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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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I suspect they use a sequential method where they weakly drive the pin high and low and sample the states. There are four possibilities in the truth able, one of which should never happen. Since they do not seem, to be too concerned about floating outputs connected to SA0, a reasonable CMOS input should be okay- but I'd be concerned about long conductors ...


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You can achieve exactly that effect using two CD4015 Serial Input, Parallel Out shift registers. Call them IC1 and IC2. Each chip gives a pair of 4 stage shift registers. You'll only use 3 of the 4 shift registers to make a 12 stage shift register, of which you'll use 11 stages. Connect all the clock inputs of IC1 and IC2 (pin 1 and 9) together to your ...


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This sounds like a job for a CD4017 and a bunch of or gates. (you have to provide the clock.) I used a 4017 in a circuit, fairly recently. Re-reading I'm not sure about the going out part.... more parts or learn a uC. (Arduino is good because it's easy.)



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