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From the original post: "On the other hand, NMOS shares a common substrate, so if you were to connect the source and bulk, you will have to do so for all NMOS." However, I still don't see the reason doing this in NMOS will cause any problems at all. The reason this causes problems in a bulk process is that all the NMOS devices' "wells" (or ...


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Regarding the split in the diagrams, ST has separated the crystal equivalent circuit from the load capacitors with which it resonates, whereas TI have chosen to keep all the resonant components together for their description. As LvW points out, the circuit needs a resistance in the gate output to raise the "amplifier's" output impedance. Logic gates with ...


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Perhaps looking at it another way, a traditional bulk CMOS process is built on a P-type wafer. This allows NFETs to be built directly in the substrate without "extra" steps. To get PFETs in the same circuit, they need to add N-type wells to the substrate. Since this well is an added feature, you can place it wherever you want. It forms a diode to the ...


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Diverger - I think some confusion can arise because there are always two different ways to explain the functioning of oscillator circuits (models). Normally, we discriminate between two-pole and four-pole oscillators - however, it is important to know that this is nothing else than a different way to describe the oscillation principle. a) Two-pole ...


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It works when the source is at the same potential as the substrate. But not all NMOS transistors will have their sources connected to the substrate. Sometimes they are put in series, where the lowest one is connected to the substrate, and then the drain is shared with the source of the next transistor and so on. It's also possible to use an NMOS in a ...


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If you connect the input and output of a CMOS inverter, both the PMOS and NMOS FETs will be drawing current, and the exact voltage it stabilizes at is determined by the relative "strengths" (or current gains, really) of the FETs. It's not necessarily half-supply, but typically it's close. The output voltage stabilizes at the point where the current through ...


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The diodes in a MOSFET are from the source to the bulk and from the drain to the bulk (often called source/drain diodes for short). These are not really part of the ideal MOS transistor structure. Rather, these are parasitic devices that are not helpful to MOS operation, and they exist in all "bulk" CMOS processes. They add undesired parasitic capacitance ...


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The Diodes are forming from p- to both n+ Areas. The Gate contact is connected via the gate capacitor to the middle of the diodes in this model.


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Logic gates like AND, OR, NOT, and so on are not discretised in time, they don't handle time in steps. At every moment they reflect the input at some earlier moment (a few nanoseconds ago, usually). For various reasons signals can get delayed for a tiny moment of time (line length, capacitance, etc), and as a result for similarly tiny lengths of time your ...


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@sanjay: A standard CMOS NAND gate has two PMOS devices in parallel to pull the output (to supply) and two series NAND gates to pull the output down (to ground). If either input is zero, then at least one PMOS is on pulling the output up and at least one of the series NMOS FETs is off preventing the output from pulling down. So the output will not "lose ...


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With the output of a CMOS inverting gate connected to its input through a resistor, a rising output voltage driving the input past a certain point of equilibrium will cause the output to servo its voltage down until the input gets to that happy point. Conversely, if the output voltage falls below that point it'll be servoed up until the input rises to the ...


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Diverger - I recommend to study the V(in)-V(out) transfer characteristic to be found in the relevant CMOS data sheets. As you will see - the output voltage will be at Vdd/2 in case the input also is Vdd/2. Because the transfer curve of the inverter has a negative slope (rising input causes falling output) you can find a stable operating point at ...


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Consider a simple CMOS inverter composed of one n-channel MOSFET and one p-channel MOSFET: With feedback IN and OUT are connected through a resistor. Suppose IN is at about 0V. The p-channel MOSFET is on and the n-channel MOSFET is off, so the voltage at OUT is high (at \$V_{DD}\$). Due to the feedback resistor current flows from OUT to IN and charges up ...


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An inverter is just a rather nonlinear amplifier. It is possible to use 'digital' inverters to build some simple analog circuits. Generally, the initial oscillation of a crystal oscillator will be very small, much too small to get to the logic-level threshold of the open-loop amplifier. So adding some feedback allows the small signal to be amplified and ...


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Assuming an n-type substrate, and applying a positive voltage to the gate, the following will happen: The positive voltage on the gate will produce an electric field between the gate and the substrate. Hence, the positive charged particles "holes" will be pushed down-ward, leaving behind it uncovered negatively charged acceptor atoms. (they won't ...


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Building on @Peter Bennett's answer, the voltage level at the source matters very much. The transistor, along with many other electrical components, could care less about the voltage level of an individual terminal. What matters is the relationship between the different terminals. For example, if the source was at 10 volts, the gate at 15 volts, and the ...


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It's the mobile electrons that form the channel. When a positive voltage is applied to the gate, the electrons (which are minority carriers in a p-type substrate) are drawn to the surface. When you then apply a voltage between source and drain, an electric field is formed and a drift current flows. The ions could not do this, since, as you say, they are ...


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The transistor only knows about the voltage between its terminals. It doesn't know or care what the voltage is between any of its terminals and what you consider "zero volts" or circuit ground.


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The substrate (body) is (almost always) connected to the source on discrete MOSFETs. See this diagram.


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Pin 3 (CLK) of each driver should be tied to pin 3 of the 555. Pin 4 (LE) should be tied to Vdd. Pin 2 (SDI) of each driver is connected to pin 14 (SDO) of the previous driver. However between the first and last chip the signal can be inverted to create a 'first on, first off' sequence. If you assign the LEDs to random rooms then this may give a more ...


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The circuit below should do what you want. It works by inverting the output of the 40 bit shift register chain, a bit at a time, and then feeding it back to the input so that the data in the register recirculates forever, inverted each time it goes through. One caveat is that since you're not using an MCU and the chip has no RESET, if the data in the ...



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