How do I convert 16v AC to 12v DC?

[bertiedog]

Well the clipper must have a transformer pushing out at least 24 volts AC peak, or thereabouts to get a 20 volt output across the cap, (and there's no resistor mentioned).

I only mentioned the circuit as it is there for these uses, and saves buying an extra unit in. If I built it for myself I would leave the cap and add a regulator chip, but it is more than needed here.

The use of the battery is OK as long as a fuse is added, a simple thermal trip will work, stopping the battery discharging in one go, or add a regulator, with a suitable fall back fuse as well.

For exhibitions I used to run the layouts on a sealed type car battery, no trailing mains leads needed, safer to everybody, and immune from power outage in halls will somewhat dubious mains wiring.

Stephen

[Pacific231G]

I seem to be pushed into a corner here, I am a trained electrical engineer with long experience and I can measure AC on a scope quite accurately. If I published the exact formula here it would be a massive turn off! RMS values are approximations to save calculations.

 

 

__________________________________________________

 

On electromagnets AC works fine, the poles alternate at the frequency of the AC wave, and have a phase lag due to the build up on each pulse and the collapse of the field, till the reverse wave takes over. An iron core helps keep the magnet steady, as it gets saturated by the field, and maintains a pull despite the alternating wave form. AC is a bit more efficient at maintaining the field against pure DC which being constant is producing a bit more heat, and produces a saturated iron core with steady poles. When shut off the DC one will produce a back EMF kick, and if the core is iron or steel it retains some magnetism.

 

With AC, chance comes into shutdown, the wave form may be anything between 0 and full voltage, but the saturated field is still there, so on shutdown you get the back emf but it will vary in effect, and usually leaves the iron un-magnetised or with a weak field.

 

Generally if you are attracting steel to an electromagnet, then AC or DC will both work, and if you are attracting a magnet then DC is better, as AC fields may demagnetise the magnet. With DC you have the choice to attract or repel the magnet by reversing the coil.

 

What is a game changer are Neodymium magnets, which resist quite strong AC fields trying to demagnetise them. Simpler Alnico will suffer in the same field. An AC solenoid will over come the Neo magnet and attract it, despite the poles. Also a DC solonoid will overcome the repulsion of a Neo if they are close enough.  But it is safer to use DC to work with magnets, no risk to the magnets and repulsion or attraction works.

 

The idea of using magnets as couplers in pairs and then using an AC field to release them sounds interesting, but may have a flaw in that the alternating field would at any one wave only affect one magnet, and the other maintains the grip.

 

However if one wagon has a magnet on the end and the next has steel instead, then they would part in an AC field. They would not part in a DC field as all that happens is pole reversal, and steel will still be attracted.

 

So tiny neo magnets on one end of a wagon, and steel on the other in an alternating rake, sounds like a good experiment to try out. No way to get advanced uncoupling, but you could use a hand held electromagnet to wave over the coupling to release it for remote situations.

 

Stephen

 

Thanks for this Stephen. No we probably don't want to get into integral calculus with imaginary numbers. Electrical engineering was only one part of my OND and being rude mechanicals we called it magic. I did a bit more for my degree but not a lot. 

 

The coupler I'm looking at was invented by a French modeller in the 1970s and apparently used successfully by a number of others.

It was single ended and relied on a permanent magnet to hold a link in engagement with a hook and a track mounted DC electromagnet to  create a repulsive field for uncoupling. It looked like something that would never work but according to the people who used it (not the inventor!) apparently did and with virtually no uncommanded uncouplings. Its main feature was that it looked like a real screw link coupler.

 

The magnetic field only held things in places so didn't carry any drawbar forces. This is different from neodymium based couplers currently used by some O gauge modellers (and the ordinary magnets used by Brio for the same purpose in their wooden track toy trains)

Edited April 16, 2017 by Pacific231G

[Ruston]

A wall socket 12 V supply would be cheaper and self contained, as I said earler.

 

David Green, Diploma in Engineering (Electrical and Electronic)

Thanks but the whole point is that I don't want an extra cable from an extra socket. Fitting this gadget means one socket and one cable into the baseboard where everything, points, uncouplers and controller can be powered.

[Junctionmad]

For plain AC to Dc full wave bridge rectification

 

 

AC ( rms) = (DC + 2.7 + Vripple ) /1.414

 

I.e. For a DC voltage of 12 V , this results in a Vacrms of 11.5 rms or 16 V peak for a ripple voltage of 2v pk

 

In order to generate a ripple voltage of that the formula

 

c= I load /( F x Vripple ) where F is 100 Hz normally

 

c = 1/100 * 2 , is 5000uf for a load of 1A

 

 

Note there's more to this , includeding ripple current and load regulation calculations.

Edited April 16, 2017 by Junctionmad

[Nick G]

For plain AC to Dc full wave bridge rectification

 

 

AC ( rms) = (DC + 2.7 + Vripple ) /1.414

 

I.e. For a DC voltage of 12 V , this results in a Vacrms of 11.5 rms or 16 V peak for a ripple voltage of 2v pk

 

In order to generate a ripple voltage of that the formula

c = I load / 100 * 2 , is 5000uf for a load of 1A

 

 

 

Pardon?

[Junctionmad]

Pardon?

That's quite all right

[Crosland]

Transformers are rated as RMS, so the 16 V transformer is 16V RMS, however you measure it.

 

If you try to measure it with a scope you will see the positive and negative peaks greater than 22V and would need to calculate the RMS for yourself, unless the 'scope includes that function.

 

When rectified it will produce positive only peaks (i.e. DC) of around 22V, depending on the load and diode voltage drops. With no load the voltage will be highre due to the poor regulation from a transformer.

 

You need to add a suitably rated smoothing capacitor to "iron out" the peaks, see post #29.

 

For 12V DC you add a suitably rated (for the current you need) voltage regulator, following the data sheet for any other capacitors that may be needed.

[Junctionmad]

Transformers are rated as RMS, so the 16 V transformer is 16V RMS, however you measure it.

 

If you try to measure it with a scope you will see the positive and negative peaks greater than 22V and would need to calculate the RMS for yourself, unless the 'scope includes that function.

 

When rectified it will produce positive only peaks (i.e. DC) of around 22V, depending on the load and diode voltage drops. With no load the voltage will be highre due to the poor regulation from a transformer.

 

You need to add a suitably rated smoothing capacitor to "iron out" the peaks, see post #29.

 

For 12V DC you add a suitably rated (for the current you need) voltage regulator, following the data sheet for any other capacitors that may be needed.

Just a point of definition here, DC is not the positive peak to peak of the rectified waveform. DC is the average of that waveform , the smoothing capacitor increases that average, as it releases " power " during the decreasing edge of the cycle, hence the DC Voltage rises when you add a suitable cap

Edited April 16, 2017 by Junctionmad

[bertiedog]

When I worked for a company that made transformers, they were rated as RMS and peak output, but the issue here is not that, but the conversion of 16 to 12, and I have not seen a railway unit marked 16 VAC that produced 22 volts peak. As I said the HM was about 15 volts AC and produces 12 volts.

 

As an experiment I tried a bridge rectifier on my bench Variac, and measured the DC till it read 12 volts and then checked the Variac, it was 15.5 volts. This was checked on an AC meter(frequency compensated) and on the scope.

 

We were taught not to use RMS but use calculus to calculate the area of the sinewave so that imperfect waves could be measured as RMS only works with a perfect sine wave. RMS is the value of AC current that produces the same heating as a DC current, and is only true on a perfect sine curve.

 

Stephen.

Edited April 16, 2017 by bertiedog

[Suzie]

...I have not seen a railway unit marked 16 VAC that produced 22 volts peak. As I said the HM was about 15 volts AC and produces 12 volts...

 

That is because unlike in your diagram they do not have a capacitor across the output terminals. Put a 35V electrolytic across the output of your H&M 12V DC output and see what your AVO shows - it will be something in excess of 25V on 240V mains with no load.

[28XX]

I would like to point out that a dangerous generalisation has been asserted here which does not apply in all circumstances.

 

Many electromagnets will work on ac and dc, and for momentary duty in model railways, this is safe enough, (the buzz from ac is more anoying to me than a solid clunk from dc however).

 

For longer duties periods, e.g. a relay or contactor, you don't have as much freedom. The ratings given in the specifications must be followed. In a former job as a maintenance engineer I fitted a ac relay instead of an dc one in desperation to get a machine working when we had no stock of the right part. It self destructed quite spectacularly before the end of the shift.

 

I think in a dc circuit, once the armature has moved and the core is saturated, the only impedance to current flow is the resistance of windings. On ac, there is a continuous inductive impedance as long as the circuit is energised.

Edited April 17, 2017 by 28XX

[kevinlms]

I would like to point out that a dangerous generalisation has been asserted here which does not apply in all circumstances.

 

Many electromagnets will work on ac and dc, and for momentary duty in model railways, this is safe enough, (the buzz from ac is more anoying to me than a solid clunk from dc however).

 

For longer duties periods, e.g. a relay or contactor, you don't have as much freedom. The ratings given in the specifications must be followed. In a former job as a maintenance engineer I fitted a dc relay instead of an ac one in desperation to get a machine working when we had no stock of the right part. It self destructed quite spectacularly before the end of the shift.

 

I think in a dc circuit, once the armature has moved and the core is saturated, the only impedance to current flow is the resistance of windings. On ac, there is a continuous inductive impedance as long as the circuit is energised.

The above doesn't cover electronic items that simply MUST have a regulated power supply. An unsmoothed power source will play havoc with such items.

There are a few threads on RMweb, where this has caused 'mysterious' problems.

[Grovenor]

That is because unlike in your diagram they do not have a capacitor across the output terminals. Put a 35V electrolytic across the output of your H&M 12V DC output and see what your AVO shows - it will be something in excess of 25V on 240V mains with no load.

My test bench is currently powered by a Roco transformer rated at 15V according to the label, that is probably assuming a 220V or 230V input. Here on UK mains the output measured on a standard meter set to AC is 17.9V, after rectification with a 2200uF capacitor attached the DC voltage is 22.6V.

I've never come across a transformer sold for model railway use where the AC output voltage was given as the peak.

Regards

[Suzie]

My test bench is currently powered by a Roco transformer rated at 15V according to the label, that is probably assuming a 220V or 230V input. Here on UK mains the output measured on a standard meter set to AC is 17.9V, after rectification with a 2200uF capacitor attached the DC voltage is 22.6V.

I've never come across a transformer sold for model railway use where the AC output voltage was given as the peak.

Regards

 

Neither have I.

[Junctionmad]

My test bench is currently powered by a Roco transformer rated at 15V according to the label, that is probably assuming a 220V or 230V input. Here on UK mains the output measured on a standard meter set to AC is 17.9V, after rectification with a 2200uF capacitor attached the DC voltage is 22.6V.

I've never come across a transformer sold for model railway use where the AC output voltage was given as the peak.

Regards

 

just taking your example , an assuming you are measuring the output with effectively very little load ( i.e. the meter ) , we can assume the ripple is zero for that low load ( i.e. I load tends to zero )  ( the Vripple at 1mA with that  capacitor is approx 4mV) 

 

hence using my formula AC =( DC + 2.7 + Vripple ( pk)) /1.414 , gives (  22.6 + 2.7 + 0 ) /1.414 = 25.3/1.414 = 17.89   ( well funny that )  

 

I hope Nick G is now OK 

 

PS To answer the confusion generated at the top of this thread, the DC output of a smoothed bridge rectified   AC is never higher then the peak AC  but it will be higher then the VAC RMS ( unless the load is high and the ripple voltage is significant ) , This is because the average value of of the output , which is what is taken as DC is rated considerably by the capacitor, and of course the resulting input current profile is changed completely from a sine wave to a pulse of AC current as the diodes conduct once the capacitor discharges relative to the diode input voltage  ( thats another story of course ) 

Edited April 17, 2017 by Junctionmad

[dhjgreen]

Thanks but the whole point is that I don't want an extra cable from an extra socket. Fitting this gadget means one socket and one cable into the baseboard where everything, points, uncouplers and controller can be powered.

That did cross my mind.

 

Whichever solution you choose, do not forget the back EMF diode I described earlier, or you will damage electronics, or the switches if you go for rectified AC.

Edited April 17, 2017 by dhjgreen

[Crosland]

 RMS is the value of AC current that produces the same heating as a DC current, and is only true on a perfect sine curve.

 

That statement just confirms that you do not understand the principle behind AC measurement.

 

RMS is appropriate for any waveshape.

 

RMS is calculated by squaring the original waveform, calculating the average and then taking the square root, hence RMS or Root-Mean-Square

 

It's the 1.414 approximation (square root of 2) factor that only applies to perfect sine waves.

 

For DCC, for example the peak and RMS are the same.

[AndyID]

My test bench is currently powered by a Roco transformer rated at 15V according to the label, that is probably assuming a 220V or 230V input. Here on UK mains the output measured on a standard meter set to AC is 17.9V, after rectification with a 2200uF capacitor attached the DC voltage is 22.6V.

I've never come across a transformer sold for model railway use where the AC output voltage was given as the peak.

Regards

 

Keith,

 

The output voltage marking on small transformers usually indicates the voltage at full load current. The voltage at no, or low, current will be substantially greater. The actual difference depends on the efficiency of the transformer under load. I just measured one and it was about 12% higher than the nominal voltage on the label when unloaded.

 

Andy

[Junctionmad]

Keith,

 

The output voltage marking on small transformers usually indicates the voltage at full load current. The voltage at no, or low, current will be substantially greater. The actual difference depends on the efficiency of the transformer under load. I just measured one and it was about 12% higher than the nominal voltage on the label when unloaded.

 

Andy

as you say the usual principle is to specify the output voltage at full load into a resistive load.

 

Hence it's its theory the lowest voltage you'll see

 

The load regulation percent will give an indication what the " no load " voltage will be. Cheap split bobbin could have load regulation approaching 20 %, meaning the no loss ( edit : load ) voltage could be 1/5 higher. Which is why you have to be careful over specifying transformers with too much load margin.

Edited April 18, 2017 by Junctionmad

[kevinlms]

as you say the usual principle is to specify the output voltage at full load into a resistive load.

 

Hence it's its theory the lowest voltage you'll see

 

The load regulation percent will give an indication what the " no load " voltage will be. Cheap split bobbin could have load regulation approaching 20 %, meaning the no loss voltage could be 1/5 higher. Which is why you have to be careful over specifying transformers with too much load margin.

At my club, we had some instances of some DCC chips becaming fried on the test track. On investigation, we found that the power supply for the DCC controller had been cobbled up using an old transformer, with a large VA capacity at about 18 volts. Because it was being under run with only one loco on the track, the actual voltage was well over 20 volts on the DCC waveform.

I can't remember the exact measurements, but it certainly was the cause of fried chips. Swapping to a more suitable transformer, solved the issue.

[Ruston]

I use these to power my 12v stuff from my 16cAC transformer.

 

16v AC to 12v DC Smoothed.

 

 

http://www.expressmodels.co.uk/acatalog/Regulated_Power_Supplies.html

I went and bought one of these and have just got around to installing and wiring it. The 12vDC that comes from it seems to be different to the 12v DC that comes from the battery that I was using when it comes to working the DG electromagnets.

 

With the battery, when the button to operate a magnet was pressed, the iron wire on the coupler was pulled strongly straight toward it and the coupling loops on the Dinghams lifted clean away from the hook and everything worked as it should. With the 12v from this gadget the iron wire wiggles around and vibrates - almost as if it is being repelled by the magnet, and the loop won't lift high enough to clear the hook making uncoupling impossible.

 

So I'm going to have to do what I wanted to avoid and either continue to use the battery (with wires dangling down to the floor) or have to have another mains socket plugged in with a seperate transformer.

[AndyID]

I went and bought one of these and have just got around to installing and wiring it. The 12vDC that comes from it seems to be different to the 12v DC that comes from the battery that I was using when it comes to working the DG electromagnets.

 

With the battery, when the button to operate a magnet was pressed, the iron wire on the coupler was pulled strongly straight toward it and the coupling loops on the Dinghams lifted clean away from the hook and everything worked as it should. With the 12v from this gadget the iron wire wiggles around and vibrates - almost as if it is being repelled by the magnet, and the loop won't lift high enough to clear the hook making uncoupling impossible.

 

So I'm going to have to do what I wanted to avoid and either continue to use the battery (with wires dangling down to the floor) or have to have another mains socket plugged in with a seperate transformer.

 

Apparently the Express Models regulator can supply a maximum of 1 amp. The DG electromagnet needs 1.5 amps.

[kevinlms]

Apparently the Express Models regulator can supply a maximum of 1 amp. The DG electromagnet needs 1.5 amps.

Agreed, an overloaded power supply will be erratic & may depending on design, shut down briefly.

[BG John]

I've found this one on eBay, that's supposed to be up to 3 amps.

http://www.ebay.co.uk/itm/AC-DC-to-DC-Converter-Step-Down-Power-Supply-Output-1-5V-27V-5V-9V-12V-24V-B178-/112242069026

 

There are several other types available from China, and this one is cheaper, but you have quite a long wait.

[Crosland]

I've found this one on eBay, that's supposed to be up to 3 amps.

 

 

It only quotes the maximum input current. No data for what the current rating of the output is. They are not necessarily equal.

 

To understand how switch mode converters work you need to consider the input power, output power and efficiency. For a given power budget, maximum input current will correspond to minimum input voltage, 3 V in this case.

 

Lets be generous and assume 90% efficiency at full load. That 3A input with a 3V input is 9 W so we have 9 x 90% or 8.1 W available at the output. If stepped up to 12 V that's only 675 mA.

 

It sometimes pays to take a cynical view in the absence of real data.