Very basic: how to calculate this potential difference

[j57H8I4]

Please, how to calculate the potential difference between the points +A and -A in the attached circuit?

[j57H8I4]

I am doing some experiments with my new used analog oscilloscope. I created that circuit on a breadboard, then I connected my oscilloscope to the points +A and -A and the oscilloscope showed 0.4V. Is it the correct result? How to calculate it?

[ataradov]

22.5 * (100 / 5600) = 0.4 V This is incorrect. Correct answer below.

[hamster_nz]

I am doing some experiments with my new used analog oscilloscope. I created that circuit on a breadboard, then I connected my oscilloscope to the points +A and -A and the oscilloscope showed 0.4V. Is it the correct result? How to calculate it?

Not to doubt you, but what are you using as a 22.5V source? It just seems an odd voltage to be using...

[Avacee]

It's Vout = Vin * (R2 / (R1+R2))     .... not (R2/R1)

so Vout = 22.5 / (100 / 5700) = 0.395V

Doesn't make much difference (0.401 vs 0.395) in this instance with these resistor values but other values (eg both 100 ohm) the change is pretty significant.

[j57H8I4]

Not to doubt you, but what are you using as a 22.5V source? It just seems an odd voltage to be using...

I am following this book: www.introni.it/pdf/101%20Esperimenti%20con%20oscilloscopio.pdf and I am doing the experiment number 2 on page 34.

[ataradov]

It's Vout = Vin * (R2 / (R1+R2))     not (R2/R1)

Indeed, apparently replying while distracted is not a great idea.

[hamster_nz]

Not to doubt you, but what are you using as a 22.5V source? It just seems an odd voltage to be using...

I am following this book: www.introni.it/pdf/101%20Esperimenti%20con%20oscilloscopio.pdf and I am doing the experiment number 2 on page 34.

Ah, that explains it... the book is from the early 70s, where 15-cell batteries like the "Eveready 412" were used.

[j57H8I4]

It's Vout = Vin * (R2 / (R1+R2))     .... not (R2/R1)

so Vout = 22.5 / (100 / 5700) = 0.395V

Doesn't make much difference (0.401 vs 0.395) in this instance with these resistor values but other values (eg both 100 ohm) the change is pretty significant.

WOW, for me it is very misterious why to calculate that way. I just cannot immagine what that 100 ohm resistor does to the circuit. What should I study to learn that?

[Avacee]

Search for "Voltage Divider" which is what this is.

In simple terms you have 22.5V going across a total of 5700 ohms.
 100  / 5700ths of 22.5V will be across the  100 ohm resistor .... 0.4V
5600 / 5700ths of 22.5V will be across the 5600 ohm resistor. .... 22.1V

Do NOT try and use the oscilloscope to measure the 22.1V across 5600 ohm resistor as that will mean your oscilloscope ground lead isn't attached to ground and you'll blow up your scope.

[ataradov]

Ohm's law: U = R*I. I is the same for both resistors, and it is equal to I = U / (R1 + R2). So voltage over the R2 is U(R2) = R2 * U / (R1 + R2) = U * (R2 / (R1 + R2))

[j57H8I4]

Do NOT try and use the oscilloscope to measure the 22.1V across 5600 ohm resistor as that will mean your oscilloscope ground lead isn't attached to ground and you'll blow up your scope.

That is scary! But I limited the current on my power supply to 0.3 A maximum, would the scope blow up just the same with that amount of current?

[ataradov]

That is scary!

He says that you should not measure voltage across 5.6K resistor, as in connecting ground clip to one of its leads and probe to the other.

This is not necessarily unsafe, it depends on your power supply type. If it is isolated supply, then you are fine. And if it is not - get one, otherwise you risk to kill yourself, not just the scope.

But basically, always keep your ground clip of the probe connected to the actual circuit ground.

[j57H8I4]

He says that you should not measure voltage across 5.6K resistor, as in connecting ground clip to one of its leads and probe to the other.

This is not necessarily unsafe, it depends on your power supply type. If it is isolated supply, then you are fine. And if it is not - get one, otherwise you risk to kill yourself, not just the scope.

But basically, always keep your ground clip of the probe connected to the actual circuit ground.

Thanks a lot. Well. that is terrifing! My power supply is powered by the mains and has got the connection to the mains ground. The following of that experiment says to reverse the battery, could that kill me, with the current limited to 0.3 A on my power supply?

[ataradov]

My power supply is powered by the mains and has got the connection to the mains ground.

What matters is whether outputs of the power supply are referenced to ground. Typically you see 3 output connections on a power supply - "+", "-", and "GND". "+" and "-" are galvanically isolated from the mains plug, and "GND" is connected to the mains grounding pin.

If this is the case (and your power supply is not some dodgy Chinese piece of junk), you are safe as long as you are using only "+" and "-" terminals.

The following of that experiment says to reverse the battery, could that kill me, with the current limited to 0.3 A on my power supply?

Current limit on the PS has nothing to do with this. What will kill the scope, is the current between PS ground and scope ground.

Again, if you have fully isolated supply, then you are free to switch its outputs however you like. They behave just like the battery would in this case.

[j57H8I4]

My power supply is this one: https://www.reichelt.de/Laboratory-Power-Supplies/KD3005D/3/index.html?ACTION=3&GROUPID=4952&ARTICLE=148149&OFFSET=16&SID=10Vtv6V6wQATIAAEiIYC492e2c1acce48b75ddb602e6e2bcee198&LANGUAGE=EN and I bought it from there; but I think it is a Chinese one. I do not know if it is isolated or not.

[ataradov]

KD3005D is a decent power supply, don't worry about it.

[j57H8I4]

Ah, that explains it... the book is from the early 70s, where 15-cell batteries like the "Eveready 412" were used.

That is interesting, thanks. I did not know those 15-cell batteries, I am a beginner at electronics.

[Brumby]

Ah, that explains it... the book is from the early 70s, where 15-cell batteries like the "Eveready 412" were used.

That is interesting, thanks. I did not know those 15-cell batteries, I am a beginner at electronics.

Don't worry.  Anyone beginning in electronics these days wouldn't know about them - unless they'd sat on the knee of a grandfather who was into it.

[macboy]

My power supply is this one: https://www.reichelt.de/Laboratory-Power-Supplies/KD3005D/3/index.html?ACTION=3&GROUPID=4952&ARTICLE=148149&OFFSET=16&SID=10Vtv6V6wQATIAAEiIYC492e2c1acce48b75ddb602e6e2bcee198&LANGUAGE=EN and I bought it from there; but I think it is a Chinese one. I do not know if it is isolated or not.

KD3005D is isolated, until and unless you connect the ground connection of the power supply (or any other ground) to either the + or - output, or to another point in the circuit. "Isolated" is relative though, since there is always some leakage capacitance to ground, due to capacitive coupling between primary and secondary of the transformer (especially cheap Chinese power supply transformers) plus other paths. At higher frequencies, this small capacitance becomes a lower and lower impedance, providing (AC) current leakage to ground. It's usually not an issue, but it's there.

And don't worry, you WILL NOT kill yourself with a 30 volt power supply that is ground referenced, or floating, or otherwise. 30 V is safe for direct skin contact under all reasonable circumstances. You'll feel something through mucus membranes (mouth, or other <ahem> bodily orifices) so don't connect stick your power supply to your tongue.    IF however, you put your 30 V power supply in series with another 30 V power supply, now you have 60 V and that is becoming borderline dangerous.

You scope's ground lead is truly grounded to your AC mains cord earth/ground wire. Remember this always. This means that you can't just connect it willy-nilly to any point in any circuit. There is a good EEVBlog episode about scope grounding that you should watch, in addition to other research.