electricity 101

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jhrper

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hey guys
I have a simple question for you and i bet it has a simple answer. I was wondering how you determine the amp draw from low voltage (DC) lighting. If W=VA then A=W/V. Assuming a PF of 1 what would be the amp draw from 120 watts of low voltage lighting on a 12V DC transformer? (120V AC input, 12V DC out) Mainly wondering if you use 12V or 120V as the value for voltage in the equation. Got thinking about it at work and couldn't decide. Could have used the amp clamp but that would have been cheating.
 

jhrper

Member
alright sparky, stick with me here. I left school, went to the field and left my brain in a crawl space somewhere. So I have a 10:1 ratio. 120W/12v on the secondary is 10A so amp draw on the primary would be 10A/10 (from the ratio) so amperage would be 1?
 

480sparky

Senior Member
Location
Iowegia
Yeppur!

You get a gold star!

GoldStar.jpg
 

Jraef

Moderator, OTD
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Electrical Engineer
I know I'm being a jerk, but I can't help it sometimes...

A "transformer" only applies to AC systems. You cannot have a transformer that has 120VAC in, 12VDC out. That is technically called a "Power Supply". The reason I bring it up is that there are losses associated with running a power supply, so technically, the AC input VA will be slightly higher than the DC output watts. You cannot assume a 1.0 pf on the AC side, its likely more like a .70 pf.
 
When it comes to off the cuff calculations, watts is watts...

now the disclaimers... your milage will vary for:
-DC loads
-3 phase loads (not multiple single phase cicuits on a 3 phase feed but actual 3-phase loads)
-non-sinusoidal waveforms (i.e. dimmers in line)
-I'm sure I missed more, but hey, i'm just an entertainment guy :)
 
Jraef said:
I know I'm being a jerk, but I can't help it sometimes...

A "transformer" only applies to AC systems. You cannot have a transformer that has 120VAC in, 12VDC out. That is technically called a "Power Supply". The reason I bring it up is that there are losses associated with running a power supply, so technically, the AC input VA will be slightly higher than the DC output watts. You cannot assume a 1.0 pf on the AC side, its likely more like a .70 pf.

You probably meant eff. rather than pf.
 

K8MHZ

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Michigan. It's a beautiful peninsula, I've looked
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A "transformer" only applies to AC systems.

Let me ask you this then. How do you get 15kV to the spark plugs of your car from a coil that is fed with 12 VDC?

Transformers will step up or down any time there is a moving magnetic field. That field can be created either by AC or intermittent DC.

Lightning is DC but assumes the characteristics of approx. 100 MHz AC for the same reason.

In order to be AC the polarity must change which it does not in automotive ignition systems and lightning. I actually got an answer to a question in one of our apprenticeship tests changed by pointing this out.
 

gar

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080513-1030 EST

K8MHZ:

I would classify intermittent DC as AC, but I do not want to argue the issue.

An automotive ignition coil is a transformer in its construction, and it is a device that couples two coils together with a magnetic field.

Can we say the ignition coil is a DC transformer? Maybe in the broad sense of a general definition of the word transformer. But the ignition coil to some extent belongs in its own category.

From:
http://dictionary.reference.com/browse/transformer

1. a person or thing that transforms.
2. Electricity. an electric device consisting essentially of two or more windings wound on the same core, which by electromagnetic induction transforms electric energy from one set of one or more circuits to another set of one or more circuits such that the frequency of the energy remains unchanged while the voltage and current usually change.
The general discussion on transformers used in electrical circuits is based on steady-state conditions. Thus:

A steady-state DC input applied to a magnetic core transformer will only produce a momentary output pulse from the secondary and then zero output until the input is removed.

A steady-state AC input applied to a magnetic core transformer will produce a steady-state AC output continuously as long as the input is applied.

.
 

LarryFine

Master Electrician Electric Contractor Richmond VA
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gar said:
An automotive ignition coil is a transformer in its construction, and it is a device that couples two coils together with a magnetic field.
Actually, an ignition coil is an auto-transformer.

Come to think of it, it's an auto transformer. :grin:
 

mpoulton

Senior Member
Location
Phoenix, AZ, USA
K8MHZ said:
Let me ask you this then. How do you get 15kV to the spark plugs of your car from a coil that is fed with 12 VDC?
...
In order to be AC the polarity must change which it does not in automotive ignition systems and lightning. I actually got an answer to a question in one of our apprenticeship tests changed by pointing this out.

In all of these examples, the signal feeding the transformer actually has a large AC component. The AC and DC components of a signal can often be treated separately for circuit analysis purposes -- only the AC portion matters when a transformer (or any other reactive element) is involved, except for certain factors not relevant to this discussion (saturation, specifically). Simply subtract the steady-state DC conditions from the signal, and you are left with the AC component. Only the AC component is transformed by a transformer - the DC component does not participate in the process. The statement that the polarity must change for it to be AC is not entirely correct then. For the AC component of the signal in all these examples, the polarity does change - and that's all that matters. Also, in the case of lightning, the signal actually is "true" AC even without subtracting the DC component - lightning reverses polarity repeatedly during a strike.

It should also be noted that ignition coils are not really transformers in the sense that you are thinking of, and they do not function according to the normal rules of transformers. They are flyback transformers, and the output voltage and current are not directly related to the ratio of turns between the primary and secondary. The secondary voltage and current are determined by the rate of collapse of the magnetic field in the core, which in real life depends on how quickly the "points" (on an old vehicle) can break the primary circuit.
 

K8MHZ

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Also, in the case of lightning, the signal actually is "true" AC even without subtracting the DC component - lightning reverses polarity repeatedly during a strike.

I beg to differ with you. Lightning is the result of a transfer of electrons from one place to another with a huge difference in potential. There are several forms of lightning but for the sake of illustration I will use cloud to ground as an example. The clouds have many more electrons than the ground and the lightning brings them from the cloud (in a place called the charge center) to the ground and does not alternate in polarity. The only exception is in leaders which *may* reverse polarity for one cycle if they don't make it all the way to the ground. The strike following the lower impedance ionized trail created by the leaders is always DC. The strike pulsates for various reasons but once the potential of the charge center is depleted enough that the difference in potential is not sufficient to carry the strike it dissipates without transferring any electrons from the ground back to the charge center. Remember, lightning is a form of static electricity, it is not dynamic like true AC. I get my information from the National Weather Service. I take training every year and am a card carrying Skywarn spotter. (Not a storm chaser, we observe and report information to the NWS during storms from pre-assigned locations)

I would classify intermittent DC as AC, but I do not want to argue the issue.

Early in my career I felt the same way. I was corrected by a college professor that informed me that in order to be AC there must be a 'crossing of the zero line' and a change in polarity. If there is a rapid change in magnitude but no zero line crossing or polarity change the proper term for that would be 'pulsating DC'.

Personally, I think that only having two labels, AC or DC, for electricity is too restrictive and doe not convey the true properties of certain types of electricity. In the case of the primary of a car's ignition and in lightning, PC, or 'pulsating current' would be a much better representation, IMHO.
 

K8MHZ

Senior Member
Location
Michigan. It's a beautiful peninsula, I've looked
Occupation
Electrician
The statement that the polarity must change for it to be AC is not entirely correct then.

Indeed it is. In order to be AC there must be a change in direction of electron flow. Stopping and starting in the same direction is not AC, but does result in moving lines (flux) of magnetic force that will produce induction.

I think you are confusing the term 'AC component' with AC characteristics.

You will also find that capacitors will behave much differently when exposed to pulsating DC than they will with AC of the same frequency as the mechanics of a capacitor are dependent upon a change of the direction of electron flow, whereas an inductor produces reactance based upon moving magnetic fields which can occur without a directional change.
 

gar

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Location
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EE
080513-0724 EST

jhrper:

You do not normally use the word "transformer" for an AC to DC conversion as stated by Jraef.

I have read your question a couple times and I question if your output is DC. If I have only an incandescent lamp load at 12 V why convert from 120 V AC to 12 V DC to supply the lamps? I do not know why. My opinion is that DC excitation of tungsten filament incandescent lamps shortens their life compared to AC.


Next let's suppose you do have a DC power supply, then what is it's design?

Is it a transformer with an output bridge rectifier and nothing else? Then the average DC output voltage is 0.636 times the AC peak voltage, and the RMS voltage is still 0.707 times the AC peak voltage. However, there is about a 1.5 to 2 V drop across the bridge rectifier that must be subtracted from the input voltage. RMS current AC or DC is what determines lamp brightness.

If you have a bridge rectifier followed by a capacitor input filter, then the DC output voltage, average or RMS, is approximately V peak input to the bridge minus the bridge drop.

If you have a DC supply with some sort of regulator on the output, then there are additional power losses between AC input and DC output.

Volt-amperes (VA) and power are not necessarily the same value. VA is simply the product of V across a load times A thru the load. Power is a measure of the rate of energy use. The VA measurement of a load will be equal or greater than the power measurement of that load.

Input power to a system equals the power losses in the system plus the output power.

Transformers and distribution wiring are rated in VA because the losses in these system are are based on the current flow thru the system and not the load power. For example: consider a capacitor and resistor in series connected to an AC source. Ideally a capacitor has no power loss when a current flows thru the capacitor, and a resistor does have power loss = R*Irms^2.

If my RC circuit resistor can dissipated 1 W and the resistor is 1 ohm, then my maximum input current is 1 ampere. With an infinitly large capacitor at 1 V input I have 1 VA and 1 W input power. Decrease the capacitance and increase the input voltage to 100 V such that the current is 1 A, then I still have 1 W of dissipation, so the input power is 1 W, but the input VA is 100 VA. This may seem like a backward explanation but is intended to show the difference between power and VA.

Now consider a transformer and view it from the primary side. Whatever load exists on its secondary can be reflected to an equivalent transformed value at the input. The input impedance seen at the transformer input due to the load is approximately N^2 times the load impedance. Where N is the turns ratio. Also N^2 is N squared.

For the input impedance due to the output load you should be able to prove to yourself that input KVA = output KVA.

.
 

mivey

Senior Member
gar said:
...My opinion is that DC excitation of tungsten filament incandescent lamps shortens their life compared to AC...
I'm assuming the two voltages are rms & dc equivalents. Also, wouldn't the AC vibration stress the filament more than the DC?

What is it about the DC that causes the reduced life in tungsten? I found some info on filament notching but not a lot of detail but it seems to be prevalent in small diameter filaments. Some manufacturers said a 50% life reduction, others say the life is about the same. Seems to me you could extend the life by occasionally switching the bulb polarity.

Here is one link I found:
http://www.stormlighting.co.uk/uploaded/VC440ccb25b1a06.pdf

What info do you have?
 

mpoulton

Senior Member
Location
Phoenix, AZ, USA
K8MHZ said:
I beg to differ with you.

I'll concede that one. I just did a bunch more reading on lightning - the restrikes are all the same polarity, and thus there is no zero crossing. So, if we require actual zero crossings to call something AC, then lightning is not AC - even though the FFT of the signal would tend to indicate otherwise.
 

mpoulton

Senior Member
Location
Phoenix, AZ, USA
K8MHZ said:
You will also find that capacitors will behave much differently when exposed to pulsating DC than they will with AC of the same frequency as the mechanics of a capacitor are dependent upon a change of the direction of electron flow, whereas an inductor produces reactance based upon moving magnetic fields which can occur without a directional change.

This depends. If you are referring to the current waveform, then that is undeniably correct - capacitors "require" that net charge flow be zero through the device in the long run, thus the current flow must reverse periodically to achieve this. A capacitor exposed to a current waveform that never changes direction (in fact, any current waveform with a nonzero average current, even if it does change direction) would exhibit a continuously increasing or decreasing voltage across it, which would be impossible to sustain indefinitely.

On the other hand, the voltage waveform does not ever need to reverse, nor are voltage zero-crossings required in order for a capacitor to behave "normally". In fact, one of the common applications of capacitors in electronic devices is to separate an AC signal from a DC bias -- as in the microphone circuit of a radio, where the PTT signal (DC) and mic audio (AC) are carried on the same line, or between stages of an RF amplifier. These uses are referred to as DC blocking capacitors, because they stop the DC component of the waveform while passing the AC component. In most applications, the DC bias is much larger than the AC signal, so there is never a zero crossing.

In the case of an inductor, these rules are reversed -- net voltage must equal zero over time (else it saturates), and current may flow in one direction only even though the voltage waveform may switch polarity repeatedly.

So now we have a new question: We may accept that in order for something to be called AC, the waveform must have zero crossings. Is that the current waveform, or the voltage waveform, or both?
 
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