Lets say we had a circuit with a bridge, a resistor and an LED.
If we power it from 5V DC, we calculate X current draw, based on the resistor.
If we power the same circuit from 5V AC instead, is it exactly the same current?
Since the 5V AC is RMS, is it the exact same ohms law calculations as for DC? Same for wattage?
db2db:
Lets say we had a circuit with a bridge, a resistor and an LED.
If we power it from 5V DC, we calculate X current draw, based on the resistor.If we power the same circuit from 5V AC instead, is it exactly the same current?
Since the 5V AC is RMS, is it the exact same ohms law calculations as for DC? Same for wattage?
No, and for 3 separate reasons:
An LED+resistor is a non-linear system, so will not respond proportional
to applied voltage.
A bridge loses 2 diode drops, its again a non-linear system with losses.
"Exactly the same current" is meaningless since with DC the current is constant,
with AC it is varying. You have to use a careful definition of "same current",
such as "root mean squared" or "average".
With 5V the diode losses are noticeable. With a higher supply voltage, just a
resistor as load and "same current" defined as "same rms current", the currents
would basically be the same. With a much higher voltage (so that the forward
voltage of the LED is negligible) then its true for the LED too.
Is this a high power LED? For a simple indicator LED the current isn't very
important, the eye's reponse is logarithmic.
If you would have left the LED out, the power dissipated in the resistor would be the same for 5 V DC and 5 V rms across the resistor.
If you are supplying the 5 volts at the input of a bridge rectifier, and have a capacitor filter, you can have more voltage at the load with the AC case since the filter will start out at the peak voltage and drop depending on the load and capacitance.
It is important to realize that we need to be specific when talking about AC.
5VRMS AC will cause the same amount of power dissipation in a given linear resistance as 5V DC.
For a square wave, 10V peak to peak AC has an RMS value of 5VRMS. For a sine wave, 14.14V peak to peak has an RMS value of 5VRMS. I'm talking now about symmetrical waveforms with no DC offset.
RMS is just meant to be an easier way to calculate current, voltage, and resistance. It only works for linear loads with constant resistance regardless of frequency, voltage, current, temperature, etc. If you include reactive loads like inductance and capacitance, then things change and get more complex. For that, you need to understand imaginary numbers, phase angles, etc.
We employ a lot of simplifications to avoid using calculus just to calculate the value of a resistor for an LED. Generally, it doesn't matter if the current is 10 or 20% off. So an LED that might drop 2 to 2.2V from 1mA to 50mA, we pretend it is 2V when calculating the resistor required.
KeithRB:
If you would have left the LED out, the power dissipated in the resistor would be the same for 5 V DC and 5 V rms across the resistor.
Why is it different with the LED? There's a bridge before the LED...
There's no cap.
No, and for 3 separate reasons:
An LED+resistor is a non-linear system, so will not respond proportional
to applied voltage.
A bridge loses 2 diode drops, its again a non-linear system with losses.
But the same for 12vac or 12vdc input, as the components are on the other side of a bridge?
Why is the AC current not constant?
I'm really only interested in current draw being equal as a theoretical question, not perceived brightness of an LED.
polymorph:
It is important to realize that we need to be specific when talking about AC.5VRMS AC will cause the same amount of power dissipation in a given linear resistance as 5V DC.
For a square wave, 10V peak to peak AC has an RMS value of 5VRMS. For a sine wave, 14.14V peak to peak has an RMS value of 5VRMS. I'm talking now about symmetrical waveforms with no DC offset.
Why do you say 14v p to p is 5v rms?
My original question is based on a transformer only for AC, so is a sine wave.
I thought, with a sine wave, RMS to peak was approximately RMS * 1.414, and peak to RMS was (also approx.) peak * 0.707?
With a true square wave at 100% duty cycle, isn't RMS == peak?
Is there a difference in voltage of the rectified AC right after the bridge, as compared to the same point with DC into that same bridge?
After the bridge, when using an AC input it's pulsed DC.
After the bridge, when using an DC input it's just DC.
Is that a different effective DC level?
Doesn't that in turn affect the overall current consumption?
Yeah, because a 100% duty cycle is DC.
polymorph:
It is important to realize that we need to be specific when talking about AC.5VRMS AC will cause the same amount of power dissipation in a given linear resistance as 5V DC.
To be even more specific:
5VRMS AC will cause the same amount of power dissipation averaged over a
whole number of cycles in a given linear resistance as 5V DC.
Note that for a sinusoidal AC current the power in the resistor varies as a sinuisoid
of twice the frequency, varying between zero and twice the average power. Average
this over a whole cycle or a long enough time and the power is the same as for DC.
Actually, RMS is accurate to an integer number of 1/4 cycles, as long as you begin at the zero crossing or peak and end at a zero crossing or peak.
If you measure over an integer number of whole cycles, then any timing will do. As long as it is an integer number of whole cycles.
But maybe I'm being too specific...
Note that for a sinusoidal AC current the power in the resistor varies as a sinuisoid
of twice the frequency, varying between zero and twice the average power. Average
this over a whole cycle or a long enough time and the power is the same as for DC.
Does putting a bridge into this change the equation?
Does putting a bridge into this change the equation?
Without a capacitor, and with a high enough voltage that the diode forward drop is negligible, no.
With a bridge rectifier, there are two diode drops so about 1.4V is dropped out of the "bottom" of the sine wave. So the power would be less. You could get reasonably close if you take into account the power lost in the diodes.
If you add a smoothing capacitor of sufficient size so that the ripple voltage is tiny compared to the peak voltage, then you are looking at a DC voltage just under Vpk - 1.4V.
polymorph:
Does putting a bridge into this change the equation?
Without a capacitor, and with a high enough voltage that the diode forward drop is negligible, no.
With a bridge rectifier, there are two diode drops so about 1.4V is dropped out of the "bottom" of the sine wave. So the power would be less. You could get reasonably close if you take into account the power lost in the diodes.
If you add a smoothing capacitor of sufficient size so that the ripple voltage is tiny compared to the peak voltage, then you are looking at a DC voltage just under Vpk - 1.4V.
So just to clarify my original question... With either AC or DC input, this complete circuit (with a resistor, an LED and a bridge, and no cap) will use basically about the same overall current, yes?
Slightly less current and power, since some of the voltage is dropped out by the diodes, so less voltage across the resistor means less current.
polymorph:
Slightly less current and power, since some of the voltage is dropped out by the diodes, so less voltage across the resistor means less current.
I know the DC drop is 1.4sh
Is the AC drop different then?