SHORTS, voltage across resistor converts to current

[danny davis]

What is the voltage across the resistor and the diode when the switch is open?

+Ve on one end and -Ve on the other end

Yes when there is an Open resistor or Open capacitor , when i measure i get the same potential difference on both sides

But the resistors and capacitors are not open internal, they are working and still measure resistance

I just don't understand how there is the same voltage on both sides and no voltage drop or current running through them

Also the values of the resistors or capacitors have different values , so it doesn't matter has the value it

 

[(*steve*)]

I just don't understand how there is the same voltage on both sides and no voltage drop or current running through them

You mean like in the example above?

 

[danny davis]

I just don't understand how there is the same voltage on both sides and no voltage drop or current running through them

No, I understand when there is an internal open inside a component, that it will have the same potential difference on both sides of it and no current flowing through it because there is a break inside

I don't understand how a normal operating resistor or capacitor that is working can measure good but has the same potential difference on both sides of it , no voltage drop, no current flowing through it.

It acts like there is a break or an internal open inside of the component but there isn't

That's what i don't understand

 

[danny davis]

At work , I take a good PCB board that has passed all the test

1.) I measure the voltage across each resistor and capacitor with my DVM and write those values down
2.) only 5 out of 100 resistors only have voltage drops
3.) The other 95 resistors have no voltage drop or current flowing through them, But the resistors have the same potential difference, with the same voltage on both sides of them

The only thing I can think of is that those resistors with the same potential difference with the same voltage on both sides are either controlled by MOSFETS or digital switch , The resistors seems like they are in a idle state? maybe they are waiting for a High Or Low signal?

 

[(*steve*)]

What it tells me is that if the conditions are the same, and your failure rate is less than 50% (i.e. the lower number reflects incorrect operation) then you need to determine what is causing the voltage drop (as opposed to what is causing no voltage drop on the others)

HOWEVER, if you understand the circuit and know why there is normally no voltage drop, this may give indications of what is going wrong.

Have you looked for solder bridges already?

 

[danny davis]

Its normal that there is no voltage drops on these resistors , this is a good working circuit

I just don't understand why there is no voltage drop across these resistors

Have you been across circuits that have no voltage drops around resistors or capacitors and what are they used for?

 

[danny davis]

applying Kirchhoff's current and voltage Laws to SHORTS

When applying Kirchhoff's current and voltage Laws to SHORTS

If you have a SHORT on a circuit and you're trying to narrow down where the SHORT Is coming from or where the alternative path is

You measure the voltage drop and convert it to current

Since there is a SHORT on the board the current is going to be HALF of what it is suppost to be

So Kirchhoffs Law states that current needs to Equal it's loop

So where is the other Half of the Current going? and how do you find it?

The SHORT should have the other half of the current, but how do u find it?

 

[(*steve*)]

Its normal that there is no voltage drops on these resistors , this is a good working circuit

I just don't understand why there is no voltage drop across these resistors

Have you been across circuits that have no voltage drops around resistors or capacitors and what are they used for?

Yes. As per the example above.

No voltage across the LED or resistor.

 

[(*steve*)]

The same amount of current is entering the short as is leaving it. (KCL)

The sum of voltages in a loop containing a short adds up to zero (KVL)

There is no "other half of the current".

 

[danny davis]

The Tech at my work find shorts this way:
1.) Measures the Currents Total from the power supply
2.) Then measures the current from each resistor that is tied to a Vcc supply voltage in each stage

If the Current total is 100mA out of the power supply, So Kirchhoffs Law states that current needs to Equal it's loop or Current Total

If you measure the current from each branch or resistor that is tied to Vcc then you will know how much current is being Feed to that stage of the circuit

When there is a short, the tech at my work will always reference the 100mA at the total current of the circuit. The tech will measure the current from a Resistor tied to Vcc to know what that stage is being Feed. Mostly It will be microamps because the SHORT is hogging all the current right?

So if the total current is 100mA
You measure The Resistor tied to Vcc of each stage, you will mostly get 10uA or 50uA
he will always say where is there rest of the of the current going?
100mA - 10uA =
100mA - 50uA =

I just have a hard time finding where the Rest of the current is going to lead you to the SHORT or whats SHORTED , How do you find where the Rest of the current is going?

 

[danny davis]

The stages that don't have the SHORT, will have very low current in microamps

The Stage that has the SHORT will have 97% of the Total Current referenced from the power supply Total Current?

The other 3% will be in microamps for the other stages that don't have the short?

 

[davenn]

..............
When there is a short, the tech at my work will always reference the 100mA at the total current of the circuit. The tech will measure the current from a Resistor tied to Vcc to know what that stage is being Feed. Mostly It will be microamps because the SHORT is hogging all the current right??

yes

So if the total current is 100mA
You measure The Resistor tied to Vcc of each stage, you will mostly get 10uA or 50uA
he will always say where is there rest of the of the current going?
100mA - 10uA =
100mA - 50uA =

so you have 100mA - 60uA

I just have a hard time finding where the Rest of the current is going to lead you to the SHORT or whats SHORTED , How do you find where the Rest of the current is going?

you know that the rest of the current is going through the short cct, be that a solder bridge between tracks, a blown component whatever..

so then you start breaking the circuit down into smaller sections and narrow it down to the section that is drawing all the current. Then in that section start looking for blown components etc. Its likely to be really obvious.

Ohhh one thing you didnt say and maybe the tech didnt tell you....

that 100mA, is that the current drawn when the circuit is working normally or when its in fault condition?
if its the normal current drain of the circuit, then in a short circuit or near short circuit fault condition, the current drawn will be much higher, up to whatever the PSU is able to supply to the circuit, assuming it doesnt blow a fuse ( if one exists) first

cheers
Dave

 

[danny davis]

that 100mA, is that the current drawn when the circuit is working normally or when its in fault condition?

Can a short draw more current than the total power supply current?

Because if the batterys are 4.5 volts and puts out 100mA total , how can a short draw more current?

so you have 100mA - 60uA

No, What I mean is that the SHORT has the all the rest of the current

1.) Total power supply current is 100mA output
2.) if you have 5 stages, each stage of the circuit is 10uA when you measure, that's
50uA total
3.) 100mA - 50uA is whatever number result, but that result is the SHORTED current Total right? The short is hogging all the rest of the Current

 

[BobK]

Batteries do not supply a current. They supply a voltage. They also have an internal resistance that limits the current, or actually drops the voltage as the current increases. A battery designed to provide 100ma continuously might be able to provide 10A for a short time.

Bob

 

[danny davis]

An open circuit may cause the total power supply current to go up or down.

Why would an open circuit cause the total power supply current to go up or down?

Does an open circuit cause the Total power supply Voltage to go up or down?

Does a short circuit cause the total power supply voltage to go up or down or stay the same?

Because from my experience , A short circuit only changes the total power supplys current either up or down and the voltage stays the same

When there is a short circuit somewhere on the board, all the voltage drops are in millivolts and the currents are in microamps or nanoamps

When the circuits are in normal operations, the voltage drops are in volts not millivolts and the current is in either milli-amps or microamps

For Open Circuits, There is no current flowing through them. But their is DC Voltage on both sides of the component with the same potential?

For Open Circuits, The current

kirchhoff's never made it clear as to make his current and voltage laws applied to open circuits and short circuits

Because Kirchhoffs KCL states that the total power supply currents will equal whatever the whole circuit is Drawing from it.

But Kirchhoff never explains what happens to his KCL and KVL laws when there is a Open circuit or short circuit

Yes, when you have a open circuit or short circuit somewhere in a circuit or network the KCL and KVL will have to equal the total power supply voltage and current

The KCL and KVL doesn't explain what a open circuit or short circuit does to the current or voltage LOOP because the open or short takes a percentage of the total power supply current.
0.) All the Currents and voltages should Equal the Total power supply current and voltage
1.) Each SHORT is different and will take from a range of 1% to 99% of the Total power supply current.
2.) A Short will change the circuits Drawing current value of normal operation.
3.) An Open Circuit will change the circuits current or voltage drawing value of normal operation right?
4.) An Open Circuit should draw LESS current from the power supply, So the Total powers supply current will be Lower it, will go down?
5.) A SHORT has no voltage drop across it, but has very high current
6.) The Component around the SHORT circuit will have Big or lower voltage drops compared to normal operation?
7.) A Open Circuit changes the voltage drops only in that faulty stage but doesn't affect the other stages or networks?
8.) An Open Circuit Has the same DC potential voltage on both side of the component?

 

[(*steve*)]

I think this may be the fourth or fifth time I've said this, but you need to have an understanding of the circuit. If you don't have one then you must develop one.

I mention this again.

It is clear that you have no fundamental knowledge of electronics nor particular knowledge of the circuit you are trying to troubleshoot.

You are asking for hard and fast rules in areas where there are none (or few).

As an example. If the driver of a car has a heart attack, will the car go faster or slower?

You still apparently don't understand what the voltage across, the current through, or the potential at each end of the components in this circuit when the switch is open.

This is absolutely fundamental. And if you can't answer the following questions, you really do need to go back to basics:

Assume you are trying to find a fault in the following circuit

1) With the switch open, what is the voltage at point B with reference to the ground?

2) What is the voltage across the resistor?

3) What is the current through the resistor?

4) What voltage would we expect if we placed the red probe of our voltmeter at point B and the black one at point A?

5) if we closed the switch, what would the answer to the question above be?

 

[CocaCola]

If the driver of a car has a heart attack, will the car go faster or slower?

Neither, the car will proceed at the same speed but will veer 13.4° to the right for 232 feet, and then make an abrupt 87.9° left turn at which point it will impact a 4.27 foot tall dark red brick wall ejecting the dead driver that will revive/reincarnate mid flight for 5.37 seconds only to meet his real death when he drowns in the 367 foot deep river he lands in, and will be found 16.30 miles downstream

 

[danny davis]

1) With the switch open, what is the voltage at point B with reference to the ground? 9 Volts

2) What is the voltage across the resistor? ZERO, cause it's not a complete circuit

3) What is the current through the resistor? ZERO, cause it's not a complete circuit

4) What voltage would we expect if we placed the red probe of our voltmeter at point B and the black one at point A? Zero Volts

5) if we closed the switch, what would the answer to the question above be?

I would need the value of the resistor to answer the other questions

 

[(*steve*)]

1) Correct

2) Correct

3) Correct

4) WRONG

5) WRONG

You do not need the resistor value, but you may assume it is 3.14159 kOhms and your meter's impedance is 2.7182 MOhms. The answer is required to within 1% of the actual value. Assume the battery's internal impedance is 1.618033987 ohms.

HINT: you can ignore *all* of the values above. (for question 4 -- which is the only one I asked you about, sure, it's needed for Q3)

For extra bonus marks, you can show why, but I expect that is well beyond you at this stage.

 

[(*steve*)]

Neither, the car will proceed at the same speed but will veer 13.4° to the right for 232 feet, and then make an abrupt 87.9° left turn at which point it will impact a 4.27 foot tall dark red brick wall ejecting the dead driver that will revive/reincarnate mid flight for 5.37 seconds only to meet his real death when he drowns in the 367 foot deep river he lands in, and will be found 16.30 miles downstream

WRONG. You have failed to notice that car was 3.64 metres in length and thus the answers are out by a small factor that involves the speed of light in an imperial universe.

And a bonus question: What is the driver's blood group?