test load without a 1 ohm resistor?

bbshamsa · Sep 12, 2010

Hi guys.

After searching all morning, I couldn't figure this out.

Can I get away with making an effective test-load for red (4x1N4001's) and blue (6x1N4001's) diodes, without using a 1 ohm resistor?

Is there anything else I could use instead, to serve the same purpose?

This is my first ever move into anything electrical, let alone circuits, and I'm trying to build my first driver (DDL).

I haven't got a 1 ohm resistor to hand, and am eager to see if I've built my driver properly, before I plug in my laser diode.

I will order some anyway, but would love to be able to confirm my first build now

Thanks, and atb.

HIMNL9 · Sep 12, 2010

Well, you need a resistor, for measure the current, and the 1 ohm value is the better choice cause you measure directly the current as voltage dropout on the resistor (with 1 ohm, is 1 mA for each mV measured, no calculations needed)

You can use different values, but this require that you calculate the current from the voltage measured, after.

Anyway, if you cannot find 1 ohm 2 or 3 W resistors, you can always take 10 resistors with 10 ohm value. and place all them in parallel, this give you an 1 ohm 2,5W resistor.

Or ask Jerry (lasersbee) to send you some of them, paying him the shipping, in this thread

bbshamsa · Sep 12, 2010

I'd considered 10 x 10 ohm in parallel, my beginners guide to electronics taught me that at least

, but I don't have ten of those either.

Thanks for pointing out the reason for the resistor.

I'll read up more on that while I wait for the ones I order to arrive.

Toaster · Sep 19, 2010

You could try different Ohm but you would need to convert the numbers to get the mA .

Toke · Sep 19, 2010

You could also do without if you have a meter that can handle mA DC.

Some ready made driver die if run without a load, but if you are building something with a LM317 it is not a problem to connect the load through your meter with the inherent risk of loose connection.

You would likely want to measure both voltages in the circuit as well as the current, that requires moving the meter around.
As in powering up, measuring the voltage across the load, disconnecting one wire from the load, placing meter pin where the wire was, hold the other to the wire.
(Remember to change settings and maybe wiring on the meter)

Warning: The mA setting on a multimeter is usually a different hole for the red wire.
That hole come with a fuse to protect you, as the mA setting is basically a short circuit that measure how much current goes through.

It you e.g. try to measure the mA in a wall socket you will get sparks, and depending on quality your meter may blow up. A good one will just blow it's very expensive mA fuse.

HIMNL9 · Sep 19, 2010

Toke said:
..... A good one will just blow it's very expensive mA fuse.

YIKES

..... 10$ each ? :wtf:

This is another example about how companies (like Fluke, in this case, but is not the only one) takes advantage from the ignorance of the most part of the peoples (and no, is not an offense ..... ignorance as "ignoring a fact", not stupidity)

Those fuses are nothing other than common 5x20 glass fuses, that 90% of the multimeters uses as safety internal fuse (sometimes small meters uses 5x15 ones) ..... and these fuses are normally priced like 1$ for each 10 pcs pack ..... but they sell you them at those absurd prices, only cause they placed a tag on them that say "specially made for DMM" ? ..... ROTFL ! :banned:

Toke · Sep 19, 2010

Not quite, the Fluke fuses are a special type. Called something like busbar.
(Yes, $10 is still a ripoff)

The 5 x 20 mm are used is some meters, and I have a box sortiment of them.

Cyparagon · Sep 19, 2010

There's no need for $300 equipment and $10 fuses. A $10 multimeter that uses 5¢ fuses will work just as well for this application.

Toke · Sep 19, 2010

Cyparagon said:
There's no need for $300 equipment and $10 fuses. A $10 multimeter that uses 5¢ fuses will work just as well for this application.

You are absolutely right, I am trying to explain how to avoid blowing a $10 fuse.

bbshamsa · Sep 19, 2010

Thanks for all the help guys.

It's more useful than you probably realise.

I've just started out in electronics, and all the books I read seem to assume I know a lot already; wrong books for me obviously.

Anyway, I just learned first-hand about the multimeter and fuses as I just popped my first fuse when I got curious and tried to measure the current directly from a 9V battery.

I guess I'm not supposed to do that.

I looked at the .5A glass body fuse, and yes there's a nice new gap in the middle of the wire.

More on the way though, so happy days.

I got the DDL driver working, btw, I couldn't wait to build the test load properly, and it works a treat!

Now I want to put it all on a small PCB and then move onto a 445nm DIY driver.

Toke · Sep 19, 2010

I want to write a lengthy post about how professional multimeters are proof against the mistakes only professionals can make.
It involves a video from trade school with a busbar, multimeter and dummy that ended up burning.

In short, as an electrician I get access to switchboards carrying several KA, and my tools should therefore be proof against idiotic mistakes that could kill me.

An amateur can at most screw up with the house fusebox.

A $10 Fluke fuse can break a really big short circuit, a regular 2 x 20 mm glass fuse cannot.

AMooMooCow · Mar 26, 2013

Good thread. I just put together a laser and it seemed to run horribly underpowered. So I tested my 1 ohm resistor (made in Mexico) at 1.6 ohms. So Instead of running the test load at 120mv, I've gotta get it to read about 192mv.

Cyparagon · Mar 26, 2013

I think you forgot so subtract your probe lead resistance from that reading.

AMooMooCow · Mar 26, 2013

Forgive me, but I'm a total noob. Do I just test resistance of the probes on my DMM by bridging the probes together?
BTW, thank you so much. I might have totally fried my PHR805s

Cyparagon · Mar 26, 2013

Correct. Even the crappiest of resistors will be +/- 20%. 1.6 isn't going to happen.

AMooMooCow · Mar 27, 2013

That is a crucial detail that I never even heard about, though I read through countless threads to make sure I was doing it right. I +1'd you

test load without a 1 ohm resistor?

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