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FrozenGate by Avery

DRIVER SCREW UP EL HELP-O PLEASE-O

look at my reply #8

and

Which scale did you use on your DMM? 10A or the mA one?

If you broke the load connection to insert the DMM, the driver output would have swung high and you would have offloaded all that energy into the LD.

Assuming you cut power before you inserted the DMM using the mA scale, you just inserted a couple of ohms of resistance in series with your load - what happens then? If your battery overhead is great enough the driver will stay in regulation, if not, it will drop out and the current will fall. The 10A scale will have the least (notice I did not say zero) effect on the circuit.

You mentioned that you pur the DMM (while on current mode) across (parallel to) the LD. If you were in 10A scale, you just shorted out the LD.

As you can see, there were lots of times the train could have left the tracks. Don't get too torqued over this - anyone who has experimented with electronics has "released the magic smoke". These are invaluable learning events.
 





Gazoo said:
Do it like this:

Positive wire-------------------meter probe  meter probe---------------------------------------Positive wire.

Got it.. ;D

You can do the same thing on the negative side and get the same results. Make sure your meter is hooked up and set correctly.


ummm thats what ive been doing, its called DMM in series... how do I hook it up on battery side. My whole post was saying that when hooked up to DMM in series with cicuit, it doesnt show 250mA... WHY? id like to learn how to do battery side, or what it means, because I killed my diode my attaching in series with positive wire leading to LD, disconnected and connected for a split second, and BOOM, instant LED...

thanks,

amk
 
chimo said:
look at my reply #8

and

Which scale did you use on your DMM?  10A or the mA one?

If you broke the load connection to insert the DMM, the driver output would have swung high and you would have offloaded all that energy into the LD.

Assuming you cut power before you inserted the DMM using the mA scale, you just inserted a couple of ohms of resistance in series with your load - what happens then?  If your battery overhead is great enough the driver will stay in regulation, if not, it will drop out and the current will fall.  The 10A scale will have the least (notice I did not say zero) effect on the circuit.

You mentioned that you pur the DMM (while on current mode) across (parallel to) the LD.  If you were in 10A scale, you just shorted out the LD.

As you can see, there were lots of times the train could have left the tracks.  Don't get too torqued over this - anyone who has experimented with electronics has "released the magic smoke".  These are invaluable learning events.

i did not short the LD, this is with motor load. i would never short the LD. I killed it in another way. (as stated above). on 10A scale, i read 180mA or so, not 250... my question is: HOW do I calculate the current being dropped cuz of my DMM, and how do I calculate that out to find REAL current?
 
amkdeath said:
i did not short the LD, this is with motor load. i would never short the LD. I killed it in another way. (as stated above). on 10A scale, i read 180mA or so, not 250... my question is: HOW do I calculate the current being dropped cuz of my DMM, and how do I calculate that out to find REAL current?

Just measure the voltage drop across a known resistance in the circuit. That would be the sense resistor. Then calculate the current using Ohm's Law. A negligible amount of current actually flows into the Adj pin (~50uA) - almost all of it goes to the load.

It doesn't get much easier than that. You don't even have to disconnect anything. The V/mV ranges on DMMs usually have very high impedences (usually in the meg-ohm range) so they hardly affect the circuit at all.

If you insist on using the current range, use it on the battery side of the driver on the 10A range.
 
okay so, i put the DMM on 200m in the DC amp range, while probe is in 10Amax port, and did this:


|---====(10ohms)===---|
----DMM---|----====(10ohms)===---|----DMM

these were the two resistors in parallel that are connected to pot and LM and ETC. I got 025, and sometimes 026. does this mean .25A, aka 250mA? thanks,

amk
 
You are supposed to be measuring the voltage across the resistors..You said you had your meter set to the 10 amp scale when you took the measurement. Set your meter to measure voltage and let us know what the voltage is across the resistors. Make sure your test leads are in the right ports on the meter.
 
my V/mA/ohm port has high resistance as stated, and shows 1.0 volts. thats 100mA i believe? not true, under mA setting I also get lower mA's thatn on 10A port.
 
If you read 1 volt across the resistors then the input voltage is too low. The voltage between the adjust and the output of the regulator should always be 1.25 volts.

1.25 divided by 5 ohms = 250ma's. Since you are measuring 1 volt it equals 200 ma's. What is your input voltage???
 
Gazoo said:
If you read 1 volt across the resistors then the input voltage is too low. The voltage between the adjust and the output of the regulator should always be 1.25 volts.

1.25 divided by 5 ohms = 250ma's.  Since you are measuring 1 volt it equals 200 ma's. What is your input voltage???


ADJ and OUT are 1.25 resistors are 1.15-1.19 or so.
 
a_pyro_is said:
[quote author=amkdeath link=1196824307/15#24 date=1196913461]ADJ and OUT are 1.25  resistors are 1.15-1.19 or so.

:-? come again?[/quote]

Although not much is written, I would think that he has a pot in series with the resistors. It's hard to help someone when they are not forthcoming with details.
 
amkdeath said:
what be wrond wit driver?::

When DMM in series with motor load reads: 187 or so mA MAX (this is as high as I can go)
When DMM in parallel with load reads: 250 or so mA, can go to about an amp
when DMM attatched directly to leads from driver: Exactly like when parallel with load

When cricuit output leads are attached with a 10 ohm ressitor in parallel with DMM I read .2 volts

What is going on!!?!?!?

regards,

amk

So, you are substituting the LD load with another load (a motor).

Are you familiar with how the linear regulator works? - In particular the voltage overhead *required* by the regulator in order for it to maintain regulation? If not, you should read up on it, because I suspect that is why you got the results you did.

It looks like the LM317 dropped out of regulation because the voltage across the motor at 250mA was too high. When you shorted the output (DMM in 10A scale across the motor) the LM317 could re-enter regulation because the output voltage dropped (considerably) - if you kept measuring for enough time the LM317 would probably have shut down because it would have had to dissipate too much power [(Vbatt-1.25)*0.25 -> for a 6V battery that would have been 1.2Watts, for a 9V battery, 1.9Watts]. These are *dynamic* circuits - you cannot change one element and expect everthing else to remain the same.

This is important, remember it: When you have your meter in 10A mode (using the proper DMM sockets), the leads are essentially a short circuit. Many people have blown expensive fuses in their DMMs figuring this out. For yours, a replacement fuse may be as expensive as your DMM.
 
I am running on 4 AAA batteries, which, as i now remember, sag VERY fast. I shud make the upgrade.

are there any AAA size batteries with like 3 or more volts output? I know a 14500 is AA sized, but what about AAA sized? thanks

and yes I have a pot in there. its 100 ohm cement something trimpot.

regards,

amk

I guess with the 220-230 mA I have I wont fry the diode. Thatl do untill I pimp it out lol.
 
a_pyro_is said:
10440s are the AAA size Li-ION cells. but they aren't protected so be careful.


i no have to worryings about not protectedings if I have driver rightings?

thanksings,

amkings

w00t fonejacker!
 
He meant the BATTERIES are not protected cells. Take my previous recommendation and use 6 nimh or alkaline batteries...problem solved. Nimh will be much better.
 





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