Re: International standards

thinking about it, this is unlikely to be the problem - the capacitor would also be keeping the supply rails high while it discharged, so it wouldn't be driving the 4538 past the rails.

a 12V 1.2Ah lead acid battery.

a 4538.

That could well be it - when the battery is low, the supply gets pulled down to around 7V. Maybe need a bigger battery.

Yes, thanks. It looks like it would work OK, but I'd rather stick with the design I have now.

Hi, I have a bunch of silver (colour), axial lead capacitors. They are about the size of a 1/4 - 1/2 W carbon resistor. I don't understand how to read the capacitor values from the markings on them, and webpages I've looked up don't help too much. I don't have a meter that reads capacitance, and I don't really want to make mistakes with the values. Hoping you can help me to understand how to read these. Here are some examples. Thank you in advance for any help you can offer. Keith creekchubbAThotmailDOTcom (replace AT with @, and DOT with .)

--example1--

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--example2--

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--example3--

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--example4-- M39003

--(there is a crown mark between the 7 and the Y on this one-----------

--example5-- M39003

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--example6-- C513B J683K

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Probably, yes, but.....

The direct connections between two "RS232" equipments with the Send Data of each connected to the Recive Data of the other is called a "Null Modem."

I am aware of two different "RS232" specs, however.

One, is the RS232 recommended standard (thus the RS prefix), and its interface is at +and - 12V. This has always been accepted as a Standard.

The other is the ANSI/EIA 232, which IS an industry Standard, and its interface is at + and - 5V.

I've mixed the two occasionally with trouble free results, but I usually use a pull-down resistor on the Receive Data pins to drop the voltage a bit.

Don

ah, the days of ECE290. I'd suggest looking around for course websites for "Computer Engineering I" or similar; I happened to find this at Rice with a little googling:

Reference to

for

can

I want to build the one I worked out for myself, not just copy another design. I'm doing this to practice/learn how to, as much as to have a working system.

and mine has 2 chips to your 3, so nerrr. ;)

Get a variable load like headlight bulbs and keep loading it up till it gets hot after several hours so the temp will stabilize.. If you can comfortably hold your hand on it you've found a load that it'll be happy with and should work for years. Remember that half wave vs full wave recitifiers will have double utilization factor.

--

You would be better off to mount a rather small fan on the end of a shaft, then drill a suitably sized hold at the stern of your boat, take proper steps to prevent water from leaking around the shaft, and connect the shaft to the motor in the belly of the boat. Make sure the fan remains under the surface of the water and rotates such that it is pushing the water backwards. This arrangement would be more compatible with Newton's laws of motion, since the fan would be pushing against the water rather then blowing air into the sails.

Good luck!

I assume you've double-checked that your 4538 (and other) connections are wired up exactly as intended? And that the chip is inserted with correct polarity! Have you swapped it for a fresh 4538, with same result? Have you used the 4538 in a basic test circuit to check it's not already zapped?

BTW, I got this message on trying that JPG link: ==================== Notification Too Many Users There are more clients using this WebWasher than your license allows. Please upgrade your license to support more users.

-------------------------------------------------------------------------------- generated 11/Aug/2004 10:00:39 +0100 by bncfms04 (WebWasher 4.4.1 Build 1003) ====================

Thanks Dan,

Looks like I need a little more research and study. I will look into the regulator issue and try to find the book you mentioned.

The alternators are the same and will be driven off the same belt so they will be in sync. Sounds like I should let them gen DC and then invert instead of the complicated disconnect of that alternator function.

Thanks again Jay

I have hand-wound hundreds of power transformers for personal use, and

Sorry, that's my careless description. There's only *one* primary, tapped in the familiar way (presumably to accommodate mains in the range 220-240V, although there are no markings). I simply measured the three combinations, and used the full winding.

orange o---------o | C| C| 4.5 ohms brown C| o---------o C| Total winding 47 ohms C| C| C|42.5 ohms C| C| C| white | o---------o

created by Andy´s ASCII-Circuit v1.23.080803 Beta

Thanks, but I mentioned what IMO is the snag with that up-thread in Message-ID:

I've since implemented the alternative I suggested. Currently (sorry ) I have the secondary delivering 2.2A (an arbitrary first choice), while I monitor transformer case temperature with my DVM.

But I still don't see how even a protracted series of such tests is going to tell me with any accuracy what I can expect using DC loads at various voltages in the range of my supply.

Ah! Ok now, thank you.

Yes

Yes

No - need to get another one.

It was working before, so don't see the need for that.

I read in sci.electronics.design that Robert Baer wrote (in ) about 'Estimating transfomer current rating?', on Wed, 11 Aug 2004:

Your windings have zero resistance? Maybe you are using a different definition of copper loss than I^2R?

This depends on the ratio of window height to limb width. For scrapless and even semi-scrapless laminations, the copper loss considerably exceeds the iron loss if proper allowance on maximum induction is made fro high mains voltage. With modern silicon iron, an iron (hysteresis) loss of 5 W/kg at 1.5 T is typical.

I read in sci.electronics.design that Terry Pinnell wrote (in ) about 'Estimating transfomer current rating?', on Wed, 11 Aug 2004:

If you are using a series regulator following a bridge rectifier with a large filter/reservoir capacitor, the a.c. secondary current will be between 1.6 and 1.8 times the d.c. load current. It has a peaky waveform, so you need a true r.m.s. meter to measure it, or use your scope. The waveform is close to repeated half-cycles of a higher frequency than 50 Hz, maybe 150 Hz, interleaved by zero-current periods. So you can take the r.m.s. value as the peak value divided by sqrt(2) and then divided by the ratio of the duration of the pulse to the whole half-period, i.e. 3, if the pulse looks like 150 Hz.

The load voltage is irrelevant, because the rectifier always produces the full voltage across the capacitor.

All you need to check is that the transformer doesn't get too hot with the maximum current you want to draw from it.