Please help I need to understand Dummy Test Load.

[jatsingh]

Hi everyone. Hope all is good.

I've been reading nearly every page I can find on dummy test loads/diode test load.

I think I have a little understanding about it, but I still feel that im unclear on some parts of it.

I usually just buy ready set drivers, but this time I felt as though I need to learn more about various other drivers so I ordered an adjustable diode driver.

I need help setting it up because I don't have any confidence in my newly discovered calculations.

I've ordered a 3A Selectable Test Load

I also ordered a BlitzBuckV4

I want to set the Driver to 1.8A to use with a Single 18650 battery.

I've read posts by many members on LPF, some members have quoted that 5 - 6 diodes are a good setting while others have said its 4 diodes.
I recently read another post where a member was selling these test load kits. BLORD said we have to minus the voltage drop of the 1Ohm resister (which is 1.8) from 4.5V which gives us 2.7V which would be around 4 diodes. Here is a link to the post that I read.
http://laserpointerforums.com/f39/fs-selectable-test-load-up-3a-assembled-diy-kit-stock-63450-7.html

I am not too clear on how I can set this up. I've got a DMM. If im wrong please correct me and help answer this.

Do I need to
Set the jumper to 4 Diodes (0.9 x 4 = 3.6) V
Plus 1V drop from resister (3.6 + 1 = 4.6) V
Measure the pins around the resister and set my DMM to Volts (2000)

Then the read out on DMM gives me mV = mA?

If this correct can someone please go into depth of why? Please if you can, give examples of other setups so everyone who reads this in the future can visualise the setup and make their own calculations.

I appreciate any responses I get. I'd like to thank you all for spending the time to read this.
If I have misquoted anyone I apologise.

 

[ARG]

That will work. The voltage of the diodes isn't particularly important. Within a voltish will do, I've even set some drivers just by shorting the output across an ammeter. Most will not be that picky.

mV does equal mA in your case. The reason for this is ohms law, V = IR and R =1 therefore V = I*1 = I

 

[jatsingh]

ARG, thank you for your reply. I will try this as soon as my delivery is here. I'll upload my new build as soon as its complete. I wasn't too sure if I was right. Thanks to you I now know that I was kinda right. Thank you once again.

 

[Cyparagon]

I think you forgot the fundamental requirement of buck drivers: the load voltage needs to be lower than the supply voltage.

 

[jatsingh]

Hi Cyparagon, thanks for your reply. I don't understand what you mean. Could you please explain it in detail?
How would you go about setting this up using a single 18650?
I appreciate any help I can get. Thank you.

 

[Bionic-Badger]

He means that a buck driver's output voltage (across the load) must be less than the input voltage. In your case, the voltage at the output V_out = ~4.6V, and input (supply) voltage from your battery maxes out at about V_in = 4.2V fully charged.

V_out > V_in ==> Your buck driver doesn't work.

You'll need a boost driver to make the above situation work, or add another battery in series to make it so V_in > V_out.

 

[jatsingh]

Bionic-Badger, thank you for your response. If I add 2x 16340 cells to power my buck driver will that be ok? I'm sure you already know the voltage but for everyone else who will read this in the future that's 2x 3.7v cells (16340)? Will that work better? If so, what would be the correct jumper setting? The voltage would go up to 7.4v on fully charged Batteries.

 

[Bionic-Badger]

Those LiPo batteries, like your 16340s (assuming they're rechargeable), are about 4.2V when fully charged, and about 3.6V when depleted (don't let them drain less than that or they may be damaged). Your setup should assume that the input voltage of each LiPo battery is 4.2V, not 3.7V.

Your dummy load doesn't affect the batteries. The batteries are your input whereas your dummy load and its jumpers are for your output. Your dummy load should reflect roughly the voltage drop of the laser you'll be powering.

A 445nm laser will drop about 5V. What jumper do you choose to make the dummy load produce a ~5V drop? Let's say you want to use that 1.6Amps. That one-ohm resistor will produce 1.6V at 1.6A. The remaining 3.4V needs to be produced by the diodes of the dummy load. Each diode will drop about 0.7-1V. It'll depend on their specs and how much current is going through them (look it up on the diode's datasheet). Assuming it's about 1V per diode, you need about 3-4x diodes in addition to your resistor to make about a 4.6-5.6V drop. 4x will work for 5.6V. It doesn't make a huge difference to mimic the exact voltage of the laser diode with your dummy load anyway.

So there you go. Set the jumper to 4x diodes.

What about your batteries? The driver will use whatever power it needs from the input to power your output. As a buck driver, the input voltage must always be greater than the output voltage. Opposite for boost drivers. Boost+buck drivers don't care.

 

[jatsingh]

Bionic-Badger. Thank you for spending the time to answer my questions. I've managed to completely understand what I need to do. Thank you.

Rep +1 for you. For helping me understand this.

Everyone who has responded to this post thank you all.

 

[Bionic-Badger]

No prob, good luck with your laser building!