I live in a country with 120VAC rms mains. I intend to build a high voltage/current dc power supply using an inline fuse, 1:1 isolation transformer, bridge rectifier, and filtering cap(s).

I'm new to ac and have a great deal of respect for it so I would like to iron out a few questions I had before continuing:

1. If I understand theory correctly, under no-load I should expect to see 120*sqrt(2) = ~170VDC. Under load, I understand this value will drop by a certain amount. Is there an equation that would help me identify what drop I would expect to see with say.. a 0.5A load vs a 1A load?
Additionally, is there a point where this drop "stops" per-se? I've read something about calculating under-load values with the equation VDC = (2*[VAC peak])/pi and am trying to figure out where this comes into play.

2. Is there a common way to to apply a load or drop the voltage so for a moderate range of loads, the voltage will remain somewhat steady?

3. What can be done to save the circuit in the event of a power surge?


I have a headache from reading about this all day, but at least I feel somewhat accomplished now.

 

The loading effect is dependent on the impedance of the transformer. Performing voltage measurements at 1/2 amp and 1 amp load is a method to determine the amount of the loading effect and the impedance of the transformer. (1/2 amp is not a very large load.) There is no limit to the lowering of voltage because the load increases, only a change in the rate of voltage sag. Your equasion seems irrelavent.

DC voltage regulators are commonly used to assure a steady voltage, but it works by wasting a few volts to achieve stability.

Power surges on the input are absorbed by a device called a MOV. Fuses are used to protect the components in case of excess power on either side of the transformer.

Gotcha started?

 

Yes, this is very helpful. Thank you. I was unfamiliar with MOVs.

That equation came from wikipedia's rectifier page:
http://en.wikipedia.org/wiki/Rectifier

In that context, what does that formula represent? It's listed as the "average" of the AC power, but even that seems strange since that should be 0!

 

That formula (2Vpeak/pi) represents the Volts (average), not power, of a rectified AC wave without capacitive filtering. It has nothing to do with voltage sag due to loading effects. ps, this form of energy is very rarely used in electronics.

You seem to be confusing voltage with power. The average voltage of a sine wave is zero, but when current flows, power happens, and it is not zero. Power is voltage times current. Once some current flows through a load, that power has been inflicted on the load. When the AC voltage wave reverses polarity and current flows in the other direction, more power is inflicted on the load. Thus, the average power is not zero.

Any more questions?

 

I misspoke, I intended to say voltage instead of power in respect to that equation. With that said, you did help me understand the use of that formula

I think that's it for the questions for now until I do a little more reading.

Thanks again

 

I live in a country with 120VAC rms mains. Does this mean that what you are pulging your power suplly into a 240VAC outlet?

 

There has been no mention of 240 volts in this thread.

 

Another voltage drop to be considered is the ripple voltage. The reservoir capacitor in your power supply will become partially discharged between the peaks of the AC waveform when it recharges. Getting an accurate figure for this is a bit complex, these days probably best done using simulation software.

A (pessimistic) approximation to the voltage drop ΔV which may be good enough is to assume that a constant load current I is drawn out of the capacitor of C Farads for half a cycle of the supply running at F Hertz, assuming a full-wave rectifier.

Then IT=CΔV, so ΔV=IT/C = I/2FC

 

Both adjuster and #12 have good points.

A little more detail.

When considering the transformer, look for the % regulation.

This figure is give by good manufacturers. It is

% regulation = 100(E - V)/V

Where E is transformer secondary voltage off load (open circuit)
V is the transformer voltage at rated load.

A straight line to get the figure for some intermediate load is usually adequate.

So a small 120 volt power transformer rated at 15% regulation at 2 amps would drop to 104 volts at 2 amps and about 112 volts at 1 amp.

All voltages are RMS voltages.

The output from your transformer feeds your rectifier(s) and reservoir capacitor(s) which will charge up to the peak (√2) value of the waveform half cycle on no load.
You will suffer a small voltage drop of a few volts in the rectifiers.
You will also suffer ripple voltage depending upon the type of rectification you employ.
This is because whenever the ac input from the transformer is less than the capacitor voltage (which is most of the time in its cycle) , the capacitor will actually be supplying the current to the load so its own voltage will fall. The capacitor voltage is replenished the next time the ac input is greater.

You can minimise ripple by choosing large enough reservoir capacitors, and using a bridge rectifier rather than a single one.

Rectifiers are specified by their inverse voltage rating (PIV). This is the maximum reverse or blocking voltage you can apply and corresponds to the peak of the applied ac waveform not the RMS value.

go well

 

Note that the required PIV will be at least double the maximum peak voltage you ever expect to come out of the transformer. This arises because the reservoir capacitor voltage can remain at almost one peak input voltage, even while the input to a diode reverses to the opposite peak value. A generous factor of safety is advisable on top of foreseeable variations.

Note that some good manufacturers of transformers make their products to deliver their full rated voltages at full load, so the off-load voltage may be surprisingly high. The mains supply is also subject to variation, with short-term surges possible on top.

Don't get neurotic about it, but since you can get easily get devices rated at several hundred volts upwards, why take chances?

 

Note that some good manufacturers of transformers make their products to deliver their full rated voltages at full load, so the off-load voltage may be surprisingly high. The mains supply is also subject to variation, with short-term surges possible on top.

This is normally the case for mains power supplies (often mistakenly called 'transformers') otherwise known as wall warts, mains eliminators, mains adapters etc.

A bare transformer rated at X volts should measure X volts off load. Otherwise your 1:1, 120 volt transformer would be a step up transformer!

 

This is normally the case for mains power supplies (often mistakenly called 'transformers') otherwise known as wall warts, mains eliminators, mains adapters etc.

A bare transformer rated at X volts should measure X volts off load. Otherwise your 1:1, 120 volt transformer would be a step up transformer!

True sometimes, but not always. Sometimes the on-load voltage is critical, and so is strictly adhered to. Such transformers are quoted in terms of on-load voltage rather than ratio. This is particularly true for transformers used for lighting applications, or for heating the filaments of electron tubes, magnetrons etc. (In such cases, it may be unwise for instance to operate a small lamp or tube from a grossly-over-rated transformer.)

Here is an example specification: note the round figure rated on-load voltages, and the "odd" off-load values. http://www.farnell.com/datasheets/74757.pdf

Since we do not know from where the OP will be getting his transformer, or what service it might have been intended for, I thought to cover all the bases.

 

If you are curious why I am making this vs. buying something off the shelf, let me give some background:

I'm building a neon news ticker. These neon bulbs have a firing voltage of 140v, but a keep-up voltage of 99v. To control this, I'm using a simple 5v pic.

The 5v stuff will probably require about 500mA of current tops and I've got over 200 of these tubes that are rated for 0.5mA (I plan to drive them at a little less than that).

At 170VDC, That's 17 watts... but that's only if all the bulbs are on at the same time. Typically only 10 or so will be on.

I have had a few ideas for power supplies (and concerns):
1. Power everything with a AC-DC 5v wall wart and use a step-up controller like the MAX1771 (I use a off-the-shelf circuit for testing based on this)
- This thing makes an irritable whine
- It is not rated to supply as much power as I need

2. Use a 5v dc wall wart and villard cascade voltage multiplier
- I'm thinking the bulbs will noticeably dim as more and more are turned on.


3. Use a transformer/bridge rectifier+ smoothing caps for neon tubes, and another transformer to get the 5vdc
- I had a little concern that the bulbs will dim
- I want to make sure my bulbs/circuit doesn't blow up in the event of a power surge and that means having to engineer these avoidance measures into place. (On the upside, I'm learning something new!)

So right now, I'm on #3 as you might have guessed

And thank you all for your continued thoughts. They are helpful.

 

Past my bedtime really, but a few observations about neons:

1. Make a bomb-proof interface to link from your logic, so that there is no easy way of having a high-voltage into logic accident.
2. Most neons require resistive ballast, or some other current limitation. Some "bulbs" have this internally, many do not. If you try to do without it, you are liable for a high-voltage version of the sort of accident that many newbies have with LEDs.

Of course, you probably know all this, but you can't be too careful!

 

Yup, I'm aware that I need current limiting resistors and I'm using optocouplers to isolate this from everything else.

I will say I blew up one of these bulbs, but that was more of a "I wonder what would happen if I didn't current limit these." It was glorious.

 

@Adjuster

I live in a country with 120VAC rms mains. I intend to build a high voltage/current dc power supply using an inline fuse, 1:1 isolation transformer, bridge rectifier, and filtering cap(s).

It would be nice to know how one could get more than 120 volts out with one of these.

 

@Adjuster

It would be nice to know how one could get more than 120 volts out with one of these.

Ok, I admit that I had not checked back to the OPs original description. If he definitely uses a transformer is described as "1:1 ratio isolating", you are right. If it is described as "120V at 1A (say)", that could be another matter.

 

This question is a good one and has been discussed many times before here, for example

http://forum.allaboutcircuits.com/showthread.php?t=17850

http://forum.allaboutcircuits.com/showthread.php?t=23251

The current discussion is also helpful and productive with many good points being made.

go well

 

Just chiming in again to point out a nice read on linear power supplies:
http://mysite.du.edu/~etuttle/electron/elect9.htm

What's the confusion with the 1:1 transformer now?