Building Test Load - Help

Hey everyone,

I just went down to my local Jaycar to buy resistor and diodes to build my test load for my reddie I’m building.

I’ve purchased:
10 x 1R0 5% 0.25w Resistor (Just incase 0.25 isnt enough)
10 x 1N4004 Diode – 400V 1A
Copper chip board

Will this stuff work? Will I need to do more or less diodes/resistors?

Please help me out 

Thanks guys! (they didn’t have 1N4001 diodes)


Tom

Yes, that should do fine. put 3 groups of 3 parallel resistors in series and then attach that to 4 to 6 diodes in series and you'll be ready to rock.

I don't understand. I'm very new to building something like this. Can you show me what you mean by group of 3 in paralell? Image? Paint? Something...

Tom

Thanks drlava.

You take 3 resistors and connect them end to end (like train box cars)...
You do that 3 times to get 3 the same as the first one (that will give you 3 Ohms ea)

Now you take one end of all 3 "trains" and connect them together...
Then take the other end of all 3 "trains" and connect them together...

That will give you 9 resistors connected in Series/Parallel (1.0 Ohms
but higher wattage)

(A)-----[1ohm]-----+-----[1ohm]-----+-----[1ohm]-------(B)
(A)-----[1ohm]-----+-----[1ohm]-----+-----[1ohm]-------(B)
(A)-----[1ohm]-----+-----[1ohm]-----+-----[1ohm]-------(B)

Connect like this...
All (A) are connected together
All (B) are connected together

Between (A) and (B) you will get aprox 1.0 Ohms

[EDIT]
Thanks HIMNL9.... didn't have my coffee yet... It is corrected

Jerry

i didnt really understand that either.. lets do this in steps

his trying to say is use three groups of three parrallel resistors.. the picture below shows a parrallel circuit!



next he said to put them in series.. so the group of 3 resistors in parrellel are then placed in series.. like this



Sorry for the bodgey pics.. but hope you get the idea..

if not check out this link.. http://laserpointerforums.com/f38/my-phr-dummy-load-43327.html

but remember.... that PHR and red diodes need differnet set ups!

EDIT:

nevermind Lasersbee got it covered! A++ explination!

hope that helped -Adrian

Laser - are the "+" solder points? Joints between them?

And did you see that my resistors are 0.25w.

And where do the diodes go??

Tom

lasersbee said:

You take 3 resistors and connect them end to end (like train box cars)...
You do that 3 times to get 3 the same as the first one (that will give you 30 Ohms ea)

Now you take one end of all 3 "trains" and connect them together...
Then take the oter end of all 3 "trains" and connect them together...

That will give you 9 resistors connected in Series/Parallel (33.3 Ohms
but higher wattage)

(A)-----[1ohm]-----+-----[1ohm]-----+-----[1ohm]-------(B)
(A)-----[1ohm]-----+-----[1ohm]-----+-----[1ohm]-------(B)
(A)-----[1ohm]-----+-----[1ohm]-----+-----[1ohm]-------(B)

Connect like this...
All (A) are connected together
All (B) are connected together

Between (A) and (B) you will get aprox 33.3 Ohms


Jerry

Click to expand...

I'm sorry i have to correct you (and i think you probably just typed it wrong)

The 3 resistors of 1 ohm in serie, give you 3 ohm each one, and if you connect these series in parallel, you get again 1 ohm ..... but now the assembly can hold 2,25W max ..... you can also connect the "middle" points of each serie all together, the thing don't change, like this

(A)-----[1ohm]-----(x)-----[1ohm]----------[1ohm]-------(B)
(A)-----[1ohm]-----(x)-----[1ohm]----------[1ohm]-------(B)
(A)-----[1ohm]-----(x)-----[1ohm]----------[1ohm]-------(B)

All (A) are connected together
All (B) are connected together
All (x) are connected together
All are connected together

measuring from A and B, still 1 ohm (not 33,3) ..... anyway, the only advantage of solder also these points in common, is just a little bit more mechanical stability if you build it "flying" ..... using a proto board, no advantages


@tommii89: like this:



Is not important if you connect the resistor (or group of resistors), on the positive or on the negative output of the module, the important part is that you do not revert the diodes, otherwise the load don't work.

All the diodes (4 for simulate red diodes, 6 or 7 for simulate BR diodes) must be in the same "direction", anode-cathode connection, with the anode that must go at the positive output, and the cathode at the negative one.

Im using a proto board, but my resistors at 1/4 (0.25) WATT - Can you confirm this is still what I need to do???

?? I'm so confused..

Is this for a red diode? I am making a red laser.

Tom

Wait a minute, i make a pic for your setup

Edit: you can connect them in this way:



or in this way:



The result is always the same, the assembly of the resistors always give you 1 ohm 2,25W at the sides, and you just need to take your measure at the sides of the resistors assembly ..... the only important thing, as i've said, is to not revert any of the diodes, and to connect the load always in the way that the cathode of the diodes are connected on the negative output of the module, and the anode at the positive one.

The draw shows you the load set up for simulate a red laser diode, with 4 x 1N diodes ..... For simulate the blu-ray ones, you just need to use 6 (or 7, as you prefer) diodes, instead 4, in this way:

Thank you!!!!

Tom

Thanks HIMNL9.... It was too early...


@tommii89.... why didn't you get some 1 Ohm 1/2 Watt resistors here....

http://laserpointerforums.com/f39/free-1-ohm-test-load-resistors-42409.html


Jerry

They didn't sell them, and I dont want to wait 7-10 days for the delivery lol.

I have 8 1/4 watt diodes, can someone help me with THIS type of diode????

Tom

We need a part Number or a picture near a ruler...

Diodes are not rated at watts..
Diodes are basically rated at Amperage and Voltage...

BTW...
it would have only require one (1) of the resistors I offered not nine (9)....
Too bad you are is such a hurry...

Jerry

Hello


how many diodes are needed to simulate a 808nm diode im using the 1N4004 diodes and a shunt

Thanks

IR diodes are usually from 1,8 to 2,3 (at least, the ones i tried, don't know if there are other types much high), so, the usual 4 diodes are more than enough.

HIDL said:

Hello


how many diodes are needed to simulate a 808nm diode im using the 1N4004 diodes and a shunt

Thanks

Click to expand...

As HIMNL9 posted.... for RED LDs the standard Test Load uses
4 silicone rectifier diodes...
and for Blu-Ray LDs it is 6 silicone rectifier diodes...

BTW.... to the Forum....

You could have easily found this info on the forum if you
would have done some research on your own...

Don't forget to read the Newcomers section and the Forum Rules...
Enjoy the Forum...


Jerry