# Tag Info

1

How can I accomplish this? I would buy or build a small buck converter that takes a range of input voltages (say 18 V to 36 V) and delivers your logic supply (5 V or 3V3) for the Arduino. Maybe one of these: - The NPN transistor would probably have its emitter grounded with the motor wired between collector and 24V - you will need a flyback diode across ...

2

My preferred input capacitors for SMPS circuits are ceramic (X7R); generally speaking that is sufficient for most designs but it does depend on the load current to some extent (because that determines how large a 'gulp' of current is taken from the input during switching). I always specify X7R devices with Vin * 2 as a rated voltage as that gives me some ...

0

As each relay will dissipate 140mW, the power dissipation will be the same for either part. You mention in your second paragraph 'current dissipation'. That is a nonsense, you perhaps mean current consumption? Your two suggested power supply configurations are both reasonable, but are not directly comparable, as one uses an LDO to get to 3.3v, the other a ...

1

... that current dissipation in 5 volt would be higher. Current isn't "dissipated" - it flows through the relay coils. The power is dissipated in the coils according to $P = VI$. Since the power required to pull in the relay is 140 mW (a constant, whatever coil voltage is selected) then if the voltage is reduced the current must go up. If 12 V is ...

2

That is called via stitching, and is used for shielding or thermal purposes (connecting the copper planes of the PCB together to better shield the internal signal layers from interference, or connecting them together to make a larger heat sink to keep the connected components cool). This is very common in high-power or high-speed (high-sensitivity) circuits.

0

Use 7411 and gate to stop counter and reset back to zero. This link may help you understand better. http://hassam794.weebly.com/digital-clock-using-4026-ic.html

2

To meet the criteria for an oscillator, you need to have positive feedback at a gain of one or greater. Your op amp as configured will act as an integrator at higher frequencies, so you have a 90-degree phase shift (in addition to the 180 degrees of the negative feedback) and with the instrument amp's higher gain at the high end, you have met the criterion ...

1

Cost (money and time), and availability. There are several "standard" transformers that can be bought ready, for example it is likely that you can buy 1:1 with different cores and wire gauges. If you need an uncommon ratio, multiple secondaries, off-center taps or a combination of these, then you find yourself in the "long tail" of the market, where ...

2

Yes you did! NAB=Vt for any core saturated topology including all 50/60 Hz transformers, and Vrms=4.44fNAB for sinusodial waves. You can assume 1 for B for laminated iron core or perhaps 0.8 for a conservateive design (you are trading no-load losses for copper losses here). Wire gauge is normal thermal heating for all low frequency transformers or rule of ...

4

But zener diode is just a clipper that regulates the output voltage In this case it is not, it just drops about 18 V. Suppose 100V AC came out of the transformer there would be 18 V across the zener but around 80 V across the 100R and the 470R resistors. There would still be almost 100 V at the output. The output voltage is regulated by the zener and the ...

5

The zener, here, is not used as a "clipper", as you say. In this situation, it is used as kind of voltage reference, used to provide feedback to the TNY267 IC and maintain regulation. When the voltage on the output rises above the zener voltage, the zener starts to conduct. This allows current to flow through the two resistors (100R and 470R). This creates ...

7

The zener DOES control the output voltage - at about 19V across the output terminals, the zener begins to conduct and starts to turn on the the opto-isolator. This in turn signals the TNY267 that the correct level of output voltage has been met and it's time to start backing off dumping too much energy into the transformer. If loading increases and the ...

3

Assuming VCC5 is 5V and VCC12 is 12V, it won't work because the PMOS FETs won't turn off. To turn off they need +12V on the Gate, but the 555 only puts out +4V (leaving 8V on the Gate). Another potential problem is that you don't have any 'dead time' when switching between the upper and lower FET. During the transition both FETs will be on and a large '...

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