A thermoelectric (peltier) generator turns heat flow into DC electrical power. Here is a typical schematic:

traditional thermoelectric

(image from here). Notice that there are two thermoelectric materials required: p-type (red) and n-type (yellow).

Why can't you make a thermoelectric generator with just one material (say, n-type), with a wire to return the current? As follows:

One-element generator

If the wire is long and thin, ~100% of the heat will travel through the thermoelectric, not the wire. It seems to me that this is simpler, equally efficient, and has the advantage that you only need to find one good thermoelectric material instead of two.

Since nobody does this, I figure there must be some problem with this design. What is it?


3 Answers 3


You don't.

However, if you use a wire to return the current, your efficiency will drop. The efficiency of a thermometric generator increases with:

  • the difference in Seebeck coefficient between your two element types (Note you could even use two N types, or two P types as long as their magnitude is different)
  • the thermal resistance, or insulation value between the hot side and cold side.
  • the electrical conductivity of the circuit
  • the temperature difference

Using metal for one of the elements would reduce the difference in the Seebeck coefficient, as well as the thermal resistance, though it would increase the electrical conductivity. Overall it usually reduces the efficiency.

However, it some instances it is much easier to manufacture devices with only one type, for example, printing on flexible substrates. In these cases metals are used for their excellent electrical conductivity.


When a material is heated at one end, the atoms at the 'hot' end have more kinetic energy than the atoms at the 'cold' end.

For an n-type material, free electrons are able to diffuse from the hot side to the cold side, thereby generating a potential difference (voltage) across the material which is proportional to the temperature difference across the device.

enter image description here

A similar thing happens to an p-type material, however the signs of the charge carriers and hence voltages are opposite.

The charge carriers in both materials, in effect, carry heat with them from the hot end to the cold end, whilst setting up a voltage at the same time.

If you were to connect an external circuit across the n-type material, heat will be conducted away from the hot junction by the wire itself, so no electricity will flow.

To create a thermoelectric generator, what is needed is to connect both the n-type and p-type material 'back-to-back' with the hot junction, so their voltages are added together, whilst the wires to the external circuit are both connected to terminals at the cold junction, thereby allowing electricity to flow through the external circuit. As long as there is a temperature gradient across the device, there will be a voltage given out across the terminals at the cold junction. Of course, if an external circuit is connected and an electric current flows, heat will need to be added to the hot junction to maintain the flow of electric current.

enter image description here

In practice, there may be hundreds of such junctions connected (electrically in series, thermally in parallel) to generate a 'useable' quantity of electricity.

enter image description here

  • $\begingroup$ You say "heat will be conducted away from the hot junction by the wire itself, so no electricity will flow." (I assume you mean "electrical current" not "electricity".) I don't understand how you reach that conclusion. Some heat will surely flow through the wire, but not all of it, indeed probably almost none of it. A long thin copper wire has a very low thermal conductance. In the design I drew, it is entirely possible to have 99% of the heat flow through the thermoelectric material and 1% flow through the wire. Then why should heat flow through the wire make any difference at all? $\endgroup$ Dec 4, 2014 at 0:30
  • $\begingroup$ Metals conduct electricity and heat better than semiconductors. For thermoelectric generators in particular, it is important to use semiconductors with low thermal conductivities to maintain a temperature gradient. A single semiconductor 'pellet' only generates a tiny voltage (~60mV) so a 'long thin wire' wont draw much current unfortunately. To increase the voltage requires the 'pellets' to be connected electrically in series. Using just n-type material would mean connecting hot side of one to the cold side of the next (thermal 'short') which compromises performance. $\endgroup$
    – theo
    Dec 4, 2014 at 0:51
  • $\begingroup$ Copper has 400X larger thermal conductivity than BiTe. So if you compare the heat conductance of a 1-meter-long, 1mm-diameter copper wire with the heat conductance of a 10cm^2, 1mm-thick BiTe, the BiTe will conduct more heat than the copper wire ... by a factor of 2000. $\endgroup$ Dec 4, 2014 at 0:57
  • $\begingroup$ I don't see any reason that ohmic losses are an inevitibly large problem in the design I drew, if you make reasonable choices about geometry etc. I agree that a large voltage is more practically useful in the final output, but it's not a big deal, you can always put a step-up voltage converter at the device output. $\endgroup$ Dec 4, 2014 at 1:09
  • 2
    $\begingroup$ Current will still flow, it's just that there will be no added voltage, so you'd need twice* as many elements to get the same voltage. Along with the increased thermal conductivity efficiency is generally reduced. (* as n-type materials generally have better Seebeck coefficients they actually usually produce more than half the voltage so you wouldn't need quite twice as many elements if you replaced the p-type elements with metal) $\endgroup$
    – Rick
    Dec 8, 2014 at 17:18

Current flows only when there is difference in electric potential or commonly reffered to as voltage, First the second part of your Question, about thickness of wire, if wire is too thin then according to the following relation [(R=ρxl/A), where R is the resistance of wire, ρ is the resistivity of the wire, l is the length of the wire, and A is the cross-sectional area of the wire.] if cross-sectional area is reduced the resistance increases a lot, which hinders the current flow.

Now the first part of your Question, Why 2 different Semiconductors are used. To make the current flow, there needs to be some non-uniformity in the distribution of electrons to cause a potential difference(which is very essential for current to flow),which is provided by two differently doped silicon or germanium substrates, where n-type is at a relative higher potential and p-type is at a relative lower potential. After these 2 semiconductors are bought into contact, the electrons begin to flow intrinsically and after few microseconds a potential barrier is formed which prevents further flow of electrons. To cross this potential barrier we require biasing voltage which depends on semiconductor being used and the component being made(e.g, for silicon diodes it is nearly 0.7 Volts and for Germanium diodes it is 0.3 Volts, while as for Transistors it is even lower.). This biasing Voltage can be attained by many ways, by using electric field, by using photons in photo-voltaics, or by using heat as an energy for electrons in n-region to allow them to move in an external circuit and then cross the potential barrier. Which is actually the flow of current.

If you would use just n-type or p-type, current will still flow but ntype is what actually holds surplus electrons, which aid in conduction and similary ptype helps electrons flow faster increasing the rate of flow of charge or simply increasing current.

Here current I refer to as recognisable current above 20mA, otherwise the electrons are enoughly motile at room temperature in any of the semiconductors whether doped or not to move in an external circuit to make current happen, but that would be very very small, and that's currently of no use in my opinion.


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