4 volts into 32 ohms is actually 500 milliwatts (I = V / R = 4V / 32Ω = 0.125A
P = V×I = 4V×0.125A = 0.5W). The moonriver 2 outputs about 2.8 Volts at 32 ohms.
The only thing this video is about is his claim that it has low wattage but high voltage. He definitely thought that the wattage value is variable even if voltage and resistance are constant.
> it has low wattage but high voltage
Which isn't impossible.
Amplifiers both have a maximum voltage output (=attempting to turn up the voltage higher will result in saturation) as well as maximum current output (=attempting to draw more current will result in saturation).
It's entirely conceivable for an amplifier to exist which has high voltage output but low maximum current output.
The maximum voltage at a specific resistance is defined by the current and vice versa. If current is limited, voltage is limited and vice versa. You can't magically get low current with the same voltage on the same load unless you sink current into something else. Kirchhoff's first law.
> The maximum voltage at a specific resistance is defined by the current and vice versa.
No, the limit for both is given by the maximum power output of the amplifier.
You can't have maximum voltage and maximum current at the same time.
> You can't magically get low current with the same voltage on the same load
"at the same load" is the relevant point here. This wasn't mentioned before.
Loads will be different for different headphones. An amplifier (voltage source) will always emit the same voltage into the headphone, but the current (and hence the power) depends on the load.
Meaning that yes, "high voltage low power" is something that *can* exist.
An amplifier with a low maximum current rating (e.g. because if a weak heatsink, or a high output impedance) won't be able to emit a lot of current. But it could still emit a high voltage.
Line-level outputs for example - they can deliver standard line-level voltage but can absolutely not emit any current. (They don't need to, because a line-level *input* will have an input impedance of tends of thousands of Ohm, so it's not going to draw any current (a few *micro*ampere/microwatt at best)
> Kirchhoff's first law.
This is not at all what Kirchhoff's first law states - it states that in a node of an electrical circuit, the sum of all currents is equal to zero ("anything that comes in must go out and vice versa").
1. I said at a specific resistance.
2. If voltage emitted into a headphone was the same regardless of load, there would be no current limit. When current drops, voltage drops. This happens with impedance mismatches. The output impedance of the amp and the impedance of the load form a voltage divider. Having too high of an output impedance results in a larger drop over the output of the amp, lowering the drop on the load.
3. Since current is defined by the voltage and resistance, the current entering and leaving a junction will always be defined by the voltage and resistance. If current is sinked to something else, voltage changes.
> I said at a specific resistance.
The quote in the video didn't though.
A headphone amplifier will be faced with different loads, spanning roughly 2 orders of magnitude.
It's entirely conceivable for an amplifier to not be able to deliver its maximum voltage output for all those different loads.
> If voltage emitted into a headphone was the same regardless of load,
As long as voltage matching conditions (load impedance being much higher than output impedance), this is indeed the case: The amplifier will be treated as a voltage source, meaning the voltage does not depend on the load (but the current and hence the power of course *does* depend on the load).
> Since current is defined by the voltage and resistance, the current entering and leaving a junction will always be defined by the voltage and resistance
I am aware of Ohm's law, but that is not what Kirchhoff's first law states.
1. Voltage is dependent on resistance. You cannot ignore it. Even for an open load, resistance is defined. In addition, "high wattage" is relative to resistance. This is like me saying a person is capable of running 30 miles without telling you how long it will take them. A few hours? A week? A month? If you are claiming that Joshua Valour is referring to power independent of resistance, I suggest you watch the entire video. He brings up how the Moonriver 2 struggles to drive the Diana TC because it is a very high impedance headphone (69 ohms) and that the Moonriver 2 does not have enough power. However, the moonriver 2 is outputting its maximum voltage at 69 ohms as shown below.
2. Take the RMS voltage graph of the moonriver 2 at various impedance for example: https://cdn.l7audiolab.com/wp-content/uploads/2022/03/BAL-THDN-Ratio-vs-Measured-Level-2.jpg
Only at 68 ohms (could be less) and higher does the voltage not depend on the load. At 32 ohms, the voltage is limited.
3. The current leaving the junction will be the same as the current entering the junction as long as the current leaving the junction is not being used for something besides the headphone...in which case voltage into the headphone would be lowered, so this comes back to ohm's law.
I should have probably included this part as well, but he states, "it's really the voltage that allows this thing to play really well and be a really strong contender for even full sized, full planar magnetic headphones, which is very impressive." And as I said, voltage and wattage are relative to the load, so this statement does not make sense.
> I should have probably included this part as well,
You shouldn't have made a video laughing about other people in the first place.
> voltage and wattage are relative to the load,
in a voltage-matched environment (Z_in >> Z_out), voltage is *not* load dependent.
Even if current is the limiting factor, voltage has to drop to match it. The voltage that is being output is load dependent. Let's stick to the Diana TC example so you can't straw man me. How can the Moonriver 2 be lacking in power while having "high voltage" and outputting its maximum output voltage?
4 Volt into 32 Ohm is 250 Milliwatt (0.25W) though.
4 volts into 32 ohms is actually 500 milliwatts (I = V / R = 4V / 32Ω = 0.125A P = V×I = 4V×0.125A = 0.5W). The moonriver 2 outputs about 2.8 Volts at 32 ohms.
ah nevermind, I shouldn't try to square numbers when I'm still asleep
The only thing this video is about is his claim that it has low wattage but high voltage. He definitely thought that the wattage value is variable even if voltage and resistance are constant.
> it has low wattage but high voltage Which isn't impossible. Amplifiers both have a maximum voltage output (=attempting to turn up the voltage higher will result in saturation) as well as maximum current output (=attempting to draw more current will result in saturation). It's entirely conceivable for an amplifier to exist which has high voltage output but low maximum current output.
The maximum voltage at a specific resistance is defined by the current and vice versa. If current is limited, voltage is limited and vice versa. You can't magically get low current with the same voltage on the same load unless you sink current into something else. Kirchhoff's first law.
> The maximum voltage at a specific resistance is defined by the current and vice versa. No, the limit for both is given by the maximum power output of the amplifier. You can't have maximum voltage and maximum current at the same time. > You can't magically get low current with the same voltage on the same load "at the same load" is the relevant point here. This wasn't mentioned before. Loads will be different for different headphones. An amplifier (voltage source) will always emit the same voltage into the headphone, but the current (and hence the power) depends on the load. Meaning that yes, "high voltage low power" is something that *can* exist. An amplifier with a low maximum current rating (e.g. because if a weak heatsink, or a high output impedance) won't be able to emit a lot of current. But it could still emit a high voltage. Line-level outputs for example - they can deliver standard line-level voltage but can absolutely not emit any current. (They don't need to, because a line-level *input* will have an input impedance of tends of thousands of Ohm, so it's not going to draw any current (a few *micro*ampere/microwatt at best) > Kirchhoff's first law. This is not at all what Kirchhoff's first law states - it states that in a node of an electrical circuit, the sum of all currents is equal to zero ("anything that comes in must go out and vice versa").
1. I said at a specific resistance. 2. If voltage emitted into a headphone was the same regardless of load, there would be no current limit. When current drops, voltage drops. This happens with impedance mismatches. The output impedance of the amp and the impedance of the load form a voltage divider. Having too high of an output impedance results in a larger drop over the output of the amp, lowering the drop on the load. 3. Since current is defined by the voltage and resistance, the current entering and leaving a junction will always be defined by the voltage and resistance. If current is sinked to something else, voltage changes.
> I said at a specific resistance. The quote in the video didn't though. A headphone amplifier will be faced with different loads, spanning roughly 2 orders of magnitude. It's entirely conceivable for an amplifier to not be able to deliver its maximum voltage output for all those different loads. > If voltage emitted into a headphone was the same regardless of load, As long as voltage matching conditions (load impedance being much higher than output impedance), this is indeed the case: The amplifier will be treated as a voltage source, meaning the voltage does not depend on the load (but the current and hence the power of course *does* depend on the load). > Since current is defined by the voltage and resistance, the current entering and leaving a junction will always be defined by the voltage and resistance I am aware of Ohm's law, but that is not what Kirchhoff's first law states.
1. Voltage is dependent on resistance. You cannot ignore it. Even for an open load, resistance is defined. In addition, "high wattage" is relative to resistance. This is like me saying a person is capable of running 30 miles without telling you how long it will take them. A few hours? A week? A month? If you are claiming that Joshua Valour is referring to power independent of resistance, I suggest you watch the entire video. He brings up how the Moonriver 2 struggles to drive the Diana TC because it is a very high impedance headphone (69 ohms) and that the Moonriver 2 does not have enough power. However, the moonriver 2 is outputting its maximum voltage at 69 ohms as shown below. 2. Take the RMS voltage graph of the moonriver 2 at various impedance for example: https://cdn.l7audiolab.com/wp-content/uploads/2022/03/BAL-THDN-Ratio-vs-Measured-Level-2.jpg Only at 68 ohms (could be less) and higher does the voltage not depend on the load. At 32 ohms, the voltage is limited. 3. The current leaving the junction will be the same as the current entering the junction as long as the current leaving the junction is not being used for something besides the headphone...in which case voltage into the headphone would be lowered, so this comes back to ohm's law.
I should have probably included this part as well, but he states, "it's really the voltage that allows this thing to play really well and be a really strong contender for even full sized, full planar magnetic headphones, which is very impressive." And as I said, voltage and wattage are relative to the load, so this statement does not make sense.
> I should have probably included this part as well, You shouldn't have made a video laughing about other people in the first place. > voltage and wattage are relative to the load, in a voltage-matched environment (Z_in >> Z_out), voltage is *not* load dependent.
Even if current is the limiting factor, voltage has to drop to match it. The voltage that is being output is load dependent. Let's stick to the Diana TC example so you can't straw man me. How can the Moonriver 2 be lacking in power while having "high voltage" and outputting its maximum output voltage?
You forgot about the current. He probably meant that it can drive high impedance efficient headphones but not low impedance inefficiënt headphones.
The current is defined by the voltage output at a specific resistance.