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I explain my self: we're going to make a photovoltaic cell with severals photodiods in series. Problem: the current isn't really high (I'm in photovoltaic mode, so no external biais source).

I've already looked at the transimpedance amplifier , but the thing is that for me... it just converts the current into voltage (Vout=Rf*I), but it doesn't amplify my input current ....

Are there any other way to amplify the output current ?

I just want to add that these photodiods are used in photovoltaic mode, and that i'm going to load a battery (12V maybe), with the output voltage from the PV cell . The thing is that I don't know if I really need to amplify the current (I have not so much knowledges in this field), but i thought that it would be great to do it...

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  • \$\begingroup\$ using a buffer, a simple op. amp. with a directly connected output to the - input. \$\endgroup\$
    – Jakey
    Commented May 14, 2018 at 14:46
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    \$\begingroup\$ @Jakey, if this is part of a photovoltaic cell, the goal is to extract power from the the photodiode's output. Routing it through an op-amp won't do that. \$\endgroup\$
    – The Photon
    Commented May 14, 2018 at 14:51
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    \$\begingroup\$ Why arrange the diodes in series? What do you expect to happen if they aren't illuminated equally (for example, when one part of the surface gets dirty)? \$\endgroup\$
    – The Photon
    Commented May 14, 2018 at 14:53
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    \$\begingroup\$ Why do you want to amplify the output current? You could just use a push-pull follower, but if you're wanting to get power out of it that won't help. \$\endgroup\$
    – Hearth
    Commented May 14, 2018 at 15:00
  • \$\begingroup\$ I dont know if I understood your question but, if you want to amplify the output from a photodiode without adding external sources you can: a) Choose a better photodiode b) Improve your optical mechanism c) Add more photodiodes d) Reduce the losses \$\endgroup\$
    – Jose
    Commented May 14, 2018 at 15:32

2 Answers 2

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The current capability of a PV cell is proportional to its area. If you need more current, you need bigger cells. Or put multiple cells in parallel.

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  • \$\begingroup\$ Right or more light. (~ every photon makes an e-h pair.) \$\endgroup\$ Commented May 14, 2018 at 16:58
  • \$\begingroup\$ @GeorgeHerold: Yes, within limits. PV cells aren't all that efficient, and any light not converted simply becomes heat. Once the cells get hot enough, the efficiency goes way down. \$\endgroup\$
    – Dave Tweed
    Commented May 14, 2018 at 17:09
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The only way to 'amplify current' is to use an amplifier, and an external power source.

It appears that you want these photodiodes to generate power, to charge your battery, without any other power source.

Unfortunately there is no way to increase the power they generate, from the terminal side at least. You could increase the amount of incident light, or use cells with a larger area.

As your question is specifically about current, you can increase the current, at the expense of voltage, using a switch-mode power converter. However, that will lose power, so is probably not what you want for battery charging.

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