11

The KTC noise is actually the total RMS noise power at the output of the low pass RC filter. The noise power is the product of the noise power spectral density (PSD) and the bandwidth (more accurately "effective bandwidth") of the system. So, actually the total noise power does depend on the bandwidth. Interestingly, for the case of RC filter although the ...


9

The MOSFET with the minimal Rdson is not always the most suitable. Those with a very low Rdson have a large Qg gate charge. It means that they need a powerful gate driver if you want to do PWM. A higher gate charge makes turn on/off slower, thus higher switching losses. The good MOSFET is the one with the lowest product: Rdson * Qg The total loss of a ...


7

I like Russell's suggestion of optical methods. Another approach is a capacitance meter : water has relative permittivity of about 80, therefore measuring the capacitance between two insulated plates a fixed distance apart will show if there is some high permittivity liquid between them. It's fairly independent of water purity. Couple of drawbacks : ...


7

Your understanding of RDson and conduction loss is fine. Just know RDson varies with Vgs and increases with temperature, usually by 150% to 200% at its hottest, so you must account for this if you plan on running lots of current or hot weather. You can find the RDSon vs. temperature graph in the datasheet. But know there are switching losses if you are ...


5

That resistor speeds up turnoff and increases maximum allowed collector circuit voltage. The turnoff speeding is based on the fact that the charge which is stored to the BE junction is dissipated in resistor R2. The effect is substantial in pulse circuits. There's some leakage from C to B and it can cause substantial unwanted base current at high operating ...


4

It makes just about zero sense to build an Instrumentation Amp from discrete components if you really need the CMRR of an instrumentation amp. Because of resistor mismatch and temperature drift, there is almost no chance that you can match the CMRR performance.


4

But is the only reason one would use a difference amplifier over an instrumentation amplifier cost? No, one great advantage that a difference amplifier provides is substantial "beyond the rails" operation. Consider the ADI difference amplifier below: - Look at +IN - it sets the scene for the voltages at the actual inputs to the internal op-amp. If "REF" ...


4

simulate this circuit – Schematic created using CircuitLab Let's make some assumptions first for making the analysis quite simple. The LED requires current of 100mA for glowing. All three transistors are identical with gain = 100 for each transistor. 3.The base-emitter (Vbe) is 0.7V for each of the transistors. I am going to give a theoretical ...


4

simulate this circuit – Schematic created using CircuitLab Figure 1. A thyristor latching circuit. Thyristors have the characteristic that once triggered by injecting enough current into the gate will latch on until the current through them falls below the hold-on value - typically a few tens of mA. Pretty much any regular thyristor should do the ...


4

The coil inductance is generally not important. In the case of your SMPS the inductance will result in a more gradual increase in current when the relay is connected than a resistor of the same value as the coil's resistance would. The reason to care about inductance is generally not about switch-on but about switch off when you need to have snubbers fitted ...


4

It is useful to set output DC voltage. \$V_{REF1} and V_{REF2} \$ are DC voltage and at the input there are two resistors: by doing some calculations: \$V_{ref out}=\frac{V_{REF1}+V_{REF2}}{2} \$ In the first image, \$ V_{ref1}=V_{ref2}=5 V\$ that means \$V_{ref out}=5 V \$ and this voltage will be amplified by the OpAmp and the output DC voltage will be ...


4

My first suggestion here is to place at least 100nF ceramic disc capacitors across the motors to reduce the EMI emissions. You could also place additional 1-10µF capacitors in parallel with the ceramic capacitors. If that doesn't help, you could add small inductors (one to the positive, and one to the negative side to each motor) and the 100nF caps towards ...


3

Ideally, a transistor should have ZERO resistance when turned on, but we don't live in an ideal world, so we're stuck with just getting the lowest resistance we can. I don't understand why would you even ask such a question, seeing that you understand that higher RdsON means more losses. OF COURSE you will try to get the lowest RdsON available. There are ...


3

Slope detection: If your AM tuner is tuned to one side of the FM carrier frequency, the received signal strength will vary with the amplitude of the instantaneous frequency delta. Thus it will look a lot like an AM signal.


3

A shunt resistor value is a trade-off between how many millivolts you can afford to lose, and how easy it is to measure the voltage. Common values for the voltage drop at the rated current are 50 mV or 100 mV. If you want 50 mV for 5 A, then the resistor would be 0.05V/5A = 0.01 ohms. If you can afford to lose a higher voltage, then it will be easier to ...


3

the ktC noise formula simply happens when you insert the noise bandwidth of an RC filter, B=1/(4RC), into the formula for voltage square over a resistor. The R cancels out. So, no, there's correctly no bandwidth in that formula, because the bandwidth is "hidden" in the C. The wikipedia article on Johnson-Nyquist Noise actually containst this explanation, ...


3

Any combinatorial logic function can be implemented with just NAND gates. And you can make a NAND gate from an AND and a NOT. Therefore any combinatorial logic function can be implemented with AND and NOT gates. OR is an example of a combinatorial logic function. Therefore an OR gate can be implemented with AND and NOT gates. can all gates be built ...


3

Yes, that's a legitimate way with AND gates. NAND or NOR gates are usually the primitives available so an XOR would be built from those in practice.


3

You should study why open-ended and closed-ended pipes resonate: - Picture from here It's the same mechanism as an electrical stub (open or short). So, in the top example the closed-end pipe will resonate when the applied frequency is such that the length (L) is a quarter wavelength of the frequency. When the pipe is resonant it acts like a parallel ...


3

that was an easy search: Analog devices has an application note on "Op Amp Precision Positive & Negative Clipper" using one of their opamp models, but you can basically use any rail-to-rail opamp (or an opamp who's got enough headroom given your power supply). The schematic is dead simple. It's just a comparator with a diode. (OUTBUFFERED is optional, ...


3

In the context of assembly language programming, a "register" is a primitive object in the ISA (instruction set architecture), which is what defines the interface between software and hardware. A register stores a single value, and the number of bits determines how many distinct values it can hold: N bits can hold 2N values. Some instructions will allow you ...


2

Here's one way to do it. simulate this circuit – Schematic created using CircuitLab I've used one more switch than is strictly necessary, but seeing as you have an infinite number that shouldn't be a problem :)


2

The problem is that low voltage Zener diodes have a soft 'knee' and relatively high leakage current - too high for your sensitive analog signal. The easiest fix is to simply use a higher voltage Zener which has lower leakage current (eg. BZX84C12 passes < 100 nA at 8 V). However this will cause the protection diode in the ADC input to inject some ...


2

Here is the type of sensor that @RussellMcMahon mentions (photo from here) Other methods include capacitance, heat loss, and a simple mechanical float switch (usually a floating magnet operates a sealed reed capsule). There is also the ancient one-shot flood detection method of putting an aspirin tablet in a clothespin. When the tablet dissolves, the ...


2

You are literally shortening the gate and source of your PMOS via a resistor, meaning that if you had only a DC signal as the supply voltage the MOSFET would be always OFF (open). The small signal that you see across R1, arises from the fact the MOSFET gate capacitance takes some time to charge and discharge, therefore for a very short moment, the voltage ...


2

up to where you estimated the voltage I understand what you do and it is correct. Now, the As such, we can set Vout=(1 V/μT)B and solve for the ratio of R1/R2 that solves properly calibrates the signal, no? I don't understand what you mean by that. However, I think you're on the right track. You have $$V_{coil} = N\pi r^2 B_0$$ and you want $$ A\cdot ...


2

As another answer says, you can't get it with your proposed design. You need a source of 10 V. But you can get that by adding a single additional resistor: simulate this circuit – Schematic created using CircuitLab Now when IN is high, you'll get about 0.2 V at the MOSFET gate. And when IN is low, you'll get about 10.4 V at the gate. There is a ...


2

Is this the sort of thing you are after? simulate this circuit – Schematic created using CircuitLab


2

The feedback from each NOR gate should go back to the AND gate on the same side. Also the flip flop needs to be triggered with a very narrow clock pulse. Master-Slave flipflops for completeness


2

Yes and no. simulate this circuit – Schematic created using CircuitLab Figure 1. The wiring circuit. The problem is that motors don't behave like resistors. The more load you put on the motor the lower its resistance will become and the more current it will try to draw. If you got the right lamp it could protect the motor enough but when the motor ...


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