Looking for Reduced Voltage From..........

[eZAK]

.......Your Transformer?

If your like me, who uses only one transformer for the entire layout,
Or you may have a transformer that is always set to full power because TMCC or DCS, And now you need various voltages to control your accessories.

Click here http://home.cogeco.ca/~rpaisley4/xACdropper.html

[spankybird]

HI Pat,

I use one channel of my ZW to control switches and a KW, which already has taps at 6 volt and 14 volts.

[lionelsoni]

When you need an odd number of diode drops, it doesn't hurt to connect the ~ (ac) terminals of the last bridge together, to share the current among the four internal diodes.

I have been using bridges wired this way to control the speed of a locomotive that I want to run attended. It works particularly well with a rheostat, which keeps the field current pretty constant.

You can get a 30-percent voltage reduction just by putting a diode in series with the transformer. This works for loads, like lamps, which don't care what waveform they are getting. If used with large loads, they should be balanced, some with the diode one way, some the other way, to keep the dc current from saturating the transformer.

[cmrj]

This is a great site, have been looking for something like this.

[lionelsoni]

I have been able to figure out that the reduction in rms voltage that the diode trick gives you is about 90 percent (sqrt(8)/pi) of the total forward drop of the diodes, when the total reduction is reasonably small compared to the original voltage. As you add more diodes, the reduction does increase, but each diode that you add has less of an effect that the previous one.

The waveform that results from this trick is of course not strictly sinusoidal; but for modest reductions it is close.

[eZAK]

Bob,

I'm not as technical as you so let me see if I'm understanding you correctly.

I can use a rheostat with the bridges in order to fine tune the voltage?

I should balance the load by putting a set of bridges on the hot as well as the neutral?

For the full story see my post entitled Electrical / Electronic project.

Thanks

[lionelsoni]

Yes on the rheostat. No need to "balance" anything.

[Anonymous]

lionelsoni: when you use a diode on ac, doesn;t it convert the ac to pulsating dc at half voltage?
So if you have 18 volts on the track and you want your passenger car lights dimmer just put in a diode in series to the lights and they will see 9 volts.
My question, is this what happens to the current draw when the diode cuts the voltage?
Will it cut in half also. Dave

[lionelsoni]

Dave, it is pulsating; but the voltage drops to 71 percent, not half. The way voltages of all kinds are conventionally measured ("rms" voltage) is by the power that the voltage causes to be delivered to a resistor. The rms voltage of a varying voltage waveform, like a sine wave or half-cycles of a sine wave, is the dc voltage that would deliver the same power into the load. With the diode, the power does drop by half, since the supply is completely shut off half the time. However, cutting the power in half corresponds to reducing the same dc voltage by a factor of the square-root of one-half, or .71.

If the load is a fixed resistor, the current simply drops in proportion to the voltage, to 71 percent. However, an incandescent lamp will cool off somewhat; and its resistance will drop. In this case, the current will drop to 83 percent, not 71 percent, because of the reduced resistance. The brightness will however drop dramatically, since the cooler filament produces much less light.

[underworld]

Rheostats are a good idea. Later if you need to you can always adjust the voltage.

underworld

[:D][:D][:D][:D][:D]

[lionelsoni]

Dave posted this on another thread; but I think he meant it for this discussion:

"Bob: I know the sine 45 degrees is .707 for rms power, but doesn;t the diode besides haveing 1/2 wave for the ac also require 7/10 of a volt to turn on. So wouldn;t the ac to DC be voltage be half voltage plus an additional .7 volts. Then each time another diode is added it is getting hit by dc so just the extra .7 volts would be lost.
I know you can expect a 1.4 volt drop across a bridge rectifier as you have .7 in 3 forward biased diodes and .7 on the one reversed biased.
Anyway this is getting technical and I respect you knowledge on it all."

The half-wave output rms voltage is 71 percent of the input rms voltage for a perfect diode (no forward voltage). This is because the diode halves the power into a fixed resistive load, which would have the same effect as lowering the full waveform to 71 percent. A real diode will lower (not raise) the output voltage somewhat more, but not enough to make much difference in the 24-volt ballpark. (The exact amount is not easy to calculate.)

The dc component of a full-wave-rectified sine wave is 90 percent of the input rms voltage (the square-root of 8, divided by pi). This is just the average over time of the voltage waveform. Therefore, the dc component of a half-wave output is just half this, or 45 percent of the input rms voltage. However, if you connected this half-wave output to a lamp rated at this voltage value, it would burn out quickly, because the rms voltage is what counts as far as power dissipated is concerned and it would be 57 percent too high for that lamp.

[eZAK]

For an update and further clarification on this subject see the thread entitled

Electronic problem / In Summary!