# Guess power from board?

I ordered a small <1mW laser off of ebay that comes with a small controller board. I tested it briefly with a 9V battery and it works fine, but I don't want to use it extendedly without knowing what the design voltage of the thing is. Spec sheet seems to be unavailable as it appears to be a custom design salved by the eBayers off of a larger end products.

Is there any way to guess from this picture what the correct voltage is? I'm a software dev dabbling in electronics, so I can waddle my way through such things but I still lack a lot of actual knowledge. Thanks!!

Description: The black and red wires on the left go to the negative and positive poles of the laser unit, the red and brown wires in the center go to the positive and negative poles of the power source, and the white wire on the right shorts the switch connector.

The device marked as "U3" is a LM2940 voltage regulator which, without a schematic or seeing the back of the board, I am going to assume is connected to the input. The input voltage range of the LM2940 is 6-26V but is further bounded by two other factors.

On the low side, the LM2940 has a dropout voltage, which is the difference between the input voltage and the regulated output of 5V. This dropout voltage is maximum 1V at an output current of 1A, and somewhat less at lower currents. This means that a minimum input voltage of 6V is required to maintain the output voltage under all load conditions.

On the high side, linear regulators like the LM2940 work by dissipating the extra energy between the input and output voltages as heat. That little metal tab with the hole in it? It's for attaching a heatsink to get rid of the excess heat. At the maximum output current of 1A, the heat produced is 1W for every volt the input is above 5V. The datasheet mentions a junction temperature rise of 23C for every watt produced, but note 2 implies that there is a heatsink attached. I think a rise of 65C/W, the thermal resistance with no heatsink attached, is probably closer to the mark. This can quickly become a problem since the maximum junction temperature is 125C, which would mean that you would burn up the regulator at an input voltage of 7V. You are probably drawing far less than 1A, so you can get away with no heatsink but eBay is not known for having well-designed electronic parts.

Finally, those yellow blobs are tantalum capacitors and will probably be the limiting factor since voltage costs \$. 10 or 16V are common voltages, but the exact value will be printed on the side. There's only one you need to worry about - the input capacitor - since everything downstream of the regulator only sees the 5V output.

• That was some impressive spec reading. Thank you very much!!! Accepted as definitive especially for warning about lack of heatsink. C3 Capacitor says: 4.7 16 +- So I assume this means the capacitor is ok with 4.7 to 16 volts
– cmc
Feb 21, 2022 at 21:30
• The back of the board indicates that indeed the regulator is connected to input, and the C3 capacitor is upstream from the regulator while the others are downstream.
– cmc
Feb 21, 2022 at 21:33
• 4.7 is the capacitance in microfarads. 16 is the maximum working voltage so in theory you're fine in the range of 6-16V though 16V is not recommend since if the input isn't well regulated you'll exceed the maximum voltage and also for the thermal reasons mentioned. A very general rule of thumb is to select component voltages of twice the expected voltage but for some things (like tantalum capacitors) that have large, expensive steps between voltage ratings, this isn't always possible. E.g. the price difference between 16V and 25V 4.7uF capacitors from the same manufacturer is ~20%.
– vir
Feb 21, 2022 at 21:45

Looks like the chip on the right hand side is LM2940 linear regulator, so literally from the datasheet "Input Voltage Range = 6 V to 26 V", there is a low risk that the capacitors, especially C3 limit it. Can you read what's on them?

It looks to be a 5 volt regulator, so the lower limit is 6 volts. With longer cables to the board you might experience voltage drop which you need to compensate for. 9 volt power supplies are widely available and those should fit your purpose. Linear regulator power dissipation is proportional to input voltage and current drawn, so any excess voltage will result in more thermal losses. Witout a heat sink connected the TO-220 package the device will heat a lot. Even with a heat sink it will heat up 23 C for every watt dissipated, so with a 9 volt battery and 200 mA current draw (guess) the regulator would heat to 20 C above ambient temperature.

(9 - 5) V * 0.2 A * 23 C/W = 18.4 C

You might have more accurate current specification available.

The circuit on the left hand side is a famous "555 timer" that's used to provide pulsed power to the diode to ensure it's driven with high enough current, but keeping the average current moderate, so it doesn't overheat.

• Thank you for the great example formula and extra board explanation!! Learned a lot right there.
– cmc
Feb 21, 2022 at 21:18

That TO-220 device with three legs is a low-dropout regulator (LDO). It is marked LM2940CT-5.0. The data sheet for that part number says that it outputs 5V (as the part number suggests) and suggest that it is designed for supply voltages that can be significantly higher than the 9V you tried. If i read it correctly the dropout is 0.5-1 V so the supply probably shouldn't be less than 6V.

Common sense suggests you consider the power dissipation in the regulator especially since it appears to have no heatsink.

• The datasheet states that under 6 volts is not acceptable, even if the dropout voltage would allow it. So pretty much what you said, but without the probably :) Feb 21, 2022 at 20:38
• Thanks especially for pointing out the heatsink issue!
– cmc
Feb 21, 2022 at 21:31

Based on the voltage ratings on the capacitors and the type of regulator used, and the power of the laser diode - this is a board that runs on nominal 9V or 12V. It’ll probably run fine from 6V to 12V. Anything over 12V is asking for trouble from the input voltage tantalum capacitor. Those need to be derated due to inrush current and the inductance of the power supply connection. Feeding that board anything over 12V is not recommended.