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When implemented in CMOS, simple inverting logic gates take one stage, that includes inverter, NAND, NOR. Non-inverting logic gates take two stages. For example, a buffer would actually be two inverters back to back. An AND gate would actually be a NAND gate plus an inverter... I assume your professor did the following: \$Y = A + B * C = ...


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Standard CMOS gates used in IC design are inverting for a variety of reasons, including lower area and delay compared to the non inverting versions. See this Question. This also needs to be taken into account; not just the number of gates. It would be inefficient to use an inverter when the inversion can be optimised away in the logic by using an inverting ...


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Both transistors would have Vgs below their respective threshold voltage, so only leakage current would flow Drain-to-source. Consider this circuit as two complementary source-followers connected together, each with 20K source resistors.


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After Studying my circuit I was able to incorporate one extra component and changed a resistor to a diode. I hooked a signal diode cathode to the output of my IC and tied a resistor to the 15 volt supply. Joined the resistor and anode together and tied it to my output circuit that I needed to drive, giving me plenty of drive voltage. I had to increase my ...


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No, you need a voltage translator (maybe discrete transistors or a CD40109). Worst-case 11 volts is required according to the datasheet with a 15V Vdd. 5V is also not enough to guarantee operation with an 8V supply- it will "usually" work, but that's not good design.


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You are exactly right. And any resistor value less than about 10k will work fine, as long as it is more than about 330 ohms (Look at the data sheet for the maximum current into a low output, Iol(max). Assume 15 mA. Then a low output will nominally put 5 volts across the resistor, and R = V/I = 5/.015 = 333.) 2.2 k is a perfectly good value. If you're doing ...


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I can't remember reading this in a book, but I can try to derive it. (Here considering an open-loop inverter without anything else loading it) Transconductance (\$g_m\$) is defined as: $$ g_m = \frac{\partial I_{out}}{\partial V_{in}} $$ The transconductance of an NMOS is known as: $$ g_{mn} = \frac{\partial I_{ds}}{\partial V_{gs}} $$ According to this ...


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Speaking about "transconductance" you are referring to a circuit in which a CMOS inverter is used as a linear amplifier. This is possible if we fix a suitable dc operating point in the middle part of the transfer characteristic Vout=f(Vin). This can be simply done with a feedback resistor RF between output and input. This feedback resistor RF acts as a load ...


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Speaking about the general case, IR explains how gate series resistance can be important for controlling turn on time. http://irf.custhelp.com/app/answers/detail/a_id/215


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The 50mA draw indicates that the chip has been damaged, and ESD is a likely cause, or trigger, for such damage. A damaged chip could exhibit strange behavior as you indicate, perhaps because of power supply bypassing that is insufficient for such a high fault current (it's possible that if you shunted the 100nF cap with 1000uF it would appear to work ...



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