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I have designed my own PCB and it doesn't work correctly.

I needed to have a PCB that has only one push button to switch on and off a laser diode. The circuit does work kind of, but the laser diode only shines a little, it doesn't give the amount of light it supposed to.

Can somebody help me with solve this problem?

I have added the schematic and PCB design as images.

I used a 555 timer to design this PCB and I have already added a 220 ohm resistor to ground and pin 3 of the laser diode.

enter image description here enter image description here enter image description here

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  • \$\begingroup\$ For clarity; It works in general as intended, but the light is not on the right intensity? You say you've already added the 220 ohms between pin 3 and the GND, could you provide an updated schematic? Have you measured the voltage to the LED? \$\endgroup\$ – Capt. Frost Nov 24 '20 at 9:10
  • \$\begingroup\$ Yeah exaclty, the light is at is minimum. But when i connect the laser directly to the battery, it is giving it's maximum light. When i was designing the pcb, i had taken a normal LED as reference (2 pins) but then i found out a laser diode had 3. So pin 1 of the laser is connected to pin 3 of the 555 timer, pin 2 of the laser is connected to ground and pin 3 is connected trough an 220 ohm resistor to ground \$\endgroup\$ – leon bouman Nov 24 '20 at 9:18
  • \$\begingroup\$ Did you measure the Timer's output voltage, before and after R2 (400 ohms resistor)? From what I understand is that you use PD anode and LD cathode (you wired the output of the timer to pin 1 of the laser) different than usual. If I remember correctly you should connect + (positive) to PD anode, - (GND) to LD cathode. Built-in photodiodes in lasers are used to regulate the voltage to the LD. This could be one cause of your problem, as low voltage is supplied to the LD. \$\endgroup\$ – Capt. Frost Nov 24 '20 at 10:05
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    \$\begingroup\$ See CR2025 Battery datasheet, typical rated drain current is only 0.19mA (at 2.9V), and pulsed load is rated 6.8 mA @2.7V, so the laser driver may be too much for a coin cell. \$\endgroup\$ – MarkU Nov 24 '20 at 11:01
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    \$\begingroup\$ You are lucky. If you tested the laser diode across a decent battery you would have killed it. It's designed to be current driven, not voltage driven. Luckily the battery was too weak to kill it (though if it supplied more than 40mA momentarily it may have done some damage) \$\endgroup\$ – Brian Drummond Nov 24 '20 at 12:52
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\$V_f\$ for the laser diode is 2.5V. Battery voltage of 3.0 leaves 0.5 dropped across R2. That's \$ \frac {0.5}{400}= 1.25 mA\$. The laser is rated for about 33mA.

Your circuit is limiting the current to the laser diode too much.

Then there's the current rating of the battery to consider. The CR2025 isn't rated to deliver the current you need, and probably struggles to deliver the current your circuit consumes. (Just a couple of milliamperes is a heavy load for the CR2025.)

Now, take a look at the 555.

Its datasheet says the NE555 can operate on 4.5V to 16V, and that it will draw several milliamperes even at low voltage:

enter image description here

enter image description here

  1. You need a higher operating voltage for the 555.
  2. You need a battery capable of delivering the required current.
  3. You need to re-calculate R2, taking into account the voltage at the output pin of the 555 (pin 3) and the current the laser diode needs (at least 24 mA, but no more than 40 mA.)
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    \$\begingroup\$ Note: you'll need a different battery anyway, but using a ca 10 mA-swallowing NE555 with a 4.5 V minimum voltage is madness, anyway, in a battery-powered application. There's CMOS derivatives of the 555 that use far less power, and can work at far lower voltages. \$\endgroup\$ – Marcus Müller Nov 24 '20 at 12:00

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