow and decay is limited by the voltage ...


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Scale your pull-up PMOS circuitry to have the same conductance as the pull-down NMOS circuitry. For an inverter, this means scaling the PMOS to be 2-3× wider than the NMOS, because holes are 2-3× less mobile than electrons.


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When reading an SRAM bit, both column wires can be driven high (precharged) before raising the row wire high; one of them will then be pulled low, while the other one won't and will remain in the precharged state. To write an SRAM bit, one of the column wires should be pulled low while the other is either precharged or pulled high. Turning on the access ...


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The bit line is the source. To write a zero the bit line is driven low and M6 is turned on. M6 is designed to just sufficiently overdrive M4. Feedback of the latch completes the latching action. M5 operates in a like manner to write the one state.


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I think you had a power supply dropout or overload. When a spinning motor is suddenly stopped, as I think it would when it hit the ground, there is a sudden surge in current into the motor as the controller tried to get it back up to the desired speed (the controller does not know it hit the ground). That will make it increase the voltage sent into the ...



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