I'm doing a project that basically transfer the power from a laser and light up an led using it.

The power output we are getting from the laser is at most 0.45V and 0.6mA (we used photodiode to receive the power.)

We used a buck booster that requires 0.7V and 3mA to make an led glow. As we are only allowed to use BJTs that's the best buck booster circuit we can make.

So my problem how I boost 0.45V and 0.6mA to 0.7V and 3mA so that I can feed that into the buck booster?

If you have any other solution to transfer power from the laser and make an led glow, please suggest them.


closed as too broad by JRE, Warren Hill, Finbarr, PeterJ, Nick Alexeev May 5 at 3:29

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    \$\begingroup\$ Please specify more clearly. Are you trying to power the laser from a buck boost converter? what are its specs? If you say 'we used' then that implys you already have a buck booster so what is it you need? Please take time in writing a quality question or else you'll get no or poor quality answers. Id also add on a free forum you cant really ask 'reply as quick as possible'.... as long as it takes is as quick as possible. A better quality question will get a quicker response! \$\endgroup\$ – TheAndyEngineer Apr 17 at 6:46
  • \$\begingroup\$ IMPORTANT: Are you allowed to use a battery or power supply in your circuit or must the power to operate the LED come from the LASER \$\endgroup\$ – Russell McMahon Apr 17 at 8:39
  • \$\begingroup\$ Your circuit has an input power of 270 uW and output power of 2100 uW. No circuit is able to do that, unless you invent an over unity machine. You need to increase input power, so that requires a power supply. \$\endgroup\$ – Bart Apr 17 at 9:32
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    \$\begingroup\$ If you do not answer questions you will probably not get the answer that you want. \$\endgroup\$ – Russell McMahon Apr 17 at 13:28
  • \$\begingroup\$ Photodiodes are basically teeny solar panels, so I think they would have a maximum power point. If you hook your photodiode up to just a capacitor and no load, what does the voltage get to? If you draw less current, you may be able to get more voltage and more total power. \$\endgroup\$ – K H Apr 18 at 4:38

That's impossible. The total amount of power you're receiving from the laser is 0.27 mW, and your buck/boost requires a minimum of 2.1 mW to start up.

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    \$\begingroup\$ It should be possible to have the LED flash at a 10% duty cycle \$\endgroup\$ – John Dvorak Apr 17 at 7:06
  • \$\begingroup\$ can you please tell me how should I work on that? should I charge a capacitor and discharge it ? \$\endgroup\$ – Stranger_sid Apr 17 at 14:59

The power output we are getting from the laser is at most 0.45V...

...we are only Allowed to use bjts...

Then you can FORGET about this, it is impossible. To make a BJT operate (at room temperature) you will need at least 0.6 V, at 0.45 V a BJT is not going to do anything useful.

You could heat up the BJTs, to lower the voltage they need. The 0.6 V decreases with 2 mV / K so if you heat up the BJTs to above 100 degrees C then maybe it can work at 0.45 V. This doesn't seem like a useful nor practical solution.

There do exist (integrated) circuits which can work at 0.45 V but these all use low threshold MOS transistors. You cannot buy such transistors, they're only available inside ICs.

  • \$\begingroup\$ Can you please suggest me any other idea?? \$\endgroup\$ – Stranger_sid Apr 17 at 7:54
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    \$\begingroup\$ Sure, use 2 photodiodes in series, then you have 0.9 V. Then feed that to a "Joule thief" circuit: en.wikipedia.org/wiki/Joule_thief to make a LED light up. \$\endgroup\$ – Bimpelrekkie Apr 17 at 8:06
  • \$\begingroup\$ How much current does joule theif ckt draws? Does 0.9V should be a constant just like a power source? \$\endgroup\$ – Stranger_sid Apr 17 at 9:28

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