This makes absolutely no sense to me ...
Ahh ok I get it .. however determining the Z out with Kirschoffs law and the formulas is a tad beyond my pay grade. I could probably work through it but I'll just take your word for it.
Let's assume a case where we have 40w applied to a transformer that has a 4kΩ primary, and 8Ω & 4Ω secondary taps.
Volts = √(Power x Impedance) = √(40w x 4kΩ) = 400v on the Primary side
The fundamental property of transformers is that if 1v-per-turn is applied to any winding, all windings also have 1v-per-turn. So the Turns Ratio tells use the ratio of voltages among the windings. For 4kΩ : 8Ω, 4Ω that's
8Ω Tap ---> √(4kΩ / 8Ω) = 22.36 : 1
4Ω Tap ---> √(4kΩ / 4Ω) = 31.62 : 1
So when we apply 400v to the primary, it will be stepped down to secondary voltages in proportion to the turns ratios:
8Ω Tap ---> 400v / 22.36 = 17.89 volts
4Ω Tap ---> 400v / 31.62 = 12.65 volts
Assume we have a speaker attached each tap. They will attempt to draw:
8Ω Tap ---> 17.89v / 8Ω = 2.236A
4Ω Tap ---> 12.65v / 4Ω = 3.163A
Going from Primary-to-Secondary, the Turns Ratio stepped-down Volts (because the secondary has fewer turns) and stepped-up Current. Now going back the other way from Secondary-to-Primary, Current is stepped down by the same ratio. Let's find out how much current each load attempts to draw on the primary side:
8Ω Tap ---> 2.236A / 22.36 = 0.1A = 100mA
4Ω Tap ---> 3.163A / 31.62 = 0.1A = 100mA
If we only have
one load on
one tap, then either attempts to pull 0.1A through 4kΩ. how much power is that?
4kΩ x 0.1A = 400v
400v x 0.1A = 40 watts
Now recall we started out assuming 40w applied to our 4kΩ primary, which was how we knew the applied voltage. Now assume
Applied Primary Volts remains unchanged at 400v, and the
Applied Primary Power remains unchanged at 40w. We see for the voltages that will be delivered to the secondary taps that if both are loaded, each tries to pull 0.1A (when referred to the primary) for 0.2A total.
Let's find the Primary Impedance implied when 400v results in 0.2A of current:
Volts / Current = Impedance
400v / 0.2A = 2kΩ
This shows that if we have a 8Ω load on the 8Ω ta at the same time that we have a 4Ω load on the 4Ω tap, the Primary Impedance becomes 2kΩ rather than our original 4kΩ.
This is the same result as though we used two 8Ω loads in parallel on the 8Ω tap, for a total load of 4Ω load.
Therefore we conclude that loading multiple taps is equivalent to parallel loads on a single tap. And so if we want to reflect the original-design Primary Impedance, we need to use a load of double the tap's marked impedance (for 2 taps loaded), or a load of 3x the tap's marked impedance (for 3 taps loaded), etc.
Topic: OT using multiple taps at same time (Read 3696 times)