Hot answers tagged


A Zener diode voltage regulator is highly dependent on the load current. When designing such a circuit, you have to know the current (or range of currents) you will be drawing from this regulated voltage. For proper regulation, you want the Zener diode current to be between a certain minimum value \$I_{Z,min}\$ (see the datasheet for a working curve, let's ...


It depends what level of 'design' you want to go for. If you want to design the circuits of the inverters, then it would be an interesting and long learning exercise for you, and the result, in a year or two's time, would be more expensive and almost certainly inferior to anything you could buy. If you want to choose panels, inverters etc from a catalogue, ...


If your A-to-C breakout does not have sufficient impedance control and other signal quality means to support Super-Speed signals (and even HS signals), your USB 3 drive won't work. And usually all commonly available C-C and A-C breakouts don't meet the necessary signal integrity requirements by long-long mile. Different CC logic won't fix this.


Why not operate a Zener diode at its load current? Then its voltage is spec'd.


I don't think you have any chance at all of getting this to work. The distance (1 m), multiple loads (which probably mean multiple unterminated stubs), and the fact that what you have is a single ended (as opposed) to differential interface are all working against you. This is a signal integrity (SI) nightmare. If you think otherwise, I would like to see ...


Is this even possible with a 20MHz clock signal without using a buffer on each board? It all depends on the drive-strength of the clock source, how many milli-Amperes can it source? You can approximate the trace resistance using: trace width and copper thickness on the PCBs wire gauge on cable and figure out the DC loss at the end of the chain. That ...

Only top voted, non community-wiki answers of a minimum length are eligible