AC/DC question

depends on the ratio of windings...
i
Reply to
Ignoramus30160
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Looking at Ted Edward's nice power supply graphic
formatting link
I once
again wondered something I should be able to figure out, but seem to have a
mental block. Suppose you have a transformer/rectifier AC->DC circuit. In series
with the primary of the transformer is a (perfect, ideal) AC ammeter. In series
with the DC output is a (perfect, ideal) DC ammeter. If you set things up so
that the AC current says 1000 milliamperes, what would the DC ammeter read?
GWE
Reply to
Grant Erwin
Ah, yes. OK, skip the transformer, or consider the transformer an ideal 1:1, so no current transformation is done. It's just the current change around the rectifier I'm wondering about. Good point, uh, ignoramus (why do you have such a horrible moniker?).
Grant
Reply to
Grant Erwin
The problem here is what is the AC source? Is it a real power line, the output of a transformer, or what? Assuming that the input is an ideal AC source, there is a capacitor on the DC side, and all components are ideal, the AC side will draw short pulses of INFINITE current magnitude! Ahh, well, we can't model the ideal situation, then.
With real sources and components, then, the nature of these parts affects the result.
If there is no capacitor on the output, and the meters read correctly (not always true) then the input and output will be identical. But, put that cap on the output, and suddenly the AC side is drawing short pulses, whose height depends mostly on source impedance and the value of the capacitor. Then, you have to know how the meters read short pulses. Many meters will read this kind of current pulse with large errors.
Jon
Reply to
Jon Elson
My moniker serves as an idiot detector.
If someone tries to insult me based on my choice of moniker, I know that I am dealing with an idiot.
If you measured the true RMS voltage (root mean of the square of voltage), then RMS would not change from the voltage having been rectified.
That's because square of a number is the same as the square of the opposite number. All that bridge rectifying does is it changes the "sign" of voltage to the opposite (positive) when voltage changes to negative.
instead of ^v^v^v^v
the rectified voltage is ^^^^^^^
but the absolute value does not change, and the square of voltage does not change.
i
Reply to
Ignoramus30160
4777 "Grant Erwin" wrote: Ah, yes. OK, skip the transformer, or consider the transformer an ideal 1:1, so no current transformation is done. It's just the current change around the rectifier I'm wondering about. (clip) ^^^^^^^^^^^^^^^^^ If you consider all the components to be "ideal," then conservation of energy gives you the simplest way to look at it. Whatever DC voltage comes out will have to result in the same wattage as that going in. This will depend on more than the turns ratio--you could have half-wave or full wave rectification, combined with some amount of capacitance and inductance for filtering. At no load, the output capacitors will charge up to the peak value. You now add load until the input current is 1000 ma. Depending on how much this loads the system, the output voltage will drop. In your ideal system, the input power will always equal the output power, but we don't know what the values will be.
Reply to
Leo Lichtman
Grant sez:
"...If you set things up so that the AC current says 1000 milliamperes, what would the DC ammeter read?"
If the "perfect" AC ammeter read in RMS terms, then the DC ammeter would read 1 amp.
Bob Swinney
Reply to
Robert Swinney
The question doesn't include enough info, but assuming that the AC meter in the primary circuit is showing an accurate reading of 1000 mA current (which is 1 A), and the ratio of the xfmr is 120VAC to 12VAC (factor of 10), then the output would probably be about 10 amps DC (disregarding xfmr losses and line voltage differences).
It wouldn't be a good idea to keep a sensitive amp meter in the primary circuit during use, because an accidental touch of the leads in the tank could damage it (faster than a fuse can protect it).
The same can happen with the DC output meter. Battery chargers have ruggedized meters that can withstand brief short circuits, but many other meter types just smoke.
WB ..............
Reply to
Wild Bill
If the AC ampmeter is perfect, and the DC ampmeter is perfect and the transformer is a 1 to 1 transformer and there are no core losses or any other losses , the AC ampmeter will read more than the DC ampmeter.
The reason is that the transformer has an open circuit inductance and therefore the power factor is not one. So even though there are no losses, the AC input current will be greater than the AC output current.
Dan
Reply to
dcaster
Grant's original premise of everything being perfect, would preclude any power factor anomalies, etc.
Bob Swinney
Reply to
Robert Swinney
I'm not sure this is a meaningful question considering the many unknowns in the circuit provided. It is true that one amp RMS at one volt RMS provides the same energy as one amp DC at one volt DC, namely one (1) Watt.
Reply to
Dave
Grant Are you trying to come up with Kirchoffs Law? Get your head into thinking power (volts x amps) and not amps. Figure the power being drawn from the AC source and assuming that there is no loss in the primary or the secondary of the transformer then the power in the secondary is the same as the primary. lg no neat sig line
Reply to
larry g
Of course, one must adjust for the transformer ratio and loss but assuming and ideal 1:1 lossless transformer there are a few considerations. If we are talking d'Arsenval(sp?) meter movements, deflection is proportional to average current. Such meters intended for use with AC are usually calibrated with rms values assuming a sine wave. This is fine with resistive loads but otherwise all bets are off.
Unfortunately, the question is not well defined.
Ted
Reply to
Ted Edwards
Not so unless the capacitor has infinite capacity. Assume an ideal (infinitely stiff) sine wave source and ideal diodes. Consider: During the portion of the half cycle when the capacitor is discharging, I=C(dV/dt). Integrate to find delta V. When the input voltage reaches Vpk-delta V, the capacitor will begin to charge again according to I=C(dV/dt). The slope of the sine wave is not infinite so neither will I be. The exact solution of this is an interesting exercise in numerical analasys. I do not, off hand, recall if a closed form solution is possible.
Ted
Reply to
Ted Edwards
I thought Grant's premise was very well defined. IMO, the only way he could have stated it more simply was if he had a magic lossless black box that converted AC to DC and the metering devices on both sides were lossless as well.
Bob Swinney
Reply to
Robert Swinney
And the scary part is, if you take away half of the cycles in the full-wave rectified version, ^-^-^-^-, the power only decreases to .707 of the original RMS power, not by half!
Cheers! Rich
Reply to
Rich Grise
I like the way RF ammeters function - the input is shunted (normally) and a higher impedance shunt (heating element) is placed across the terminals. At the center of this heater is a thermocouple. The thermocouple is then using a series resistor (calibrator) attached to the movement (voltmeter mode).
The heating effect in a load is said to be the same between DC and AC-RMS - and this RF type simply makes a far better device to measure with. Putting a series meter in a very high current transmission line is out of question. But these are taps on a line.
Martin
Reply to
lionslair at consolidated dot
I like the way RF ammeters function - the input is shunted (normally) and a higher impedance shunt (heating element) is placed across the terminals. At the center of this heater is a thermocouple. The thermocouple is then using a series resistor (calibrator) attached to the movement (voltmeter mode).
The heating effect in a load is said to be the same between DC and AC-RMS - and this RF type simply makes a far better device to measure with. Putting a series meter in a very high current transmission line is out of question. But these are taps on a line.
Martin
Reply to
Martin H. Eastburn
This thread is going round in circles because of loose definitions try;-
1. Perfect AC ammeter - thermal device responds to true RMS value independent of waveform.
2. Perfect 1:1 CURRENT transformer. Infinite primary inductance. Zero shunt losses.(e.g. eddy current or core losses - series losses don't matter because it's a CURRENT transformer)
3 Lossless full wave rectifier. With sine wave input, "DC" output current is a train of unipolar half sine waves.
4 Perfect DC ammeter - thermal device that responds to the true RMS value of the unipolar output current
If All of these conditions are met the measured output current is precisely equal to the input current. Everyday measurements are a bit different.
It would be common practice to measure input current with a meter calibrated in RMS current but the output meter would almost invariably be a moving coil or digital meter calibrated in the AVERAGE current value. In this case the the meter would register the MEAN value of the output current which, for true sinewave input, is 0.9 x RMS current.
If the input waveform is non-sinusoidal with a large peak to mean ration the thermal DC meter will still show the correct value but the moving coil meter errors will increase.
Jim
Reply to
pentagrid
Conservation of energy is the best way to show this.
Analog computer anyone?
:^)
Jim
Reply to
jim rozen

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