# Tag Info

5

I see at least two things: First is the capacitor that's already been pointed out to you. That's this one: It can cause many kinds of problems. It needs to be replaced, no matter what else may be wrong. The second thing I see is this joint on the back of the board: It looks to me like the left pin of C810 has come unstuck. There's what looks like a ...

4

This won't work as it required the current to be constant. In reality it is somewhere between 5nA and 550µA, depending on the state within the IC. That's what the voltage regulator is for. It permanently adjusts the voltage drop over the regulator that way the output voltage is both independent from the input voltage as well as from the current drawn.

3

At this current level, if you have 5-20mA to spare, you could regulate it with a 3.3V zener and a resistor of about 160 ohms. The zener goes from 3.3V to ground, and the resistor goes from 5V to 3.3V. If you're running other things off the 3.3V supply, take those currents into account as well. When selecting a diode, make sure that over all current and ...

2

Yes technically this is possible ... although I would not recommend bringing this to practice without a firm understanding of the basics! You say low current, but depending on who you ask this might be different by orders of magnitude. (ECG specialist vs welding engineer. You get the idea.) Can you elaborate on the current requirement or application? There ...

2

From Alternating Current Machines-Synchronous Machines If you did this with batteries, you would have a big problem (your short circuit concern), but a delta (or wye) takes advantage of the phase shift of 120$^o$. $$\vec {V_A} = 450\ \angle 0 ^o\ V$$ $$\vec {V_B} = 450\ \angle 120 ^o\ V$$ $$\vec {V_C} = 450\ \angle -120 ^o\ V$$ If you do vector ...

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Replace at least this capacitor. But you could replace all four on the other part of the heatsink.

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I believe the writer there was conflating two issues. D2 would be the likely saviour of the part due to inductive loads, since the current would continue to flow in the same direction and would pull the positive rail below ground. D1 is there to prevent the part from being reverse biased as a result of a sudden loss of input voltage.

1

simulate this circuit – Schematic created using CircuitLab The left side is shown wired for 230V operation,if you want 115V use the other wiring parttern shown on the data sheet The right side has a switch to select 3.15 or 6.3V output use a DPDT switch, you can use a centre-off DPDT switch here if you want.

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Those inputs are probably matrixed, since I don't see any obvious ground traces. Matrixing means that you drive one side of some switches low, and read the other side to see which one is pressed, then drive some others low and read them with the same input lines. Matrixing allows this particular keypad to be read with 9 I/Os instead of 20, but it means ...

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You only need two Schottky diodes. The load in the picture is 3.3 V net labelled point.

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Is the “feedback” diode acting as a “fly back” diode if there’s an inductive load on the output? No it isn't. See this note from TI in their LM317 regulator data sheet: - See this from ST: - Is D2 acting as protection to the rest of the circuit in the case of reverse voltage coming in at the output? It would do this.

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Overview I think our discussion has finally ferreted out the basics. (It took a while and I wish you'd written more at the outset, providing the entire scope and goals, as well as the parts you felt you had working and why you needed what you were asking for to finish it off.) But here's what I gather and think: You have a slow signal that is basically a \\$...

1

Have you thought of asking TTi Directly? Email: info@aimtti.com They usually respond very quickly and will pass the question onto the design engineer if nobody in sales is sure of an answer. I used to work for them and have designed some of the power supplies on that data sheet though not this particular model. But to answer your questions directly. ...

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