Academic question: Is there such a thing as an electrical constant power source, which delivers energy at a fixed rate?
Specifically … there are voltage sources and current sources, which maintain (an approximation to) a reliable fixed voltage/current respectively. A current source will deliver electric charge at a fixed rate, charging a capacitor's voltage linearly; a voltage source will deliver magnetic flux at a fixed rate, 'charging' an inductor's current linearly.
I'm trying to figure out if there is a supply that could deliver energy at a fixed rate – a power source – so its voltage changes to whatever is necessary to deliver a current such that \$VI\$ is a fixed amount.
(I've tried Googling this, but the search is swamped by power supply ads, domestic power advice etc).
Guessing at its properties…
- it would treat voltage and current symmetrically! So a fixed power source would charge up a \$1\mu F\$ at the same rate as a \$1\mu H\$ inductor.
- it would charge a capacitor's voltage at a rate of \$ V \propto \sqrt{T}\$, because the energy is increasing linearly in time and \$E = ½ CV^2\$.
- same with an inductor's current.
- any resistor placed across it would dissipate the same heat regardless of the resistance – by definition. Doubling the resistance would increase the voltage by a factor of \$\sqrt{2}\$ and decrease current by the same.
- its impedance is … undefined! Open circuit voltage is infinite (like a current source); closed circuit current is … also infinite (like a voltage source). Hmm.
For extra points: any other interesting properties? Are they actually a thing? What would they look like?
Best Answer
It's pretty simple to do with circuitry. The below circuit provides constant power at the output provided OA1 does not saturate, which limits it to something like 15A.
In reality the VCCS would also have limited voltage output compliance, which would show up in an inductor circuit.
simulate this circuit – Schematic created using CircuitLab
Here is what it looks like charging a 1F capacitor with 1W:
P.S. I used an op-amp with a multiplier because CircuitLab does not have a divider function.