Peltier plate voltage drop

Hi everyone. I’m working on a project that harvests heat energy using Peltier plates (TEC 12715) and gives me voltage of approximately 2V.
I boost that voltage to using 5V Fixed output Boost module which has a usb charger integrated in it.
I plan on using that to charge my power bank.
When the circuit is open i.e. no power bank is connected, the Peltier works perfectly generating a voltage of 2V and my booster works too and generates 5V output.
BUT as soon as i connect my load (power bank), voltage generated by the Peltier (input to the booster) DROPS to 0.5V which stops the charging action as my booster needs a minimum of 1V to work.
Please help me resolve the issue and do let me know if you need any other specifications

Thank you in advance

The boost converter is attempting to draw more current than the Peltier can supply.

If you increase the voltage by a factor of 2.5, the input current has to increase by a factor of typically 2.8 (taking into account the boost converter efficiency).

Peltier modules intended for cooling are usually rated for 80 degrees Celsius maximum temperature. If you exceed that rating the internal solder connections will melt and the module will be destroyed. Peltier modules intended for thermoelectric power generation are often rated for over 200 degrees Celsius, but are a bit more expensive.

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i can't increase the voltage factor as i'm using a fixed 5V output voltage booster,
what is the solution to reduce voltage drop?

I've never heard of generating electricity from a Peltier device... And I didn't see any information about that in the datasheet I found. So it's probably very inefficient. Of course you will need a temperature difference to extract energy from anything that converts thermal energy to electrical energy..

[quote]I plan on using that to charge my power bank.[/quote]Any idea of how much energy you need? If you are actually charging a battery you can estimate how many milliamp-hours or watt-hours you'll need. But you'll need to allow for some inefficiency in the charging process.

You'll probably just have to experiment with a variety of load resistors to see how much energy you can get out.

In case you don't know this - Ohm's Law says: Current = Voltage/Resistance (so for example, 2V across 1k Ohm is 2 mA).

And the basic power calculation is Power (Watts) = Voltage x Current.
Or you can derive Power = Voltage squared/Resistance.

Yes sir you are right
I am generating a temperature difference between both sides of the peltier plates by applying heat on one side and attaching a heat sink to the other.
Also I've applied heat sink compound to both sides.

Could you give me some tips on how to reduce voltage drop, that would great.

Thank you

You reduce the voltage drop by increasing the current source. That means more Peltier devices in parallel with your existing device.
Paul

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Basically you can’t get more power out of the module than it can generate from the heat input , and it looks like that’s where you are.

Think about where you are going to "dump" the heat as well. Typically this is a weak point in people's designs. They end up heat saturating the system and everything cooks.

-jim lee

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Your electrical knowledge is missing a very important concept, CURRENT :smiley:
How much CURRENT is the Peltier generating @ 2V?

You need a BIG heat sink, preferably with a fan. It is much easier to heat one side than cool the other.

Current output is proportional to the temperature difference between the plates.

Take less current from the Peltier. Energy harvesting generally generates very low amounts of power. Your typical USB battery bank is going to be a huge load in comparison to the paltry amounts generated by an energy harvester.

To get the most power out you need to use MPPT, which means the power out
is controlled to avoid overloading the thermoelectric generator (what you call
a Peltier device when its converting heat flow to electricity). However if your
load is a given this won't be useful - either the load is too large or it isn't.

Peltier devices are in general very inefficient, perhaps you need much larger one
for the load you are placing on it.

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Thank you for your time.
The voltage generated is 2.5V and current is 100mA.
Could you please suggest a device i could charge from these specs?

Thank you for your time.
The voltage generated is 2.5V and current is 100mA.

Thank you for your time.
The voltage generated is 2.5V and current is 100mA.
Would a MPPT work on such a low power and would it give me regulated 5V output?

i've attached a heat sink and also applied a cooling pad to keep one side cold.

The problem i am facing is to reduce the voltage drop when a load is applied.

Could you help me with that?

There is nothing in the concept of MPPT that doesn't scale - whether someone's made
a low power version is an entirely different question - it might be a case of make one
yourself which may be unreasonably complex/difficult.

You can change the load impedance so it doesn't overload the Peltier - a series resistor
after the boost converter may be enough - however that means tuning the value to one
thermal configuration.

@MarkT What did you mean by tuning of thermal config?
Also my boost and usb module are connected on an integrated board

could i connect the resistor before my booster?

Use a power supply set at 2.5 volts to run the boost reg. If the boost reg is what it claims, its output should be 5 volts.

Measure the current it takes. Without the load (power bank). You have said it appears to work.

Now attach the power bank. Your power supply will still be supplying 2.5 volts, but see now what the current required from any 2.5 volt power source connected would be!

Imma bet it will be comfortably in excess of 100 mA.

Easy experments before you start messing with the Peltiers!

a7

Thank you for your response sir
My project focuses on heat energy harvesting. That's the reason I have opted using peltier plates and cannot simply use a energy source