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In a tritium-powered betavoltaic cell, the power output obviously decreases as the tritium decays. Is it the voltage, amperage, or both that would decay?

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The voltage, in theory, is simply the energy of the betas. These don't change, they are defined by the substance.

The amperage is the number of betas, which is simply a count of the remaining tritium. This, obviously, reduces over time.

So the voltage is largely constant and the amperage falling off with the half-life.

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  • $\begingroup$ The paper cited by anna v shows a betavoltaic device with a structure and theory of operation that is nearly the same as for a photovoltaic device. I'm not an expert, but it looks to me as if the open-circuit Voltage will be determined by the forward voltage of the P-N junction (i.e., by the allowed energy levels of electrons in the semiconductor material.) I think the Voltage will be independent of the energies of the beta particles. $\endgroup$ – Solomon Slow Nov 8 '18 at 16:31
  • $\begingroup$ Also, when you say "Voltage" and "current", you should make it clear that you are talking about open circuit Voltage, and short circuit current. The actual Voltage and current when the device is making power will have different values. $\endgroup$ – Solomon Slow Nov 8 '18 at 16:33
  • $\begingroup$ Sure, but the betas that are ticking that bandgap have to have a certain energy, and that doesn't change. There could be recombination effects and defect buildup that would change V, but I think that's beyond the scope of the question. $\endgroup$ – Maury Markowitz Nov 8 '18 at 17:07
  • $\begingroup$ I'm going to step back from this discussion because I don't know what "ticking that bandgap" means. But that device still looks a lot like a solar cell to me, and maybe I'm wrong, but I don't expect the open circuit voltage of a strongly illuminated solar cell to depend on the energies of the incoming photons. $\endgroup$ – Solomon Slow Nov 8 '18 at 17:29
  • $\begingroup$ If the photons are lower energy than the bandgap, you don't get an excitation. That's why half the energy from the sun is lost, it's in the IR region below the cutoff energy of a typical cell. $\endgroup$ – Maury Markowitz Nov 8 '18 at 18:10

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