Probe Oscilloscope Ground Blow up issues USB

[tronde]

In the previous post we learnt not to connect the ground clip to LIVE.

So, how about connecting it to neutral?

This is a simple drawing showing us how the lab bench is wired from the fuse panel. It is on a branch that is shared with other things as well. Among them is a water heater. We can see that the water heater is connected to the wiring AFTER the lab bench. The wiring is made of 2,5mm² wire, and the distance from the fuse panel to the bench is about 15 meter. The water heater is a 2kW type, so it will draw about 9A from the wall socket. All wires will have some resistance, and 15 meter of this wire will have close to 0,1 ohm in the live wire and the same in both the neutral and PE wires. Because of this resistance we will have a voltage drop at the lab bench's wall socket so we will measure about 228V there.

Why care about the resistance in the wires? They are there, and it sometimes makes sense to be aware of them. I used them to calculate the expected voltage between PE and neutral. I have rounded the values. We don't need exact values.








Before we decide to connect the ground clip to neutral we measure the voltage between the enclosure of the scope and the neutral wire we intend to connect the ground clip to. We find that the voltage is about 1V, so we think this is OK. It is almost the same as nothing, so PE and neutral should be the same because of this?


As we can see from the figure we have about 1V between PE and neutral and we have a current of 9A in the circuit. The scope and the device we intend to test consume almost no energy, so I have omitted the current for them.







Since we think everything is OK, we decide to connect the ground clip to the netral wire on the device.


Shit!

Just as we connected the clip, we got a spark.

What happened???








Since the ground clip connects to PE via the scope, and PE and neutral is connected together in the fuse panel, the return path from the water heater is split between the neutral wire in the wall socket and the PE wire from the scope.






The resistance in both the neutral wire and the PE wire is the same value since they are equally long. The ground wire in the probe and the PE-wire in the power wire between the scope and the wall socket will add something to the total resistance in the PE current path, but not very much compared to the total. The result is that we will have about half the current from the water heater in each wire. That is 4,5 ampere, so nothing strange about the spark.

As we can see, we need to understand the wiring hiding in the wall too.  If this water heater had been controlled by a thermostat or disconnected we would have measured 0V between PE and neutral when it was off, and no return current had mixed with the measurement. Spend some time and consider poentially dangerous situations caused by thermostat (or automatically)  controlled equipment that can influence on our safety.

So, you think - let's make sure we have no return currents. I do really advice against that. The neutral wire is part of the power supply, and unexpected things can happen. We must try to find something that is safe to use.


Things to discuss in later posts will be about floating the scope, using an isolation transformer, and isolated differential probes. Maybe something else too.

[Electro Detective]

Good show Tronde    

Let's hope some of it sinks through to the deathwish prodders that think their single ended oscilloscopes can be used like a multimeter

or bypassing the earth/ground wire on the oscilloscope 'gets it done'

or the Ebay click turkeys that buy a secret earth ground bonded 'isolation transformer'

and 'transform' their test gear/DUT set up into a DIY fireworks display   

[tronde]

I did some testing of the ground connection on an old probe and a 1,5m long mains cord of the same type as we usually connect to the power supply input of the scope. The combined resistance in those two was slightly more than 0,1 ohm.  My guess in the previous post was that it was almost negligible. 0,1 ohm is not much, but it does increase the total value by 100%, so in this exact application it is a significant change.

If we use 0,1 ohm as the combined value for the the probe and the power cord, we will get at total resistance from the ground clip down to the fuse panel of some 0,2 ohm. The 9A current from the water heater is split between this 0,2ohm and the 0,1 ohm in the neutral wire. That gives us a current in the ground clip of some 3A, and the current in the neutral wire will be around 6A.

This is not an important difference for what we need to understand; namely that the ground clip and the probe will be part of the return path. I just make this comment if somebody think I got it wrong in the previous post.
___________________________________________

What we have done until now, is to find out that we can't connect the ground clip to any of the mains wires. This makes it difficult for us if we need to measure anything connected to the mains with an oscilloscope.

Now, you might ask: Can't we just remove the ground clip from the probe? The ground-input of the scope is connected to both neutral and PE, so why do we need a ground clip?

Yes, we can infact remove the ground clip. On most probes it is meant to be removed, and the contact is usually covered so we can't touch ground by accident. The problem by doing so, is that the ground or common contact of the scope's input will be connected to a very long wire going from our lab bench in to the fuse panel and back to the workbench. In our example this is 30 meter.  This will act as a very good antenna picking up a lot of noise. In addition the neutral wire from the lab bench will carry currents that is not from the device we test. This will add alot more noise.  This wil more or less render our measurements useless, so we must find another solution.



What all this is about, is how we can use our scope without damaging it, or ourselves, or the device we intend to test.

Mains voltages have a big damage potential. We all know it can kill given the right circumstances.
But - it is also very powerful. I mentioned that we can get 3500W from the wall socket if protected by a 16A fuse. Normal fuses have a rather large overload capacity, and many fuses are rated for 45% overload in one hour. This gives us 23A and 230V x 23A = 5300W in one hour before the fuse will trip.   

Think about all kinds of power tools we can buy and connct to this socket. Think about how much manual labour these tools can save us from. Then we understand that 3500W or even worse 5300W is a lot of power. If we release all this power in less than a second as we do if we short something, the damage can be really large. That is the reason we need to understand what can go wrong, and why.


I will wait some days before I post more.

If rajiv85 is still around, it would be nice to know if some of this makes sense to him.

[rajiv85]

@tronde: thank you a ton. I am actually saving all your detailed explanations in my notebook. I will study them in detail today. What you are explaining is extremely useful.

[tronde]

@tronde: thank you a ton. I am actually saving all your detailed explanations in my notebook. I will study them in detail today. What you are explaining is extremely useful.

Fine,

Don't hesitate to ask.
When I know you are here reading, I will wait some time now with the rest so you can mange to "digest" this first.  Do you have some kind of test equipment? If you have something, please tell. Don't worry if you don't have a scope.

You said you are a beginner. How much beginner are you? Are you familiar with ohm's law and Kirchhoffs laws about how currents and voltages are distributed in a circuit?

[rajiv85]

Thank you so much Tronde. I studied the drawings you posted. In fact as said I have an e-notebook and I saved the illustrations, specially the last ones . Really really good!!!!! I hope many others follow this and find answers. I will update my first post to point to forum readers to check in thread 2/3 for your explanations. 

[tronde]

I have weighted pro and con on this for a while, but I landed on that I will write something on wiring calculations. What I write is aimed on beginners, and I feel that they will benefit from some basic understanding about what is hiding in the wall before we look closer on what is going on on the lab bench.

In a previous post, I used 0.1 ohm as resistance in the wires. This post will be about why it is 0.1 ohm.

First we must know something about the material we use in the wiring. In normal European house wiring, we use copper. Copper is a rather good conductor. This means it has low resistance to electrical current. The price is also sensible.

The physical property that tells us about how good any material will conduct electricity is known as "conductivity".

Usually we want to know how much resistance (the property we measure in OHM) a given wire have instead of how much conductance it has. Lucky as we are, it does exist a property for that as well. It is known as "resistivity". Resistivity tells us how good a matrial is to hinder the flow of electric current.


You can read more about them here, and you will find a table listing several different materials.
 https://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity


I stole this table from wiki just in case they change it.



Here we can see that the resistivity for copper is 1.68 x 10^-8 ohm x meter if the area is 1m² . The strange little p is known as "rho" and is a Greek letter. I use to remember it as a tadpole (baby-frog).


I am usually a great fan of the SI-system, but when it comes to wiring calculations we end up with some strange values. The reason is because the SI-system uses meter as a unit. We will never buy electrical wires with an area (cross section) of something in the meter-region. We buy them in some square-milimeters instead.

One square meter is 1000mm x 1000mm = 1000000mm². We can write this as 1 x10⁶ mm² instead.

If we multiply 1.68 x 10^-8 with 1 x10⁶ we get 1.68 x 10^-2.

A more beginner-friendly way of writing 1.68 x 10^-2 is 0.0168. The meaning of this number-tweaking is to give us a more user friendly value to use in our calculations than what is listed in the wiki-table. We have now transformed the area from square meters to square millimeters as we have in our wires.

As a side note, I can tell that many tables aimed at electricians will list resistivity based on an area measured in square millimeters.

If you don't fancy the math used in the transformation, don't worry. Just remember that resistivity for copper should be something like 0.0168.  Another material used in wiring is aluminium. If we transform that into something useful we get 0.0282.

You will find slightly different values listed in different tables. One thing to look for, is the temperature used. Resistivity will increase with temperature. The different values will not differ that much, and if the temperature is around 20 degrees C, you will do nicely regardless of the exact value. Just make sure it begins with 0.01 for copper and 0.02 for aluminium if you use an area measured in square millimeters.


So, what use can we make of this 0.0168? What does it tell us?

It tells us that if we have a wire made of copper that is one meter long, and with an area of one square millimeter, it will have a resistance of 0.0168 ohm. That is, it wil be the same as a 0.0168 ohm resistor.

It does also tell us that if we have a two meter long wire made of copper with anarea of 1 square millimeter it will have a resistance of 2 x 0.0168 ohm = 0.0336 ohm.

OR

if the one meter long copper wire has an area of two square millimeters, it will have a resistance of 0.0168 ohm / 2 = 0.0084 ohm.


We can transform this knowledge into a formula.

I can't get any useful greek symbols from my keyboard on this forum, so I will write it in plain text.


Resistance of the wire = ( lenght of wire [measured in meter] x rho ) / area of wire [measured im square millimeters]


In the example, I used a lenght of each wire of 15m and the area was 2.5mm²

If we use these values, we will find that the resistance of one wire is:

Rwire = (15 x 0.0168) / 2.5

Rwire= 0.1008 ohm.

If a math teacher reads this, he will freak out becuase I give the answer with 4 decimals.

The freak out is because 4 decimals indicates a much better precision in the answer than what is possible when we use 0 decimals in the lenght and 1 decimal in the area. I will happily leave to any math teacher to write something useful about it. I know that beginners like to use as many decimals as possible, and by doing so it seems like they get some confidence in that they used the right formula. I will keep on using many decimals because of this, but now I have told that I know it is actually wrong to do so, and why.

<peace, please>


Tomorrow I hope to do more calculations on the example from 17.07

[tronde]

Last time we found that the reistance in each wire from the fuse panel and up to the lab bench was 0.1008 ohm. We will use 0.1 ohm in the calculations.

First, we need to make a drawing of how the wiring is made.
The following drawings are based on previous drawings, so I have omitted some info for clarity.




When we do calculations, we must know exactly which resistor we use in the calculations. As you can see, I have given each resistor an unique name. We can use whatever we want, but it makes sense to start with R and use something unique (and descreptive) for the rest.

I have named the resitor in the live wire for RL. This is meant to mean something like R-LIVE.

Next, I have named the resitor in the neutral fire for RN. This is meant to mean something like R-NEUTRAL.

Then I have named the resisor in the PE-wire for RPE. This is meant to mean something like R-PE-wire.

Last we have the resistor in the power cord between the scope and the wall socket. I name this for RPEC. This is meant to mean something like R-PE-wire-cord.

In this drawing, I have indicated the resistance in each wire, and I have also added a resistance in the power cord. Last time, I assumed the power cord had negligible resistance, but as I showed in the comments it did actually have some. Therefore I have added it here.




As we can see, I have also added know values. We have already calculated the resistance in teh Live, Neutral and PE-wires to be 0.1 ohm, so I add the value close to each resistor in the drawing. Teh supply voltage of 23oV is also indicated where it belongs.

This drawing shows how everything is connected before we clip the probe clip to the neutral wire.




Next drawing will show what it is like when we do so.



As we can see, I have added a resistance in the probe wire. I name it RPROBE. I have also added the water heater (RHEAT) and indicated that the current flowing through it is 9A. We usually indicate currents with I.

When we do calculations, we try to "transform" as many items as possible into resistance. A water heater is basically a resistor so nothing strange there.

The reason we try to use resistance is because we can draw a schematics with only resistors, and we will automtically (in our head) treat the item as a resistor. To do so is OK as long as the main function of the item is mostly similar to a resistor. When we deal with low frequency mains we can often do so, but sometimes we can not. It is not possible to give a perfect description of when we can, and when we can not transform into resistance for pure beginners.

This transformation to resistors is also the reason why we often use a battery as power source in our drawings. We try to make it as simple as possible, and if we can use resistors in our calculations, we can also use a battery instead of for instance AC power as we actually have in this setting.

If you find a drawing somewhere that draws a motor as a resistor, it is because it is OK in that particular setting, while it can be wery wrong in another setting. This can be confusing for beginners, but after some time you will know and understand more. Just don't think people are crazy because something is drawn like a resistor in a drawing. It has a (good) reason.


These drawings shows us in a close to life-like style how things are connected together. Quite often it is difficult to use such a drawing for calculations. Therefore, we will transform this last drawing into a normal schematic-style drawing.




The previous drawing transforms into this.

Don't just believe it because I say so. Spend some time, and follow each wire on both of them and make sure you can identify them on both drawings. This is infact important for every circuit you will work on. Try to make some real schematics out of other things you can find around you.

As you can see, I have also marked each resistor with the information we already know. All of them have got a name, and some of them have got the value as well. This is very important, because it shows us clearly what we know, and what we don't know about each item.



Now, you might ask -   where is the scope and the device I test on?

In an earlier post I said the scope and the device uses so little energy compared to everything else in this circuit, so we can omitt them in th ecalculations. That is still true. The goal for these calculations is to find out what goes wrong when we connect the probe clip to the neutral wire, and to calculate the short circuit current.



We don't know yet the value of RPEC (resistance in the power cord) and RPROBE (resistance between the ground clip and the  power cord).

The power cord is a 1.5m long cord with an area of 0.75mm².

When we know this, we can calculate the resistance just as we did for the other wires.

RPEC = (0.0168 x 1.5) / 0.75
RPEC = 0.0336 ohm

This is the same as 33.6 milli ohm, so not much.

But, just for the sake of calculations, we will use that value. I said earlier that the combined value of the PE-wire in the power cord and the ground wire in the probe was about 0,1 ohm. to make it very easy for us, I deciede that RPROBE is 0.0664 ohm.



When we add these two last values, it looks like this



I have added IN and IP to the drawing. They are meant to indicate the current in the neutral wire (IN) and the current in the ground wire of the probe (IP).

We will calculate them later on.

Enough for today.

[rajiv85]

WOW..good stuff Tronde. I would highly suggest you consider making a small electronics tutorial book on that subject. Even if it is only 25 pages you can sell it on Amazon because it addresses a specific topic in good detail that other books or tutorials do not cover. 

[tronde]

This post has been edited. The reason is the original post was based on the logical flaw that the total current in the circuit would not change when we connected the ground clip. If you read the origial text, all calculations was correct for a current of 9A. The comment from Vtile in reply #60 is based on the original text.


In the last post we saw that the current flowing from the water heater split into two paths. One path is the neutral wire, and one path is via the ground clip, and from there into the PE wire.

We have some options when we choose how to calculate the value of these two currents. Which method you choose depends on several things, among them which method you prefer, and which data are available for the items that are part of the circuit.

When we do calculations we can almost always discuss which method is "best". If we don't try to cheat the laws of physics we will always get the same answer. Sometimes we will maybe use slightly more time on the calculations, but that is not relevant for us now.

If we look at the last drawing, we can see that we have only one resistor in the neutral wire, and that there are three resistors in the path through the PE wire. The first we will do now, is to transform those three resitors in PE into one resistor.

They are connected in series, so we can just add the values together.

I mentioned earlier that we must pay attention to how we name all our parts. Since we now will transform these resistors into a new "logical" resitor, we will give it an unique name. We will name it RP

RP = RPROBE + RPEC + RPE
RP = 0.0664 ohm + 0.0336 ohm + 0.1 ohm
RP = 0.2 ohm


This gives us this circuit to calculate




I have indicated that we don't know the value of I in this circuit. Before we connected the ground clip, it was 9A, but because we changed the circuit when we connected the clip, the total current will also change.


The first thing we must do, is to find the value of this new current.

If we look at the previous drawing we can see that the resistance in RHEAT (the water heater) is important to know if we want to know the current in the circuit.


Almost all kind of heaters are made from a material that does not change its resistance when the temperature changes. This is the same as it is stable when the power or current changes. We will use this knowledge when we calculate RHEAT.

The "secret trick" is to calcultate the resistance based on the current we know the value of, namely the current we had before we connected the ground clip. We can do this, because the resistance is stable regardless of the current.

This is the circuit without the ground clip, and I have added all known values.




To find RHEAT, we must find the voltage across it.


Kirchhoffs voltage law (KVL)
https://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

tells us that the sum of all voltage drops should be the same as the source voltage.


Since we know both the resistance in RL and RN as well as the current, we can easily find the voltage across each of them.
Usually we will specify the values so we know we get it right.

U_RL = RL x I
U_RL =  0.1 X 9
U_RL = 0.9V

U_RN = RN X I
U_RN = 0.1 X 9
U_RN = 0.9V

When we add these two values to the drawing we get this.

I have also indicated the voltage across the resistors. This can be done in several ways. Most textbooks wil use an arrow. Some will use an arrow pointing down, I am used to an arrow point up. I guess some people can spend a lot of time and words on discussing pros and cons. What is most important, is that you stick to only one standard. Else you will confuse yourself. The way I see it, is that the arrow point to the higher potential. Potential is the same as voltage.
The way I see it, is that the arrow point to the higher potential. Potential is the same as voltage.









We can see from the figure that we can use Kirchhoffs voltage law to calculate the voltage across RHEAT.

U_RHEAT = U - U_RL - U_RN
U_RHEAT = 230 - 0.9 -0.9
U_RHEAT = 228,2V



The resistance of RHEAT is determined by the voltage across it, and the current flowing through it.

R = U / I

RHEAT = U_RHEAT / I_RHEAT
RHEAT = 228.2 / 9
RHEAT = 25.356 ohm

_______________________________________________






Since we now know the value of the resistance in RHEAT, we can continue with the circuit "modified" by the ground clip.







Because we transformed RPROBE, RPEC and RPE into one logcal resistor (RP),  we have only two diferent resistors to care about in that part of the circuit.




Now, we face a problem:
How can we calculate the value of the new logical R based on RN and RP?

If we look in textbooks and formula books we often see these two formulas.



I snipped this from wikipedia
https://en.wikipedia.org/wiki/Series_and_parallel_circuits



I am a strong believer in trying to understand the secret hiding in the formula instead of just looking it up.

There is also another "secret" with formulas. Many formulas does not give new knowledge. They exist just because it makes our calculations easier. This was most important in the older days before we got pocket calculators, but are not as important today.  The second formula from wiki is aspecial version only valid for two resistors. It is of the "time savings" type. Useful, but no new knowledge.

Before we continue, we will spend some time on the first formula. What does it tell us?





When we calculated the resistance of a wire, I mentioned "conductance" and "resistance".

Conductance is a value for how easy it for electrical current to flow. We measure conductance in siemens, and we use to say that G = S.
Resistance is a value for how diffilcult it is for electrical current to flow. This is what we measure in ohm, and we use to say that R = ohm.

We can see conductance and resistance as opposites.

Think about a road with a lot of cars. If we have only one lane, the flow of cars will be limited. This is because the "conductance" of the road is low. We can also see it as the road has high "resistance".

If we have the same number of cars, but increases the number of lanes to two lanes, the flow of cars increases.
One lane will still have the same "conductance", but since we have added one lane, that lane will also have the same value for "conductance".

The new two-lane road will have a total "conductance" described as this:
G_total = G1 + G2

Back to the first formula:
It says something about 1/R_total  =  1/R1  + 1/R2

What this formula actually does, is that it calculates the conductance of all resistors connected in parallel.

1/R_total is the same as if we wrote G_{total =}
1/R1 is the same as if we wrote G1
1/R2 is the same as if we wrote G2


So, how do we tell the calculator how to do this?
Amost all calculators, except the most simple types will have one button that does the "one divided-by-something" function.
Look for a button marked 1/x or X^-1
To make sure it is the right button, you can easily test it.

Enter 100, and hit the button. The answer will be 0.01. Hit the button once more, and the ansver will be 100.

And now comes something important for beginners:

If we look at the first formula, we see that it calculates the conductance.  We are more interested in the resistance. This means we must hit the 1/X - button once more after we push the =  button, else we will get the wrong answer.


Now we can continue with our calculations.
We know that the new R must be calculated form the values of RN and RP

If we use the formula we will write it as this:

1/R   =   1/ RN   +   1/RP
1/R   =   1/0.1   +   1/0.2


When we use a calculator we will enter this

0.1 [1/X -button]   +  0.2  [1/X -button] = [1/X -button] =

If we did it right, we will get 0.066666667 as the answer.

This is the same as the combined value of RN and RP is 0.0667 ohm
We name this restor just R.

We will make a new drawing to help us.








Since we now know the value of all of the resistors, and the total voltage, we can calculate the value of the current.

We will name the total resitance RTOTAL.

RTOTAL = RL + RHEAT + R
RTOTAL = 0.1 + 25.335 + 0.0667
RTOTAL = 25.502 ohm



Next ting now, is to calculate the current. We remember that the current must change from the initial 9A because we changed the circuit when we connected the ground clip.

I = U / R
I = U / RTOTAL
I = U / 25.502
I = 9.019A

The total current in the circuit changed from 9A to 9.019A because of the ground clip.
Does it make sense? Yes, the ground clip made it slightly easier for the curent to flow because it added a second wire in parallel to the neutral wire form the workbench / water heater.



Since we now know the value of this logical resitor (R) and the current flowing through it, we can calculate the voltage across it.

We will use ohm's law.
U = R x I


U_R = R x I_R

U_R = 0.0667 x 9.019 = 0.602

This means that the voltage across the resistor is 0.602V


OK,
now we know the voltage across the logical resistor R. We transformed RN and RP into this resistor only because we needed to know the voltage across RN and RP.
We know the voltage now, so let's draw a new drawing where we show where it belongs.







If we look at this last drawing, we can easily see that we know the value of RN and the voltage across it. We can also see that we know the value of RP and the voltage across it as well.

We can also see that we want to know the value of the ciurent IN and the current IP.

Back to ohm's law:

U = R x I

I = U / R

When we want to calculate the current IN, we must use the correct values for the resistor that determines the value of IN. That resistor is RN

I_N = U_RN / RN
I_N = 0.602 / 0.1
I_N = 6.02A

Then we will calculate IP

I_P = U_RP / RP
I_P = 0.602 / 0.2
I_P =  3,01A


Now, we can check our calculations.
We know that the current flowing from the water heater into the combination of RN and RP is 9.019A.

Kirchhoffs current law (KCL) tells us that the sum of all currents flowing into a node is equal to the currents flowing out of that node
https://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

Since 3.01A + 6.02A = 9.03A, we can assume that we got it right. I mentioned something about precision and number of decimals in a previous post. I have already done the calculations with more precision than what we can account for, so we must assume this value to be correct.

It does always make sense to do some simple checks like this. Sometimes we enter the wrong numbers, or forget a decimal point or something else that will give us the wrong answer.

This is what we know about the circuit at the moment

[Vtile]

Oh. superb post tronde. For conductance .. it used to be marked as mho or upside down omega (resistance symbol) before the Siemens unit came standard. As the 1/x operation is called inverse the old 'mho' made sense. Inverse is one of the most powerfull tools in calculator btw.   

[tronde]

Since I changed and added more to the previous post (#59), this wil be a complete re-write. Most of the original information is moved into the previous post.



Time to round off this part of the "how not to kill your scope" tutorial.


Since we ended up with a lot of calculations, we can spend some time on formulas.


I said I was a strong believer in trying to understand the secret hiding in the formula instead of just looking it up.

When a beginner start to learn about electricity, Ohm's law is among the first law he will meet. Quite often it will be in this form:




Then the beginner tend to say "oooh so many formulas to learn".

First ting to be aware of, is that this is actually not only Ohm's law.
Ohms law is U = R x I, but we can see something like P = U x I as well.

What is this P about? P is about power, and the formula that reads P = U x I is known as Watt's law. Watt was the guy with the steam engine.


My claim is that this "wheel" is actually only two formulas that gives us new knowledge. The formulas that gives us the knowledge, is those we must pay attention to. The rest is just a collection of "make calculation easier" formulas as I mentioned earlier.



Ohm's law is marked green, and Watt's law is yellow.




Let's see how we can turn these two formulas into twelve.

#1 to #6 are easy. That is just shuffeling the formulas around.




The "secret" is hidden in what we do to get #7 and #10.

First #7:

We start with P = U x I.
Then we know that we can write U as R x I.
R x I is #1.

# 8 and #9 are just modifications of #7.



Then we can look at #10.

We start with P = U x I once more.
We know that we can write I as U/R.
U/R is #2.

#11 and #12 are just modifications of #10.

If we understand this, we can spend the time un understanding how to use the knowledge instead of remembering a lot of formulas.


A sidenote about the "wheel".
Differnt countries and languages will use different letters for voltage. In this wheel the letter U is used for voltage. You can also find that E or V is used.
Sometimes you will find W used instead of P for power for the same reason.


So, why care about watt in this tutorial?

We have spent a lot of time calculating current. Sometimes we don't know the current consumption of something, but we know how many watt it consumes and the voltage. Then we can calculate the current.

___________________________________________________________


The last drawing I have prepared is a complete circuit, including the scope and the device we test. Click on the first thumbnail (complete.jpg) to see it.

I have indicated both as resistors, and it is possible to give them some values if you want to calculate the circuit again.
You can for instance say that the device is a 20W / 230V led driver. How many ohms will that transform into?


I have made the ground clip into a switch that can be on or off, and it is possible to see how the circuit will change. I have also indicated that the RHEAT is in fact conected to some wires. I indicated in the first post that the heater was in another room, so it is obvious it must have some wires attached. It is possible to give these wires some resistance. Try to calclulate the circuit if they are, say 10m long and have an area of 2.5mm².


What will happen if we don't have the water heater mixing itself into the circuit?
What will the result be if we just connect the ground clip to the neutral wire of the device and we power it up?

If you have some kind of ground fault protection this will trip and cut the power to the lab bench if the current in the PE wire is above 30mA (if in Europe).

Suggestions, please.

Spend some time playing with the complete circuit and do some calculations using other values than I did. It is important to be familiar with this basic stuff, and I advise you to use pen and paper instead of some of the fancy simulators until you feel really familair with "decoding" such circuits as we have here.


Next will be about floating the scope I think.

[tronde]

I post this just so those who follow will get a notice. I have added some important info to post #59 and done a complete rewrite of #61 because of this.

[IanMacdonald]

Anyway, I think we've discussed before as to why scopes are designed like this whilst multimeters never have been. Likely reason  seems to be that in the old days they were always metal cased, and being valve based had some impressive voltages inside.  So, earthing the case and inputs was the safer option, even though it creates the risk of a shock or short TO the case or probe outers from a live terminal, which would not otherwise exist.

Nowadays with plastic cases, 12v internal supplies and readily available optoisolators etc there is really no reason for them to have earthed input circuitry, but it is something which has persisted.

Thinking about this a little more, the earthing on scope probes most likely violates electrical standards, which require that any earth connection should be robust enough to blow the supply fuse should a short occur.

[tronde]

Since nobody felt they would answer my questions, I will do it.

What will happen if we don't have the water heater mixing itself into the circuit?
What will the result be if we just connect the ground clip to the neutral wire of the device and we power it up?


First thing we must do, is to make a drawing of this new situation.






or like this.




Since the water heater is disconnecected, the 9A return current from it will not interfere with our measurements. Can we now consider it safe to connect the ground clip to the neutral wire?



I said the device is a LED driver marked 20W / 230W. We can transform this in to resistance, and do a complete calculation as we did in the previous post, but it does not make any sense for what we want to find out. The transformation will depend upon a lot of things to be correct, and the resistance we calculate will most likely be wrong in a real-life situation because the device is an electronic circuit.

We are not interested in the most accurate value of the currents. What we are interested in, is to find out if we do something stupid or dangerous. For us, this will be to find the approximate value of any current and/or voltage we, or our equipment will be exposed to.

I have also included the scope into this circuit becuase it will have much larger impact on the total than when we had the 9A current mixed in. The scope is marked 25W / 230V.

If you really want to do complete calculations you can use formula 11 from the previous post anf calculate R_SCOPE to be 2116 ohm and R_DEVICE to be 2645 ohm.

Instead, I will suggest a more simpler calculation only using the current in each item.
First we will calculate the current in the device. We will use Watt's law.

P = U x I
I = P / U

For the device this gives us:

I_DEVICE = P_DEVICE / U_DEVICE
I_DEVICE = 20 / 230
I_DEVICE = 0.087A or 87mA

For the scope this gives us:

I_SCOPE = P_SCOPE / U_SCOPE
I_SCOPE = 25 / 230
I_SCOPE = 0.109A or 109mA


New drawing





We can see that the current from the scope and the device will combine into a new current of 196mA.


To calculate IN and IP we need to know the combined value of RN and RP. We have already calculated it to be 0.0667 ohm. Remember that the wiring has not changed, so we can use all relevant values calculated earlier.




Since we know the resistance of R, and he current, we can calculate the voltage across it. Remember that the current has changed, so we must use the "new current" in this calculation.

U = R x I

U_R = R x I_R
U_R = 0.0667 x 0.196A
U_R =  0.0131V






Then we can calculate each of the currents.

U = R x I
I = U / R

I_N = U_RN / RN
I_N = 0.0131 / 0.1
I_N =  0.131A or 131mA


I_P = U_RP / RP
I_N = 0.0131 / 0.2
I_N = 0.066A or 66mA

If we add 66mA and 131mA we get 197mA. This is 1mA too much, but it is because of rounding numbers.


Since the resistance in RP is twice the resistance of RN,we can just say that the current in RN must be twice the current in RP.

This is the new situation when we connect the ground clip to the neutral wire.








So, what does this tell us?

First, we can see that we still will have a current flowing through the ground wire in the probe. It has a much lower value than when the water heater was mixed in, but the probe is not meant to carry any current, so it is still wrong to connect the ground clip to neutral.


* * * * * * *


I did also mention something about ground fault protection and 30mA. If the lab bench is protected by any kind of a RCD (residual current device) it will trip and cut power to the bench because of the 66mA current flowing in the PE wire.

Many people think they will be safe if they have a RCD, but that is a very dangerous assumption. RCDs are not meant as a device that will allow you to do silly things with electricity. It is meant to be a last resort if something goes wrong with the insulation in the complete system so you will not get a dangerous current flowing through your body.

In Europe, they will most likely trip if there is a leakage current larger than 30mA. This is not the same as any current below 30mA can be considered safe. Most people wil survive an electric current of no more than 30mA, but some will not.

Since we are measuring on electronic devices, we must also be aware of something important regarding RCDs.

The normal type used in Europe is known as "type A". "Type A" will not trip reliably if there is more than a 6mA DC-current component in the leakage current. If we have a DC-component, we must use a "type B" RCD. They are expensive and not common. If you intend to use a "Type B" in an installation where you already have a "Type A" you must check the datasheet for both types. Sometimes they will influence on each other and don't trip as expected.


Some reading about RCDs
https://en.wikipedia.org/wiki/Residual-current_device

Because of size limitations, I will post two attachments about RCDs from ABB and Schneider Electric in the next post. Worth reading.


The first attachment in this post is a modified complete schematics for this post.

[tronde]

Attachments to previous post about RCDs from ABB and Schneider Electric

[rajiv85]

Great contributions Tronde.. will take some time to assimilate all of it.. I am having some exams this month. But as soon as I am done I will come back to study what you explained to make sure I got it.

Thank you!! Perhaps such posts should be pinned as it can be useful to many other people.

[swgertsch]

As a newbe what we need are some clear ABSOLUTES (guidelines) to so we do hurt ourselves or damage any test equipment.  I accept the fact that the process of learning electronics involves the death of many a component; a certain amount of sacrifice on the alter of learning is required. These "sacrifices" help cement our understanding of what is going on, how things can go bad, and most importantly how to trouble shoot and repair electronic devices.

With respect to oscilloscopes here is one thing I have learned: the simple definition of an oscilloscope being a time based graphing voltmeter causes problems for newbes. This simplistic definition causes a newbe to equate oscilloscope probes to DMM-Voltage measurement mode probes and are therefore are likely to attach the ground clip with little regard to how oscilloscopes actually work.

With respect to using any tool it is important to understand how the tool works in order to use it safely and effectively.

(I do not understand why oscilloscopes do not have better protection (fast blow fuses, etc.) on probe ground leads and between non-isolated channel BNC connectors. I suspect protection is not there because it degrades measurement accuracy.       

Right now I restrict myself to using a single probe with the ground clip always attached to the circuit's GND and using power sources with a floating ground voltages sources: <24VAC, <30VDC. 
 

[tronde]

Right now I restrict myself to using a single probe with the ground clip always attached to the circuit's GND and using power sources with a floating ground voltages sources: <24VAC, <30VDC.

This is a very wise approach. Accept that you have some limitations in your knowledge, and progress step by step as you understand more.

I can see that a list of what you call "absolutes" would be nice to have, but the problem is that no lab bench is similar. I have some more subjects to cover, and I will write more about them within not so long time.


My best and honest advice to any beginner, is to spend a lot of time calculating what is known as "combination circuits".
http://www.physicsclassroom.com/class/circuits/Lesson-4/Combination-Circuits

Use pen and paper, not a simulator. When you do this, you get familiar with how the currents and voltages distribute in a circuit, and it will be more easy to see and understand what can go wrong when you connect any kind of test equipment to a circuit.

I know many beginners find this boring, but this most basic stuff is important. It is like learning to swim or ride a bicycle. Difficult in the beginning, but you will never forget it when you have learnt it.

The problem with test instruments, is that they interact with the circuit you test on. That is why it is so importat to "see" how the circuit changes when we connect something external to it.  To see this, we must also know a lot about the instrument we connect to the circuit.

[tautech]

To see this, we must also know a lot about the instrument we connect to the circuit.

Not so much really.

We keep using the term probe ground lead and in my very strong opinion this is where misunderstandings start and grow.
Instead if we all think REFERENCE lead as the signal is referenced to the probe lead. That the probe lead is at mains ground potential too is VERY IMPORTANT.
Just these 2 things need clear understanding:
Probe lead is for signal reference AND is at mains ground potential.

Mentioned previously in the first page of this thread:

Scott, you make some valid points and if I may I'll make one that might help clear some misunderstandings.

Most of this stems from the basic knowledge that we can't just clip the ground lead of a scope probe just anywhere.
Of course the reason is that it's mains ground referenced and has the potential to short the signal/circuit out.

But let's make this simpler, the commonly used term of a probe ground lead in some way hides its real function.......it is the connection that provides a reference for the signal and need be firmly thought of as such, BUT it's also mains ground referenced and this determines where it can be connected to.

We use the term probe ground lead to remind us that it is mains ground referenced, a term I personally abhor, preferring to know it as the probe reference lead, knowing that for ALL but a very few scopes (isolated, floating etc) it is indeed mains ground referenced too.

You may've seen in posts I make that I try to take care now I use the reference/ground lead terminology in some hope that others can see in a similar way that has given me good guidance over the years.

Sure everybody, keep using the term probe ground lead at your peril, one day in a moment of inattention you'll clip it where it shouldn't be connected and either blow your scope, probes or the DUT.

Think reference..........and tied to mains ground.

[tronde]

The "problem" with ground as a term is based on history. Ground as a term was used in radio long before it became a thing for safety.

Ground is widely used for reference level, but if you work with mains, you wil see the term "protective earth" or "earth". Connecting things to the soil for mains safety was not implemented before around 1920/1930, and by then the term "ground" was already used for the reference level.

The problem is not that we call the clip for ground clip, but that we call protective earth for ground.

[alank2]

Is the problem ALWAYS that the ground lead allows a high current situation when connected to something that is not at the same potential it is?

For example, if you have a USB scope that says it can handle mains level voltages and you can connect the ground clip safely, could you then measure them without a differential probe?

[Brumby]

Is the problem ALWAYS that the ground lead allows a high current situation when connected to something that is not at the same potential it is?

That is the principle "destruction" issue.  It is most applicable to situations where there is a low impedance path in the rest of the circuit formed - which results in high current flow.

There are other equipment survival issues as well.  Ones that may not be quite as dramatic.  Just imagine a circuit that is earth/ground referenced and taking a point in that circuit which is at a different potential ... and then grounding it by using the scope reference/ground clip.  There is the possibility that this could cause stress on the circuit that might result in damage.  Such damage could be the invisible demise of a component or letting the magic smoke out.  Even if there were no damage, the circuit operation would almost certainly be affected and the measurement corrupted.

For example, if you have a USB scope that says it can handle mains level voltages and you can connect the ground clip safely, could you then measure them without a differential probe?

With those two caveats very firmly enforced, then, yes.  BUT, in the real world, there are real risks which would make it a less than wise approach.  For starters, there's no way I would (personally) run mains level voltages through a USB scope - unless it was on a computer that I was not worried about frying.

[IanMacdonald]

All oscilloscopes (except a few portable units) are basically 'Cat zero' safety rating. The have no fuses, MOVs or other input protection, and putting the probe outer onto a high current connection is going to ruin your day.

Why are things like this? Basically the reason is historical. They never have had input safety devices, so I guess the manufacturers don't see any need for them. Although when I think about it, even the old multimeters like the AVO8 had a mechanical trip that would usually save your bacon if you made a silly mistake.

[David Hess]

All oscilloscopes (except a few portable units) are basically 'Cat zero' safety rating. The have no fuses, MOVs or other input protection, and putting the probe outer onto a high current connection is going to ruin your day.

Why are things like this? Basically the reason is historical. They never have had input safety devices, so I guess the manufacturers don't see any need for them. Although when I think about it, even the old multimeters like the AVO8 had a mechanical trip that would usually save your bacon if you made a silly mistake.

It is historically this way because of its difficulty.  Making an oscilloscope with ground isolated inputs was somewhere between difficult to impossible and most applications did not require them.  It is so difficult to do with good performance that high voltage differential probes are easy in comparison.  The situation is only different today because high bandwidth digital isolation is available and this was starting to be used in oscilloscopes for this purpose starting in the late 1980s.

In rough order of development, solutions for the isolation problem were:

1. Floating the oscilloscope.
2. High voltage differential probes up to about 50 MHz.
3. Analog isolated probes up to about 20 MHz.
4. Digital isolation of the digitizer.

Oh, and another method I have used in the past is to convert the voltage to a current in the circuit and use a current probe; this commonly works up to about 50 MHz however it is feasible up to 1 GHz in some AC only applications.

For a multimeter, making a ground isolated input is trivial because the bandwidth is very limited.