Re: International standards

So from the way you talk the motor does not run at all? Gas or electric? Did you find a schematic? Sometimes taped to the back plate inside. sometimes on the back of the controll panel. Check the door interlock switch. It's usually activated by a small button operated when the door is closed. You can bypass this switch and see if it operates. This is the most common problem.

Did you check to see if the belt is continuous? Properly engaging the idler pulley?

Richard

I read in sci.electronics.design that Terry Pinnell wrote (in ) about 'Estimating transfomer current rating?', on Mon, 9 Aug 2004:

As a rough guide, for that mass of transformer, you can take the load regulation to be between 5% and 10%. Using just one secondary at once, you can probably go for 10%. So the load current that causes the voltage to drop by 10% from the no-load voltage is the maximum permissible current.

But don't forget that the AC current from the transformer is much larger than the DC output current - 1.6 to 1.8 times for a bridge rectifier.

The final test is 'how hot does it get?', which depends on where you have put it as well as its own properties. It should not get hotter than you can touch and stay touching, as, again, a rough guide.

This should be OK:

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.

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

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

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!

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

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.

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.

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.

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

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

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

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

(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

bottom

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

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.

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

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.

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

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

I think we've found it at:

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.

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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 =^.^=

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.

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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