Testing my TEC1-12706 peltier, I notice that it seems to be wasting a lot of the energy it has been supplied? - let me explain:-

I set my lab bench supply at 12v 5amps, powered it up, and the voltage dropped to 10v and my supply showed it consuming 4amps

The TEC had monster heatsinks on both sides, and I only ran the test for about 10s each.

My question:-

WHAT IF I built a high-speed switching circuit that used that "left over" 10v @ 4amps to charge up (say) 2 capacitors in parallel, then connect them in series with the peltier input (plus 12v power supply) to drive it (while, at the same time, charging a second-set of capacitors in parallel again), and keep swapping those sets of caps back and forth etc ?

(am I overthinking that? Maybe just charge 1 cap, then put it in series with the power supply to charge a second cap, and keep swapping those caps?

e.g. would that (or some other idea) be able to "reclaim" that lost energy (coming from my solar panels) to pump more of the available energy into the cooling effect?

If my idea has any merit, what kind of practical circuit would maximize the recovery of this energy (taking into account drops and losses of switching/boost circuits, the "diminishing returns" of the dynamic voltages/behavior of all that, and other stuff I've probably overlooked) ?

Chris.

 

If both those displayed voltages are accurate, you are dropping 1.25V in your wiring and connections. Simply use thicker, shorter wires with proper terminal blocks to join them together. Messing around with switched capacitors would be tricky and a complete waste of time and effort.

 

Peltier devices are inherently terribly inefficient.

There is no "left over" energy to harvest, it all turns to waste heat that must be removed by the usual monster heatsinks.

 

@KeithWalker - obviously nothing to do with what I was asking, and completely (well, exactly 99.9% to be precise) wrong suggestion.

It did raise an interesting point though - so re-testing by replacing the peltier with a short-circuit, the supply drops to 1volt at 4.995 amps, so yeah, there's some totally-insignificant loss in the wires (0.1%), but it's the peltier-effect that's using up the missing energy (of course - you cant get heat and cold for free...).

My point, since maybe I wasn't clear - is to NOT simply "short circuit" the peltier over the supply which is how everyone does this usually, but instead, put it in series with a circuit that, instead of simply wasting the energy with a short, *always* uses that energy to charge something (a capacitor), and (when that cap is full, i.e. current drops to 0 since the cap is now charged), swap that cap out for an empty one, while using the now-full one to help charge the empty again (always through the peltier)

 

Not going to work, you will simply add more inefficiency to the system.

If your PV system is producing more energy then you need for the TEC then store it in a battery.

 

@KeithWalker - obviously nothing to do with what I was asking, and completely (well, exactly 99.9% to be precise) wrong suggestion.

It did raise an interesting point though - so re-testing by replacing the peltier with a short-circuit, the supply drops to 1volt at 4.995 amps, so yeah, there's some totally-insignificant loss in the wires (0.1%), but it's the peltier-effect that's using up the missing energy (of course - you cant get heat and cold for free...).

My point, since maybe I wasn't clear - is to NOT simply "short circuit" the peltier over the supply which is how everyone does this usually, but instead, put it in series with a circuit that, instead of simply wasting the energy with a short, *always* uses that energy to charge something (a capacitor), and (when that cap is full, i.e. current drops to 0 since the cap is now charged), swap that cap out for an empty one, while using the now-full one to help charge the empty again (always through the peltier)

The power supply is not "short circuited" by the peltier device, it's delivering power to it.
Anything you do to reduce this power represents less power the device can use to cool anything.

Here is some good study material
https://www.meerstetter.ch/customer...0PAf3Mu7VGjU64aPrqurf3FaZYKFhwbxoC2lkQAvD_BwE

 

What would happen if I stuck a 500uF non-polarized electrolytic crossover capacitor in series with the 12v power supply, and switched it's polarity at 1.6khz (i.e. so it can deliver 40watts).

The input to the peltier is going to be kindof nonlinear sawtooth, right? Going from 22v to 10v before it switches again, and in theory, never wasting any input current by shorting it to ground?

Yes, I KNOW peltiers are inefficient - this is not what I'm talking about. I'm talking about re-using some of what appears to be wasted current available for re-use in their circuits (which won't fix their inefficiency, but might same some small amount of energy)

 

Sorry, but Keith is right.
Messing around with capacitors will gain you zip.
The Peltier takes a large current to operate (it's inefficient but there is no "wasted" current), and switching capacitors in and out will just reduce the efficiency, not increase it ( a fundamental electrical law is that you lose half the energy when you charge a capacitor from a power supply).
But hey, if you want to waste you time, have at it.
But no-one here who understands electronics will support it.

 

What would happen if I stuck a 500uF non-polarized electrolytic crossover capacitor in series with the 12v power supply, and switched it's polarity at 1.6khz (i.e. so it can deliver 40watts).

Then you would be delivering AC to the unit, and that's a big no no.

Unless you mean switch the polarity of the cap, which again would do nothing to help.

I'm finding it difficult to understand the problem with your reasoning other than a complete misunderstanding of basic electronics.

 

Anything you do to reduce this power represents less power the device can use to cool anything.

So if this "thing I do" is stick a capacitor in there, and when it's full, put in a fresh one, and add the charged one into the input... that's (I measured) "reducing the power" by 13%, except later on, you "Get back" 87% of the energy that you would otherwise have wasted...

If I'm wrong (I probably am), I need to know where the error in my logic is, because "logically speaking", we can adjust the input power and the thing we're charging up such that (from the peltiers point of view) it's still getting the input 12v, 4a that it wanted, so for that "thing" we're charging up to have enjoyed zero net free energy, the math has to add up. Right now, from my measurements, it's not adding up, but a significant amount. If my idea's and (F) ... What am I measuring wrong?

 

Except that there is no wasted energy in the system except that which is wasted by the TEC by default.

(Other than wiring losses and whatever)

 

you lose half the energy when you charge a capacitor

The ESR of that 500uF I mentioned cap is 0.1ohms.

If 50% of the energy is "lost" - that's a MASSIVE number!! - where did it go, or are you just exaggerating?

 

Except that there is no wasted energy in the system except that which is wasted by the TEC by default.
(Other than wiring losses and whatever)

I understand everyone's POV (that the "hot side" is disproportionately warmer than the cold side) - however - the thermoelectric effect is caused by the *flow of current* - like a water-wheel turning in a river - it's NOT removing ALL the motion of the water (electrons) - so what I'm saying, is to recycle the left-over current flow.

 

Well, if you could remove all of the motion of the electrons, Kirchoff would start spinning in his grave.

 

Peltier devices are inherently terribly inefficient.

Only if you're trying to create "cold'. They're quite efficient if you're trying to generate heat.

 

If 50% of the energy is "lost" - that's a MASSIVE number!! - where did it go, or are you just exaggerating?

No I don't exaggerate (at least not about electronics).
It's a fundamental electrical law.
(You obviously are unfamiliar with several. For example there is no such thing as "left over current flow).

The reason for only half the energy going to the cap is that the current from the power supply to charge the capacitor is delivered at a constant voltage, but the capacitor starts at 0V and then increases as it charges.
The difference between those two voltages is dissipated in the resistance (independent of its value) between the power supply and the capacitor as it charges.
If you do the calculations you will find that the energy dissipated happens to be equal to the energy stored on the capacitor.

Below is a simulation to illustrate this relation for a voltage charge source (Rs is the simulated power supply resistance).

The yellow trace shows the power generated by the supply during the time the capacitor charges.
The integral of this is the total energy supplied, which is 100mJ here.

The red trace shows the power dissipated by the resistor.
Its integral is 50mJ, half of that generated by the supply.

The blue trace shows the power delivered to the capacitor.
Its integral is also 50mJ, half of the generated by the supply.

This relation is unaffected by the sized of the resistance, except for how long it takes for the capacitor to charge.

In theory you can charge a capacitor with no loss if you place an inductor in series with the capacitor, and stop the charge when the capacitor reaches its max voltage (which will be twice the supply voltage).
Alternately, an efficient constant-current switching regulator will also charge it without that loss.

 

They're quite efficient if your trying to generate heat.

You just have to have a path for the cold side to "dissipate" its cold, that is separate from what you are trying to heat.

 

it's NOT removing ALL the motion of the water (electrons)

But it does remove all the energy in the (electron) current, even though inefficiently for transferring heat from the cold to the hot side.
In that respect it's not like a water wheel in a river.
There is no energy left.

 

You just have to have a path for the cold side to "dissipate" its cold, that is separate from what you are trying to heat.

Worst case: 100% efficiency.

 

Worst case: 100% efficiency.

True.
Not many functions have a minimum efficiency of 100%.