# Tag Info

6

It's not normally that much of a consideration. Many pots are used as potentiometers (as opposed to rheostats) with negligible wiper current and/or used way, way below their rated dissipation as trimmers. The pot dissipation rating is for the entire element, and any situation that exposes part of the element to a large proportion of the dissipation the ...

3

The left node of the inductor where the freewheeling diode connects is not the low voltage output. When the high side switch closes it connects that node to the full voltage of the input which the diode must block.

3

For that type of power supply it should be ok to parallel or series multiple devices to obtain more output. IHA0109D05 Datsheet As you can see from the graph on the top of page 4 of the datasheet the output voltage droops significantly with increase in load. This will allow the two units to share the load current. Because of unit to unit variations two ...

3

What voltage is the battery when operating? It will also be good to look at the battery voltage with a scope because the voltage may drop significantly when the switch in the converter is conducting but rise during the rest of the cycle. I find that it is essential to use a very low ESR capacitor at the input of the converter - a conventional small ...

2

The fact that there's a bar above the "SHUTDOWN" means that it's inverted. I.e. "shutdown" is active when low. If you thus pull it high, then shutdown will be disabled, and your chip will be active.

1

Yes, there's a way, with a few more components. Let's talk about your design for a moment. Note that for the load switches to work, the Vgs must enough above the threshold voltage to turn them on fully. Your circuit doesn't do that. While the N-FET gate is at 5V, the drain will never go higher than 5V-Vgs. If the drain goes above that, the FET turns off. So ...

1

My questions are: The answer to the original question recommended a metal junction box. Could a rigid plastic box be a safe alternative if the heat from the power supply is fairly low (or is it best to play safe and use a metal box)? Plastic is a thermal insulator and raises hotspot temp and lowers MTBF but could be done with a fan which eventually needs ...

1

These power supplies are not safe to leave exposed due to mains wires connections, they are meant to be inside an equipment enclosure, but they do need quite a bit of airflow. So a small enclosure with no air circulation won't cut it. If you can put it in your lathe stand, that would be practical, since that would make the whole device self-contained, easier ...

1

The pot power rating determines the maximum current that the pot can handle in any portion of its track. If the 2 kΩ pot in your diagram has a power rating of 200 mW then the maximum current the track can handle is $I = \sqrt {\frac P R } = 10 \, \mathrm {mA}$. Maximum current in your circuit will occur when the pot is closest to the top. You only need ...

1

At some point you'll realize that distributing 5V is complicated and expensive due to physics,: A 16 A supply is expensive. Connectors rated for that current are expensive, too. You need thick copper. Everything becomes bulky, and you still see voltage drops. A simple cheap Buck converter on each of your modules will allow you to do the distribution at e.g. ...

1

I could only imagine this circuit working if a resistor is added between T1 collector and base in order to bias the base of T1 (let's call it Ry), and PC being a phototransistor. Given this assumption, then R6, D4, and C1 act as a "voltage" buffer. When no signal is given to A1, PC is open, T1 conducts biased by the added resistor Ry. Z1 protects ...

1

I wonder, what I am missing. The specification says 600mA output current. Well well. These cheapo DC-DC converters always spec the biggest number they can find in the datasheet with the unit of "Amps" dangling at the end. Usually it's the big number in the title of the datasheet, which means... Datasheets usually put in the title the biggest ...

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