Power in use indicator please ?

The only switch is in the pump house is directly Wired from the fuse box to that switch and then to the pump. The switch happens to be a ground fault protected Outlet. Which I added as the original was not fault protected. Th e current setup uses the power as it comes from fuse box in house.

On slightly separate topic but related I thought I would set up the origina l diode circuit on a test board to understand and learn from the concept. I'm using 400v 3a.diodes and 100 watt light bulb. I will not use this for anything over 3a, befor you ask. However it does not seem to work. I get no ilumination of led. I have tred lowering the R value down to 10r. Still no light. I then tried to increase the load, not exeeding the diodes amps still nothi ng. I'm frustrated it's not workinh. I then found a circuit by David Johnson. Popular on this NG years ago. He has a circuit almost the same as the original giving near top BUT he uses o nly 3 diodes and 10r. I tried this and no luck. I wondering if it me or not . I know that David Like many of you are very knowlegabe. I can post David circuit if you like. But I'm not home right now.

to that switch and then to the pump. The switch happens to be a ground faul t protected Outlet. Which I added as the original was not fault protected. The current setup uses the power as it comes from fuse box in house.

nal diode circuit on a test board to understand and learn from the concept.

r anything over 3a, befor you ask.

d lowering the R value down to 10r. Still no light.

hing.

e has a circuit almost the same as the original giving near top BUT he uses only 3 diodes and 10r. I tried this and no luck. I wondering if it me or n ot. I know that David Like many of you are very knowlegabe. I can post Davi d circuit if you like. But I'm not home right now.

Add more diodes in series. The diode voltage drop (sum of all diodes) has to be bigger than the LED plus resistor. And and use a red led. (less forward voltage drop than say green or blue)

George H.

not home right now.

I agree, the LEDs often take 2 to 3 volts depending on the color and make up of the diode. As each of the power diodes only give about .6 volts, it may take more than just 2 diodes. As mentioned the red LEDs usually take less voltage than the other colors.

I do know the .6 volts is just close and can vary by several factors,but as in many electronic circuits and components you just get close when in a general discussion. There are some power diodes that have an even lower voltage drop.

Sorry about the delay. I wanted to come back with proof of concept but didn't have the time until now.

I rigged up the circuit with different diode types and different loads. Here's a combined image of two of the results. Note that we use a 230V standard here. (The soldering is poor. It was done in a hurry with diodes from a >30-year-old stock). The shots were taken on different days with different cameras under different ambient lights.

The first one is with 1A 1N4007 diodes and the load is a 100W incandescent bulb (0.435A). The glow with a 25W load is very dim. The second one uses 6-amp 6A4 diodes driving a 1000W (4.35A) resistive load. It's also very dim with a 100W load.

Thanks pimpom.

I too have created some frankenstienian circuits with my diodes. Just for fun to see how it works. What I have learned is that the diode method is load specific. For Example you have to know the specific load your using to get a good led light.

I'm testing with 4 x 3a 400v diodes 2 and 2 in series and a 10r resistor and red led and get reasonable led ilumination with 100 watt bulb.

However if I go to a new led type 10watt bulb i get little illumination.

So if just for argument you wanted a general power indicator this circuit might not work on lower power. You would presumably change the number of diodes and possibly the R.

Conclusion, if you are going to use something like this as a power indicator know the specs of your load and tune the circuit to it. To get a nice ilumination of the led. I will say this is roughly what people said I think but always good to test it out for fun.

At least that's my take away.

+----->|---->|---------+load +------|

Maybe I didn't make myself clear in my earlier posts but that's exactly what I meant from the beginning: The circuit can be used as a load indicator (not just the availability of power)

*because* the illumination depends on the load. It wouldn't work with low or no load.

Earlier you said that you didn't get any glow at all with 3A diodes on a 3A load. Now you're saying that you get a reasonable light with a load of less than 1A with the same diodes. ???

This is why I asked for the nameplate ratings of your motor before offering a solution. Your initial question was *not* about a general purpose indicator.

Now you're giving me advice based on my solution to your problem :-O No offense but I got the impression that you have a limited theoretical background. That's why I offered what I considered to be the simplest solution that would work with the parameters you specified. I can think of half a dozen more sophisticated alternatives that would work with a wider range of parameters.

I apreaciate that you went to the trouble to give advice and to actually test it. That's a lot. I'm sure your more knowledgable than me in all aspects but Particularly with the theory. I'm just tenacious and willing to try and learn. Thanks.

Ps. I should say. I tried your suggested circuit as mentioned but could not get it to work. I cannot say what was wrong exactly. I dont know if I had bad jumper wires etc. Thats when I said it was not working. Anyway I kept p laying with it and then it worked. Everything you mentioned including your circuit worked fine. As they say in computers :User Error my fault. Thanks

I felt the need to provide evidence of its working because 1) you said you could not get it to work and 2) I detected some skepticism on the part of others.

Diodes used in that somewhat unusual configuration could lead to errors in the wiring. Maybe that's why it didn't work for you at first. Glad to know that it did work eventually.

As I said before, you can add more LEDs in parallel. Each LED takes up only a tiny fraction of the load current and will not divert a significant amount of current away from the main diodes. This in turn will not significantly reduce the voltage drop across the diodes which is what enables the LED(s) to light up.

shunts are going to be bad for powering LEDs. current transformers make voltage much more efficiently than shunts do. (less power is cosumed by the sensor so more power gets to the pump)

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  Jasen.