Electronic problem / In Summary!

[lionelsoni]

I just did the experiment of dropping 18 to 14, using a number 53 lamp and 4-ampere bridges. It seemed that 3 1/2 bridges were about right. So I imagine you would need at least 7 bridges for 10 volts. However, you can share about half of the bridges. Just make a string long enough to get to 10 volts and tap into it about 3 1/2 bridges from the top for the 14-volt circuit. You could also get the 12 volts from it too, if you want, as well as any other voltage between 10 and 18 that you should need in the future.

[lionelsoni]

One diode in series doesn't cut the rms voltage in half--it reduces it to 70 percent, from 18 to about 12, as Wayne showed us above. The rms voltage is what counts here.

The Simpson doesn't measure true rms voltage. It measures ac by full-wave rectifying it to dc. The scale is calibrated to show the rms value, but only if the input is a sine wave. I would expect it to do just what you saw; but it doesn't mean anything in the case of powering the lights.

If you can set the transformer to 18 volts rms, just put a diode in series and you will have 12 volts rms for the lights. The Simpson is okay for measuring the 18 volts, since that is a pure sine wave. But you'll just have to trust Wayne and me that the half-wave voltage on the lamps is 12 volts rms.

As for the bridge-rectifier trick, the drop that you get from each bridge is in the neighborhood of two diode drops, so more like 1.5 volts. But it is complicated to calculate and not the same for each additional bridge that you add. If you had a true-rms meter you could simply measure it. But you could use the trick that Wayne did and add bridges until a lamp looks the same on 18 volts with the rectifiers and on 10 or 14 from an adjustable transformer. Remember that you can add half a bridge by tapping the + and - terminals.

[eZAK]

[:0]OOOOOOOOUUUUUUUUUUCCCCCCHHHHHH!!!!!!!!!!!![:0]

You guys are killing me!!!! [:D][;)]

Now I really got a head ache![:p]

Now for some seriousness,
I Think I'm getting the hang of this[^]

For the lights;
I could use ONE or Two if I wanted to balance it out?
1 diode per circuit cuts the voltage in half,
(This I found to be true. Using a N50 6A l diode, it dropped down 16v to 8v on an analog Simpson meter)
BUT Not really b/c......[?][?][?][%-)]

14v & 10v for Acc's;
I could use 5 bridge rectifiers, wired in series, to give me 14.5v, And
Add 6 more to give me 10.3?

Bear with me! I may be a little slow but I'll get it some day.

Thanks!

[lionelsoni]

QUOTE: Originally posted by lionelsoni
The only concern that I can see is balancing the load to avoid transformer saturation. This might not be important with your big transformer; but it is easy to do anyway, by dividing the lighting load between two such circuits, with the diodes reversed.

[Anonymous]

QUOTE: Originally posted by lionelsoni
Daniel, I suggested the single diode for 12 volts when I read, "I'll use a rheostat for all low voltage layout lighting. Should be a fairly fixed load. (12v sourse)". Incandescent lamps don't much care what the waveform is. (Although the dc component does reduce the lamps' lifetime somewhat.)

I agree that the single diode is a good solution for incandescent lamps. I would suggest splitting the load and using 2 diodes. Hook the cathode of one diode to the 18V and the anode to one set of bulbs. Hook the anode ot the other diode to the 18V and the cathode to the rest of the bulbs. Hook the other end of all the bulbs to common. This mimimizes the DC current flow through the transformer.

Daniel Lang

[lionelsoni]

Nice demonstration, Wayne!

Daniel, I suggested the single diode for 12 volts when I read, "I'll use a rheostat for all low voltage layout lighting. Should be a fairly fixed load. (12v sourse)". Incandescent lamps don't much care what the waveform is. (Although the dc component does reduce the lamps' lifetime somewhat.)

[webenda]

Here is the one diode solution to the 12V circuit implemented.

One diode, four lights, connected to 18VAC tap on transformer.


Three lights turned off. Good voltage regulation indicated. Notice that the voltmeter does not indicate 12V as predicted. That is because the voltmeter is affected by wave form. It is calibrated for a crest factor of 1.414. A half sine wave has a crest factor of 2.

Since the lights are not affected by waveform, I used one of the lights to measure the voltage on the others. Notice in Figure 3 that the upper lamp has been wired to the variable voltage output of the transformer. I adjusted the transformer until the upper light had the same brightness as the other three. The eye is a good judge of brightness differences. Since the waveform on the upper light is a sine wave, the voltmeter accurately reads 12.21V.

[Anonymous]

QUOTE: Originally posted by eZAK

Well I think I'm starting to understand this now!

Daniel Lang,
You suggest using a 10a or even a 15a or 25a rated bridge.
Why may I ask?
With a load of .75a and possible slightly higher, I figured on a 4a rating.

The first bridge rectifiers will be handling the combined load for all the accessories (You only need one chain for everything). The higher current bridge rectifiers give a greater safety margin in case of an accidental short. You could use the 4 amp bridges & place a 4 amp fuse inbetween the transformer and the first bridge. You may also have to add some heat sinking on the 4 amp bridges (The larger bridges can handle a couple of amps without heat sinking but do need heat sinking for the full current rating). Here is a place that sells bridge rectifiers:
http://www.allelectronics.com/cgi-bin/category.cgi?category=110&type=store

Also, the single diode in series with the power supply produces half wave rectified DC. Some accessories will be fine with that but some may balk at it.

Daniel Lang

[webenda]

QUOTE: Originally posted by eZAK

I want to use the most cost effective method that gives me the best result!

eZAK,

Thank Lionelsoni for reminding us of the diode.

I gave you information on using the rheostat because you asked how to figure out which one to use. Diodes are the most cost effective method giving the best results. We were concerned with the half wave waveform, but now that you mention the 12V is for lamps, waveform is not a concern. Lamps are not sensitive to waveform (at least not at 60 Hz.) As Lionelsoni said, you only need a single diode, not a bridge.

The diode solution has good voltage regulation. With a diode, you can even wire your lights to switches so that you can turn them on and off. The voltage will remain fairly constant.

With a rheostat, turning one light off (or having one burn out) will raise the voltage to all that are left. That is, if they are in parallel. If in series, they would all go off.

[lionelsoni]

Pat, excuse me if I muddy the water again.

I remember sometime in the recent past suggesting a single rectifier diode to get a 30-percent voltage reduction. I notice that you are now working with 18 volts instead of 24. This would drop to just about 12 volts with a diode. The half-wave waveform is screwy, but fine for lights. The only concern that I can see is balancing the load to avoid transformer saturation. This might not be important with your big transformer; but it is easy to do anyway, by dividing the lighting load between two such circuits, with the diodes reversed.

On the size of the bridge rectifiers: They needn't be so large as long as they are protected from fault current. A fuse is probably faster and therefore better than a circuit breaker for this.

[eZAK]

Well I think I'm starting to understand this now!

But the more I understand the more questions I have![%-)]
(Talk about negative resistance [sigh] )

Ok,
I'll use a rheostat for all low voltage layout lighting.
Should be a fairly fixed load. (12v sourse)

Now for the 14volts. I will use the diodes.

Daniel Lang,
You suggest using a 10a or even a 15a or 25a rated bridge.
Why may I ask?
With a load of .75a and possible slightly higher, I figured on a 4a rating.

Wayne [C):-)],
You said "It seems as if you wanted to use rheostats. They have a couple of advantages, the output waveform looks like the input (a sine wave) and you already have two."

I want to use the most cost effective method that gives me the best result!
I found rheostats for under $4. Bridge rectifiers are slightly more.
The only one I have could be the variac I suppose.
Thanks for the math!

[webenda]

eZAK,

You gave the parameters needed for some math in this post. So here it is:

There is more than one way to do the math. This is one...

RHEOSTATS
11 ohm/ 25w/ rated at 1.25a,
OR a 100 ohm/ 25w/ rated at .5a.

VOLTAGE TO BE DROPPED
18vac

Need: 14V @ .75A, 12V @ .5A and 10V @ .5A

14vac @ .75a max
RHEOSTAT--LOAD
--/\/\/\/--/\/\/\/--.75A
-- 18v -- 14v -- 0V
LOAD RESISTANCE R = V / I = 14 / 0.75 = 18.67 OHM
TOTAL RESISTANCE R = V / I = 18 / 0.75 = 24 OHM
RHEOSTAT RESISTANCE = TOTAL - LOAD = 24 - 18.67 = 5.33 OHM
RHEOSTAT VOLTAGE = TOTAL - LOAD = 18 - 14 = 4 VOLTS
RHEOSTAT WATTS = VI = 4 x 0.75 = 3 WATTS
THE 11 OHM RHEOSTAT WILL WORK.

12 vac @ .5a max
RHEOSTAT--LOAD
--/\/\/\/--/\/\/\/--.5A
18v -- 12v -- 0V
LOAD RESISTANCE = V / I = 12 / 0.5 = 24 OHM
TOTAL RESISTANCE R = V / I = 18 / 0.5 = 36 OHM
RHEOSTAT RESISTANCE = TOTAL - LOAD = 36 - 24 = 12 OHM
RHEOSTAT VOLTAGE = TOTAL - LOAD = 18 - 12 = 6 VOLTS
RHEOSTAT WATTS = VI = 6 x 0.5 = 3 WATTS
THE 11 OR 100 OHM RHEOSTAT WILL WORK.

10vac @ .5a max
RHEOSTAT--LOAD
--/\/\/\/--/\/\/\/--.5A
18v -- 10v -- 0V
LOAD RESISTANCE = V / I= 10 / 0.5 = 20 OHM
TOTAL RESISTANCE R = V / I = 18 / 0.5 = 36 OHM
RHEOSTAT RESISTANCE = TOTAL - LOAD = 36 - 20 = 16 OHM
RHEOSTAT VOLTAGE = TOTAL - LOAD = 18 - 10 = 8 VOLTS
RHEOSTAT WATTS = VI = 8 x 0.5 = 4 WATTS
THE 11 OR 100 OHM RHEOSTAT WILL WORK.

These values are only for the parameters given. If you change the load, the rheostat must change to maintain the voltage. As Lionelsoni said, "If you can be sure the load is constant..."

It seems as if you wanted to use rheostats. They have a couple of advantages, the output waveform looks like the input (a sine wave) and you already have two.

Show your math and someone will tell you how you got a negative resistance.

[Anonymous]

eZAK,

What you can do is this:

Get 6 bridge rectifiers of at least 10 amps each (15 to 25 amps preferred). Connect the + and - terminals of each bridge together (but not to the + and - of the other bridges!). Connect AC1 of the first bridge to the 18VAC. Connect AC2 of the first bridge to AC1 of the second bridge, etc to form a chain of 12 taps. As you move the wire down the taps, you will lose about .7 to .8 volts per tap (starting with the +/- of the first bridge, then AC2 of the first bridge, +/- of the second, etc.). You can wire the 12 taps to terminal blocks or even rotary switches to make adjustment easier.

Daniel Lang

[lionelsoni]

Use either the rheostat or the diodes, not both (for the same load).

The diodes will give you very good load regulation. That is, your 10 volts (or whatever) will be 10 volts for pretty much any load, up to what the diodes can carry. The waveform will be somewhat distorted, but probably not enough to affect the accessory.

If you can be sure the load is constant, like a light or motor that runs all the time, you can use a rheostat, adjusted to give the correct voltage at the (constant) load current. You can also use a simple fixed resistor if you know exactly what the current is in advance and don't need the adjustment.

You can also use a combination of fixed and adjustable resistors. A previous example was 12 volts at a maximum of 1 ampere, which required 6 ohms. You could use a fixed 5-ohm resistor in series with the 11-ohm rheostat. It would be set to 1 ohm, for a total of 6, if the load were 1 ampere. However if the load turned out to be .5 ampere, the rheostat could be set to 7 ohms, for a total of 12. The maximum setting, 11 ohms, plus the 5-ohm fixed resistor would allow it to be used down to .375 amperes.

[eZAK]

Because of the way the notifications work on this forum and to better clarify & summerize my ongoing problem.

First let me thank everyone that has replied so far and to those of you who are following along.
Some great post Guys!

THANKS!

Here is the situation as it stands now.
I've installed a variac on the line side of my ind. trans.

This is adjusted to give me 18vac on the secondary.

I would still need the following voltages for ACC's
14vac @ .75a max

12 vac @ .5a max

10vac @ .5a max

So far the best way I heard to do this would be to use a rheostat And diodes in series (bridge rectifiers)
Is this correct? Would the bridge go before or after the rheostat?

In selecting a rheostat to drop voltage;
The more amps > The less ohm's needed??
The less amps > The more ohm's needed??

In selecting a bridge rect.;
Choose one that is rated over the max amps??

Please respond with any suggestions or comments.

Thank you