What would be a way to use a Ammeter current meter to track down shorts?

I was thinking since shorts are really "High current" that a Ammeter might isolate shorts better

The theory i would think is keep placing the Ammeter in each section from input to output until you read a really High current, than that would be the faulty defective stage?

SHORTS cheat sheat:
If the current is low in that stage/section than its GOOD
if the current is really high in that stage/section than its BAD
(we isolated the faulty defective stage)

Does this sound right?

Its hard to tell with a volt meter or using volts to find shorts , i was thinking a current meter would be easier to find shorts

How would u guys use a Ammeter current meter to track down shorts or isolate the area where a short might be?

If there is a Short to +VCC, does it make all the stages/sections have really high current?

If there is a short to Ground, it only affects that defective stage/section , so the current is that defective stage would be LOW while the other stages before and after are HIGH?

::Not cut traces to use a Ammeter trick::

1.) I read a trick that if you take a 1 ohm resistor plus use your Ammeter in series ACROSS any trace on a PCB in parallel that it will give you the current reading rating for that trace/path without cutting the trace

is this true or sound right?

 

Okay man, you are really trying to solve this magical shorting problem. If the short is between two nets, then how can you use an ammeter to distinguish where it is?

This concept is the same as using an ohmmeter, although maybe more accurate if using higher currents to find them.

The 1 ohm resistor will give another current path to the circuit, so current will flow through it.

Why not get a hall effect probe and probe around until you find the highest reading??

Steve

 

1.) The 1 ohm resistor will give another current path to the circuit, so current will flow through it.

Yes so then you place the Ammeter current meter in series with it you can measure the current of that path which will be the same current that the trace you wanted to cut

The 1 0hm resistor didn't ADD any resistance it was just used to convert voltage into current so we didn't have to cut the trace to measure the current so we can find the SHORT

 

How is a 1Ω resistor going to offer another path without cutting the trace? I've personally found shorts in a board full of bypass capacitors one of the most challenging problems I've ever fought, so I'll be interested in other ideas.

 

I've used this technique successfully quite a few times. BUT there are risks involved! You might end up putting high currents through places they shouldn't go, so think carefully about any components that are possibly being overloaded.

What I've done is to get a current-limited power supply and use it to drive a current--maybe 1 amp, maybe less, depending on the size of the traces--through the circuit which has the short. Then get a digital volmeter and put it on its most sensitive setting, and measure the voltage, relative to the low side of the power supply, at different points in the circuit. This works best if you have a PC board with no ground and power planes; if these both exist, you may not be able to do it. But the point is, if there is conduction there will be voltage drop. If the suspect circuit has branches, you won't see any change in the voltage in a branch which doesn't have the short, so you can gradually close in on the location of the problem. Likewise, if the voltage changes as you move along a trace, but then stops changing, you've found the place where the error exists.

 

Just use a hall probe, it will give you a fancy diagnosis method, at least until you have a truly differential path being shorted.

Steve

 

what do u mean by a differential path? being shorted.

John P-

I read about this, in the 80's they use to make a current tracer

1.) Turn off the circuit under test
2.) Turn off the voltage to the power supply, but set the current to 1 amp on the power supply
3.) Inject the current source onto the input of the stage
4.) Use a DVM set on volts to measure voltage drops

The confussing part i don't understand is

if there is conduction there will be voltage drop, i understand this

if the voltage changes as you move along a trace, but then stops changing, you've found the place where the error exists.

I don't get that part


If the short is shorted to ground than the voltage drop would be ZERO? using a current source

If the short is shorted to VCC than the voltage drop would be What?

If the short is shorted to VCC to ground than the voltage drop would be what?

 

Differential path example would be when there are two conductors, one with current going in one direction, the other with the same current going in the opposite direction. The flux density through the hall sensor would be close to null.

if the voltage changes as you move along a trace, but then stops changing, you've found the place where the error exists.

Imagine you are traveling across a highly resistive path, your voltage will be increasing with respect to some node if you are traveling away. Then, you move off of this resistive path (highly conductive), you will not experience any more voltage drop that is easily measureable.

All of your questions can be answered with ohm's law and understanding the relative resistances involved in your circuit.

Steve

 

To emphasize scubasteve's point, you have to understand the circuit you are troubleshooting. You are trying to make a checklist of infallible cures, which is not at all realistic. Say a resistor opens up. If you only check on one side of it, you may see 0 volts and think you are chasing a short. That will just waste time.

One of the hardest thing for people to learn is troubleshooting. Most adopt the Mr. Goodwrench technique, and shotgun parts until things are working again. You have to have a fundamental understanding of basic electronics as well as the particular kind of electronic equipment you are working on before you will get proficient at troubleshooting.

 

an ammeters out there that have an external shunt. An "external shunt" is basically a very low value resistor that is in series just as the ammeter was in the previous example, and then the ammeter is in parrallel to the shunt. Basically MOST of the current goes through the shunt and only a small percentage goes through the ammeter. By knowing the resistance of the shunt and the meter, the meter is able to calculate the total current by measuring only the small current that goes through the meter
Compound ammeter = I-to-V converter + voltmeter
Compound ammeter. Today's measuring instruments (DVM's, analog-to-digital converters, etc.) are mainly voltmeters. If there is a need to measure a current, a simple current-to-voltage converter (a shunt resistor) is connected before the voltmeter (Fig. 8). This ammeter is a composed device consisting of two components:
Compound ammeter = Current-to-voltage converter + voltmeter
The shunt resistor of a composed ammeter acts as a current-to-voltage converter.