Re: International standards (2023 Update)

Thanks all. On the basis of that advice I'll be cautious about anything over say 2.5A until I've used it enough to assess long term temperature rise.

--
Terry Pinnell
Hobbyist, West Sussex, UK
Reply to
Terry Pinnell
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Looks wonderful, Terry! And right about 3 seconds, or so.

(Personally, I prefer to _NOT_ bus the power around in a schematic. The extra lines detract from the thrust of the idea. I suppose they are excellent for those worrying about soldering up connections, though.)

Jon

Reply to
Jonathan Kirwan

Assuming a side-loading dryer here...

Dryers often have a narrow belt that runs around the entire drying drum that you put your clothes into. This belt is then attached to a motor. Usually, the motors last a long time, but it is possible that the motor is "shot" and needs to be replaced.

However, it's probably MORE likely that something is hanging up (or adding resistance to turning) the drum. This can be piles of lint, dirt (and if you have rats running around, wads of clothing they decided to drag into your dryer.) What happens in this case is that the drum gets harder and harder to turn, until it reaches the point that it can actually keep the motor from turning it.

The humming you hear suggests to me that you have power and that the motor is probably getting some -- hence, the humming. But that the motor is unable to turn freely for some reason. If you open up the dryer enough (top panel with the controls on it, for example) to reach the drum with your hand, try turning it yourself. If it's not "really easy" and seems to have a modest amount of friction to it, chances are that you've got garbage in there that needs cleaning out so that the drum can turn. On the other hand, if it turns very easily, it's probably your motor -- could be the bearings are worn, for example.

Even if it does "feel easy" to turn, get in there and clean out the dryer's interior. Open up the front face of the dryer where the lint catcher is at and make sure that the entire front area right down to the floor is lint free. Open up the rest of your dryer and examine everything to make sure it's basically clean of lint and wads of "other stuff." Get anything out of there that might be hanging up the drum or adding friction when it turns.

Chances are, that will do it. If not, my guess is that your motor needs work or replacement. Probably about US$100, or so.

Jon

Reply to
Jonathan Kirwan

Perhaps I have find the error: in my preceding calculation I used, to calculate beta, the DC Ic and Ib, instead of the AC values! Now infact using the new beta ( AC values ) all is OK!

--
Per rispondermi in privato togli NOSPAM.
Reply to
Francesco

A transformer is rated by power, which is directly related to the cross-sectional area, core material, and to some degree, frequency. In general, if your new line power transformer has about the same core cross-sectional area, then the power rating is essentially the same. Therefore, given a power rating (say of 100 watts), the current capability can be calculated by a simple equation P = I * E (all RMS); a

30 V winding could be made to handle about 3.3 Amps - provided the wire size used gives almost zero I*R loss. Other secondaries can eat up so much space that a smaller than adequate (for max current) wire size would be used, thereby limiting the rating for that winding.
Reply to
Robert Baer

Those gross figures suggest an overall output of about 60VA. Stretch it to say 70VA tops, which is probably pushing a bit.

Assume that every winding was designed with the same ratio of power output to copper loss. A big assumption I know, but it allows a reasonable guesstimate of the power output of each winding. As below.

Winding Current VA-out Copper Loss ~~~~~~~ ~~~~~~~ ~~~~~~ ~~~~~~~~~~~

Grey 0.08A 0.80W 0.05W

Blue 0.42 7.56 0.49

Red 1.19 13.00 0.84

Orange 1.87 48.60 3.15 ------- ------ Totals= 70 4.53 ------- ------

Those max output currents are for a resistive load, and should be appropriately reduced for a rectifier load and capacitive input filter.

That total of 4.5W copper loss on the secondary side should be matched by another 4.5W in the primary.

4.5+4.5 plus say another 2W of core loss suggests that the transformer would take about 81VA off the 240V mains at full load, ie about 0.337A full load current.

You didn't give the primary resistance, but the sums above suggests that it should be about 40 ohms. That would be a useful cross check to see if the sums are anywhere in the right ballpark.

--
Tony Williams.
Reply to
Tony Williams

Hi,

Yes, it *is* a Tait (David Tait) type programmer.

There are a number of utilities available out there to drive this, but the most popular, free one arguable is the IC-PROG.

This software is well supported, and somewhat universal in that it is configurable to drive a number of programmers.

For any further help, do write to me.

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There are a number of variations to the original, as in some versions use transistors in place of the 4066 IC, but the end result is the same.

Another favourite software of mine that would also support your programmer is

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Please note that if you are running XP on the PC, you would require the XP driver on the ICPROG site, *and* configure it to run in compatibility mode (Win95 or Win98).

For any further help, do write to me.

Regards,

Anand Dhuru

Reply to
Anand Dhuru

Why not put some Thermal Switches in the Circuit. So if anything overheats, It Shuts off. Much cheaper than New Transformers and other parts. An Old Computer fan to blow air through it is also a good idea.

Take care.......Gary

Reply to
Gary Lecomte

I suspect it doesn't; I suspect those two screw terminals are for power. Look for what goes into the big VR chip.

Yes indeed! And you're programming a PIC '84.

David Tait's TOPIC, my NOPPP, and several others fit this general description (thought what you have is definitely not NOPPP). For many relevant links see

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(And -- please -- don't start another flurry of e-mail. Designing NOPPP was one of the most e-mail-creating things I've ever done; there must have been

1000 support requests, or more.)

bottom

It might be for adjusting the programming voltage. You may have to trace the circuit.

Reply to
Michael A. Covington

sorry to take so long to get back to you, but it looks like you got plenty of good help.

as for the trick, is a matter of recognizing that a sub circuit composed of the opamp, R2 and c2 is just a standard inverting amplifier and replacing that with the gain equation for the standard inverting amplifier. Sorry that I didn't explain that.

But as a general rule that is the way I attack complicated circuits; try do identify known circuit blocks and replace then with an equivalent equation.

--
local optimization seldom leads to global optimization

my e-mail address is:   AT mmm DOT com
Reply to
Roy McCammon

Robert, Tony: Thanks, both. Reckon I'll downrate that 2.5A guess to

1.5A for time being then.

You're very close, Tony. The three primary windings measured 4.5, 42.5 and 46.9 ohms respectively. I used the last.

BTW, I was thinking of sticking my temperature sensor flush against the transformer and measuring temp with my DVM while using heavyish currents. But the heat dissipated from the 2N3716 mounted onto the rear of the case would grossly distort the result. To use that approach I'd have to disconnect the secondary, which of course means I can't use the unit for normal bench work.

--
Terry Pinnell
Hobbyist, West Sussex, UK
Reply to
Terry Pinnell

I am building a generator using multiple auto alternators and understand (I think) that wiring them in parallel will combine the amps but leave the volts the same. Is this true? I plan to then invert to AC and step up the voltage with a transformer.

However, I have read that alternators actually generate AC and then invert to DC to send the current to the battery. If so, can I disconnect this function and then skip the need for an AC inverter?

Thanks for your replies and help, Jay

Reply to
jaydee

Here's a little more, based on looking at it.

I think we've found it at:

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and in particular the pic4pgm.zip and pic84v05.zip files there.

I haven't managed to dig out the schematic yet, but it is powered by AC (probably about 12 volts) at the two screw terminals, which feed into a bridge rectifier, some capacitors, and an LM317T regulator. DC power will work just as well. You should trace the input circuit to see how it derives the voltages.

Reply to
Michael A. Covington

same

make

rise/fall

term

use

true, it doesnt, it does show a graph of the relationship between ib and ic tho and its not like an ordinary transistor at all, like i asumed it was. from the graph delta hfe goes to infinity and then goes negative. hardly much like a bipolar transistor at all its not realy a transistor anyway, its even got a pnp input stage.

i think the circuit would be highly unstable. im not sure i cld think of any use for it that actualy needed such high gain but could cope with the instability.

although thinking about it again as bias curent is so low the 10k input resistor hardly has any efect. so my original asumption i stated doesnt hold.

however it does show colector curent versus colector voltage for various base voltages.. eg, at IC = 0.825 Amp, vb = 0.825 @ vc 12.5 and vb =0.75 @33

therfore gain at that Ic seems to be 280, asuming infinite colector reistance.

Colin =^.^=

Reply to
colin

What concerns me about this one is that it requires both 5V and 13V but I only see one voltage regulator. Maybe it tries to get 5V from the parallel port, in which case I wouldn't expect it to be terribly reliable. You should trace the power supply circuit -- maybe I missed seeing a 5-volt Zener or something.

I think this is the Maplin kit described by David Tait. Are you in the UK? If so, bug Maplin for some information.

Reply to
Michael A. Covington

Hi, I am still having trouble with my amplified ear. I noticed that my headphones have a dc resistance of 16ohms per ear piece. The ones specified in amplified ear has impedances of 32 ohms wired in series to create a 64ohm load on Q4.

Could this be why my circuit does not work?

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Reply to
pil

Hi

A friend of mine recently gave me a PIC programmer that he used a few years ago as part of a course he did at university. Unfortunately, he couldn't remember much about it (the programmer or the course, which I suspect has something to do with his alcohol consumption during that time period).

The end result is that I have a PIC programmer, but no software with which to use it.

I've posted a photo of the programmer at

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If anyone can recognise this programmer and tell me what software I need to use to program PICs with it, I would be extremely grateful.

The programmer has a parallel port interface and takes its power directly from the parallel port.

The 2 IC's on the PCB are labelled: UCY7406 and CD4066BCN. When I Google for"7406 4066 PIC", the results suggest that this could be a "Tait" style programmer. If it is, what software should I use for programming?

There's a resistor soldered across the top of the IC labelled UCY7406. The one side of the resistor is soldered to pin 8 (top right), and the other one has broken loose from either pin 6 or 7 (bottom left, both have traces of solder on them). Can anyone tell me which pin to solder the resistor back on to?

Lastly, there's what looks to me like a variable resistor towards the bottom left of the PCB in the photo (the blue rectangular box with the brass screw). What is this used for?

Thanks very much for any help,

Travers

Reply to
Travers Waker

one

Sorry, that should be "pin 14", not "pin 8".

I'm guessing that the resistor should be between pin 14 and pin 6, since connecting 14(Vcc) to 7(GND) with the resistor seems pointless.

Reply to
Travers Waker

I need to read a 12v pulse from a serial port, Can I connect that 12V pulse directly to the serial cable (Receive pin) and get information from there?

Reply to
Mickael

sorry for the delay, im not sure why the capacitor made your mosfet die, but isnt the mosfet going to be running a bit close to its specs? if the mosfet is dieng it seems that the mosfet is not capable of diming the lamps as it gets too hot or exceeds its safe operating area. what power is it going to disipate when its half on ?

just thought is your rectified 38vac smotthed? if not it might explain the destruction, you wld need to put in a gate protection diode with anode to gnd and cathode to gate, u cld use a 12v zener to give even beter protection, this wld stop the negative transients on the supply cuasing the gate to be driven far too negative by the capacitance, this might be a good idea anyway.

the mosfet has an inherent parasitic capacitor from its gate to drain anyway, but this is usualy in the order of tens or hundreds of picofarads. usualy its responsible for slow turn off times but this is what u actualy want anyway.

Colin =^.^=

Reply to
colin

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