# Tag Info

0

What is guaranteed about the strength of the x3 input, and what are the output drive requirements? The most straightforward implementation using transmission gates would use two transistors for the inverter, two for inverters on x1 and x2, and four for each mux, for a total of 18. It would, however, pass through the x3 input directly without buffering in ...

0

The above problem can be solved by using TG We have 3 inputs now for sake of less number of MOS to be used lets get $x_1 x_2 x_3$ in true logic as well as inverted logic So MOS needed for 3 inverters is $3 \times 2 = 6$ now all we need are 3 TG to be implemented NOTE: since we already have inverted logic form of $x_1 x_2 x_3$ we don't need ...

1

This is essentially what commercial alarm systems do: run power, ground, and a couple of signal wires to sensors and keypads. They don't even run twisted pair, because they're not shipping data (except for the keypad, of course, and even then they rarely bother). Alarm system sensors signal with a simple voltage divider. When the sensor trips the voltage ...

2

To answer your question, both ICs (the open-drain and the "CMOS" types) have the same output characteristics: they will pull the output high when the input voltage is high, and pull the output low when the input voltage is low. See the Output vs Input voltage graphs on pg. 9 (Fig 6). In short, either IC will work for your purpose, but the open-drain one will ...

1

No I think it's telling you when the voltage exceeds a threshold the output will go from low to high. This would make a good microprocessor reset for example. It would only release reset when the voltage was above a certain level. Then it will monitor the voltage and if it goes below a threshold then it's output will go from high to low. This is good for ...

2

An open drain output refers to an output stage where the pull-up transistor is missing. This means that rather than outputting a high voltage when the output is high, it presents a high impedance to the circuit and an external pull-up resistor is used to provide the high voltage. This is useful when there is more than one device that wants to output to the ...

0

No. OR requires 6 transistors. NOR can be implemented with 4. You can't put NMOS on top in a simple digital circuit because there is no voltage available to turn it on. You can put NMOS on top on a linear analog circuit, but you will not be able to drive to the upper rail, unless there is some higher voltage available to drive the gate. If you need to ...

0

What you have to consider is the gate-source voltage needed to turn on the FET. For example, when an input is high, you want the corresponding N-FET to pull the output to VDD. But if you factor in the gate-source on-voltage requirement, it does not work as needed.

0

Interestingly, an XNOR gate turns out to be much easier to implement in NMOS than CMOS; if the signals which are feeding to an XNOR gate aren't used for anything else, an XNOR gate may be implemented using two transistors and one passive pull-up. If the input signals are also used for other things, adding inverters on both of them would increase the total ...

0

That second one is really clever! It recycles transistors from the XOR structure to invert A and B. I bet someone got a patent out of that. I'm not exactly a CMOS optimization master, so maybe it's more common than I think. It works like your first schematic, except that T6 and T9 are also used to invert A. You can almost do the same thing with T2 and T5, ...

1

The CD4050 (and its cousin the CD4049) have a special input stage that allows it to accept inputs higher than Vdd without ESD diodes conducting. So, it should work just fine in this application. Maybe you've got some kind of a wiring error or the chip is damaged.

4

There are ESD diodes on the inputs connecting the input to Vcc. You should add a resistor in the data line to allow the voltage to drop: simulate this circuit – Schematic created using CircuitLab You should then get 1.7v less the forward voltage of the ESD diode dropped across the resistor instead of heading up to the 3.3V rail.

0

You're correct; that is not the right thing to do. The key is to take advantage of the ability to implement AND and OR via wiring. For example, here's a pull-down network that ORs together the results of two AND operations: simulate this circuit – Schematic created using CircuitLab Note that the OR doesn't require any extra transistors at all. ...

0

The gm of input transistor M2 is loaded by the rds of mirror transistor M4. Remembering that rds = 1/gds, we see that the input stage gain is gm_M2 / gds_M4. I will let you figure out how to include or exclude the factor of 2 that may result from the current mirror. The second stage has amplifying transistor M7 loaded by current source transistor M8. So its ...

2

Above 180 -> 110 nm (depending upon whose process it is) you caould easily follow the scaling rules and get a rough order of magnitude. However, moving to 65 nm and below, several significant changes take place that don't allow direct scaling based estimation to work. Things like high K gate di-electric, Low K-dielectric in the BEOL (Back End Of Line). ...

0

If you really are doing CMOS analog design and you are using large resistors you're already heading in the wrong direction. There are all sorts of issues with resistors in CMOS processes like: parasitics (capacitance, non-linear capacitance), Low resistance values and extremely poor matching. NOC design (Non-overlapping clock) is trivial in comparison. At ...

0

R1 =R2=14MOhm and C1 = C2= 0.56uF. If these values were used in a sallen key filter the cut-off frequency would be: - 0.02Hz and not 20kHz. Try recalculating based on this website and then decide what you need to do.

Top 50 recent answers are included