Measuring the Power Output of a Generator

[Borek]

Sometimes a day in the library saves a week in the lab, sometimes a day in the lab saves a week in the library

 

[Total_Insomnia]

@NTL2009 wow, a week already. Sorry, work stuff pulled my attention away from this project.

Sometimes a day in the library saves a week in the lab, sometimes a day in the lab saves a week in the library

Very true. In this case, I am trying to verify some math.

Some progress was made yesterday as I wired up the circuit and took some readings. I was able to get my hands on a power resistor as well. I ended up using a 20W 8 Ohm power resistor and two 47k Ohm resistors for the voltage divider. Hopefully, all that is left is to analyze the data. I'll Let you all know how it goes.

Thank you all for your help.

 

[NTL2009]

... Some progress was made yesterday as I wired up the circuit and took some readings. I was able to get my hands on a power resistor as well. I ended up using a 20W 8 Ohm power resistor and two 47k Ohm resistors for the voltage divider. Hopefully, all that is left is to analyze the data. I'll Let you all know how it goes.

Thank you all for your help.

How did you determine the 20W and 8 Ohm values? All we've heard from you is some comment about 6 volts peak, but no answers about anything that might help us determine the approximate power you are dealing with.

If the 6 Volts ended up being continuous (but that doesn't seem to be the case, but again, you haven't really told us much at all about the actual timing and/or duty cycle of the power pulses you are trying to capture), and it didn't have reduced output with an 8 Ohm load (very unlikely), that would be a power of (6^2)/8 = 4.5 Watts. So 20 Watts is way overkill, but that's OK, other than cost and maybe time.

Without more info, we can't tell, but I won't be surprised at all if the 8 Ohm Resistor loads the output so much that you get a very low and hard to measure voltage output. This is why it helps to try to roughly characterize what you are doing. I would have started with any old 1/4 watt resistors I could get my hands on (salvaged?) for the load - start with X hundreds of K-Ohms, X tens of K-Ohms, or X K Ohms and work your way down until you see the voltage drop off. A continuous 6 volt source will not exceed 1/4 Watt until you got down to 144 Ohms. Throw in some real-world safety factor on that, so you could easily get down to the hundreds of Ohms without over-heating a 1/4 Watt R.

If a 330 Ohm R did not drop the output voltage, then you could start looking at power resistors. Or just series/parallel some 1/4 Watt Rs to get what (watt? hah-hah!) you need. If you want to experiment like this, I'd recommend getting a pre-packaged kit of various R values. Here's one I got from Amazon - nicely labelled, much handier than digging through my scrap bin:

https://www.amazon.com/dp/B00E9YQQSS/?tag=pfamazon01-20

And a handy assortment of colored LEDS as well:

https://www.amazon.com/dp/B01EWBYTF4/?tag=pfamazon01-20

 

[Total_Insomnia]

@NTL2009 I chose that power resistor because a friend of mine had one laying about so I figured I'd give it a try. The voltage did drop to about 4V but that is fine as the board can only measure up to 3.3V so w/ the voltage divider I was taking measurements of around 2V. As a result, I'm not too worried about accuracy.

Average power generated seems to be about 450mW peaking at 2.3W.

 

[Borek]

Average power generated seems to be about 450mW peaking at 2.3W.

Now we are talking some real numbers.

And it looks like you are in the range where the 12 V halogen would work as a load

 

[NTL2009]

@NTL2009 I chose that power resistor because a friend of mine had one laying about so I figured I'd give it a try. ...

OK, that makes sense to give it a try then.

... The voltage did drop to about 4V but that is fine as the board can only measure up to 3.3V so w/ the voltage divider I was taking measurements of around 2V. As a result, I'm not too worried about accuracy.

Average power generated seems to be about 450mW peaking at 2.3W.

Sounds good, glad to see some success - you got lucky with the 8 Ohm load! Sometimes it does work out for us. So average power is ~ 20% of the peak. If you replicate this design, you can probably go with far lower watt rating R's (or a combo of common 1/2 Watt R's). Hmmm, I'm actually not aware how Resistor power ratings specify peak power limits versus average, I'll need to review that to satisfy my own curiosity.

If you care to do any further tweaking, the maximum power transfer to a load from a source is generally when the source voltage has been loaded down to 1/2 it's open circuit voltage. You are pretty close to that with 6V dropping to 4, so you won't gain much with optimization. You might want to check that during the lower power output periods, it might not be linear, and you might be better off optimizing the 80% lower periods rather than the 20% peak - hard to say w/o more study, and it might not be important to you anyhow.

Could you post a schematic and some pictures? I'm curious, that's putting out some significant current and power.

And again, what sort of load do you intend to drive? That makes a big difference. And what is the driving force?