Constant controlled power source

There are circuits for constant controlled voltage, and constant controlled current as shown below:

Voltage controlled constant voltage source

simulate this circuit – Schematic created using CircuitLab

What if I wanted to create a circuit that was a voltage controlled constant power source? Here is the basic idea, I'm wondering if anyone has used a circuit that does this sort of thing, or if there are circuits that do this. I have not found any yet. I would like to control the power going into a load.

simulate this circuit

• There is a different method used in old HP RF Power meters based on thermal matching, would this latency be ok? – Tony Stewart Sunnyskyguy EE75 Jan 16 '18 at 23:14
• Probably too slow I'd need at least 100Hz, but I'm more concerned about these designs in general, it seems that it has not been done to some extent. – Voltage Spike Jan 16 '18 at 23:18
• I did a constant power source for an electrosurgical application where the scalpel had to deliver the same power as the conductivity of the tissue varied an order of magnitude. I did it using digital control, which worked out pretty well. – John D Jan 16 '18 at 23:24
• Thats cool, Digital control might be too noisy for me. I did a MPPT tracker and that worked out really well using power control, but we didn't need a lot of resolution – Voltage Spike Jan 16 '18 at 23:34
• I remember one design that did this in Electronics Magazine in 1975 using analog multipliers – Tony Stewart Sunnyskyguy EE75 Jan 16 '18 at 23:45

Here is an application note circuit that may be useful. The power monitor chip is more economical than an analog multiplier and allows the load to be grounded, which is a big advantage in many applications.

As it says in the app note, the circuit can be scaled easily and the simple linear output stage could be replaced with a SMPS block.

The circuit controls the power to the load in parallel with the divider so it tails off a bit at high load resistances, but it's still not too shabby even at the 100mW output level.

Your circuit looks conceptually okay, concerns would include the input voltage range of the instrumentation amplifier and its common-mode rejection. Stability is the most likely thing to cause headaches in many of these circuits.

One way of doing it would be to use diodes since they have logarithmic properties.

We also know that $P=V^2/R$, and we also know that R will be constant (just a resistor, right?), then we can normalize our expression along these lines, $R=1Ω$.

This means that we can go further with this expression:

\begin{align} P&=V^2/1Ω\\ P&=V^2\\ \sqrt P&=V\\ P^{1/2}&=V\\ e^{\ln(P^{1/2})}&=V\\ e^{\ln(P)/2}&=V\\ e^{(\ln(1)+\ln(P))/2}&=V\\ \\\text{Rename variables..}\\ V_{out}&=e^{(\ln(1)+\ln(V_{ref}))/2}\ \end{align}

Alright, we can go between $e^x$ and $\ln(x)$ with diodes, resistors and op-amps. I want to keep this answer short and concise, so rather than going in to how everything works I will show you the end result with voltages and you will hopefully be able to see how and why everything works.

Here's the link in case you want to mess around, say... change the number 3. Just scroll with your mouse button when it is over the number, or go to the right hand side and change the red "Voltage" slider.

The problem with this setup is the fact that the diodes have to be matched, well, you can always do some compensating with some resistors.. so it's not impossible, but it can be done purely in the analog domain.

• You could use the base-emitter junctions in a matched transistor array instead of trying to do your own matching. – vofa Jan 17 '18 at 3:16
• @vofa Hmm I was thinking about that, but I'm not 100% sure if they are entirely logarithmic like diodes. I know that the $V_{BE}$ behaves like a diode ($V_{forward}$), but I believe you mean that I should replace the diodes in the schematic above with NPN transistors that has their base shorted to their collector, right? - But hmm, the extra current that comes from the collector.. I'm not energetic enough to verify ifthe current is all logarithmic or if there is some linear term added in there. - But why can't I just use didoes in a matched diode array? – Harry Svensson Jan 17 '18 at 3:35
• Seems to work fine - tinyurl.com/yc343lk8 . Quad matched NPN arrays are commercially available. I have not seen quad matched diodes. – vofa Jan 17 '18 at 3:44
• How does this not have more upvotes? Very, very clever! – testname123 Jan 17 '18 at 7:37
• @kimstik then make an answer skipping the anti-log and use closed loop instead. I'd love to see it. Show a schematic like I did. Remember, the goal is to make a function that takes a voltage in and spits out the sqrt of that voltage. - Because that's what setting a desired power in voltage -> the voltage across the load. – Harry Svensson Sep 10 '19 at 3:57

It is certainly possible to design a constant power source, and the schematic you posted is one approach, discussed for example here

I see in the comments that you need > 100Hz of bandwidth. Depending on the nature of the load, and range of output power if adjustable, you may have some difficulty stabilizing the loop. If the load is resistive, for example, there is highly non-linear transfer function from the output voltage (or current) to the sampled variable - i.e. the output power. That means loop gain will vary based on operating point. The problem is not intractable, though, especially if you have a well characterized load and don't need to operate over a large range of output powers.

Another option is to use an energy storage device (i.e. a capacitor or inductor) to deliver discrete, controlled energy pulses at a precise rate. It could be done with a switched capacitor or switched inductor approach, but in both cases the precision will depend on how well controlled those passives are as well as the input and output voltages. More academic, this type of solution.

• Yeah, I really want to design a DC to DC converter that sinks and sources from 0 to Vin, but I'll save that for a rainy day. – Voltage Spike Jan 17 '18 at 6:23

It's certainly possible to make a constant-power source, although harder than constant voltage or constant current, especially for analog electronics. The real question is what you want it for.

To make a constant power source today, I think the easiest way is to start with a buck converter. You didn't say anything about regulation level or ripple, so a pulse on demand system would work. The processor measures the output voltage, the output current, multiplies them, then does a pulse when that is below the regulation threshold.

The electronics is the same as a normal buck converter, except for the output current sensing. The rest is firmware.

• Why would it be easiest to start from a buck converter rather than a linear regulator? – user253751 Jan 17 '18 at 0:15
• It would be the same, who cares? Just once it works, you don't need to multiply anything, the output is constant, hence you only need now to step down with constant current. And thatvstage MUST be switching to actually change current depending on output voltage. – Gregory Kornblum Jan 17 '18 at 1:30
• I try not to throw on too many requirements, it stifles creativity. I have done digital power control in the past, but it has it's own set of problems. The load, a peltier – Voltage Spike Jan 17 '18 at 6:30
• @immi: With a buck converter, its very easy to incorporate a digital processor to control it. Things happen at discrete times, which works well with digital control. Once you have the digital controller, you can have it do the multiply. That's something that is difficult, expensive, and/or inaccurate in analog electronics. – Olin Lathrop Jan 17 '18 at 11:55

Constant power load i designed was composed of a dc/dc (don't remember, probably just an isolated one, but it doesn't matter) that had a constant voltage output and then a buck stage, which closed loop on current of the first dc/dc. Was very simple and way faster than 100Hz. I would guess it was around maybe 3kHz bandwidth.

If resistance of load is constant all what you need to control power is to control voltage over load. It is easy job of LDO.

With unknown/nonlinear load current over it should be measured. Knowing current and voltage on load power can be controlled.