Very basic: how to calculate this potential difference - Page 1
Forum section: electronics / beginners

author: j57H8I4
author_rank: Contributor
timestamp: March 06, 2017, 11:20:00 pm
content: Please, how to calculate the potential difference between the points +A and -A in the attached circuit?

author: j57H8I4
author_rank: Contributor
timestamp: March 06, 2017, 11:26:45 pm
content: 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?

author: ataradov
author_rank: Super Contributor
timestamp: March 06, 2017, 11:28:18 pm
content: 22.5 * (100 / 5600) = 0.4 VThis is incorrect. Correct answer below.

author: hamster_nz
author_rank: Super Contributor
timestamp: March 06, 2017, 11:31:07 pm
content: Quote from: j57H8I4 on March 06, 2017, 11:26:45 pmNot to doubt you, but what are you using as a 22.5V source? It just seems an odd voltage to be using...

author: Avacee
author_rank: Unknown
timestamp: March 06, 2017, 11:44:12 pm
content: It's Vout = Vin * (R2 / (R1+R2)) .... not (R2/R1)so Vout = 22.5 / (100 / 5700) = 0.395VDoesn'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.

author: j57H8I4
author_rank: Contributor
timestamp: March 06, 2017, 11:45:44 pm
content: Quote from: hamster_nz on March 06, 2017, 11:31:07 pmI am following this book:www.introni.it/pdf/101%20Esperimenti%20con%20oscilloscopio.pdfand I am doing the experiment number 2 on page 34.

author: ataradov
author_rank: Super Contributor
timestamp: March 06, 2017, 11:45:59 pm
content: Quote from: Avacee on March 06, 2017, 11:44:12 pmIndeed, apparently replying while distracted is not a great idea.

author: hamster_nz
author_rank: Super Contributor
timestamp: March 06, 2017, 11:59:50 pm
content: Quote from: j57H8I4 on March 06, 2017, 11:45:44 pmAh, that explains it... the book is from the early 70s, where 15-cell batteries like the "Eveready 412" were used.

author: j57H8I4
author_rank: Contributor
timestamp: March 07, 2017, 12:04:37 am
content: Quote from: Avacee on March 06, 2017, 11:44:12 pmWOW, 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?

author: Avacee
author_rank: Unknown
timestamp: March 07, 2017, 12:10:35 am
content: 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 .... 0.4V5600 / 5700ths of 22.5V will be across the 5600 ohm resistor. .... 22.1VDo 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.

author: ataradov
author_rank: Super Contributor
timestamp: March 07, 2017, 12:10:54 am
content: 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))

author: j57H8I4
author_rank: Contributor
timestamp: March 07, 2017, 12:17:51 am
content: Quote from: Avacee on March 07, 2017, 12:10:35 amThat 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?

author: ataradov
author_rank: Super Contributor
timestamp: March 07, 2017, 12:31:46 am
content: Quote from: j57H8I4 on March 07, 2017, 12:17:51 amHe 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.

author: j57H8I4
author_rank: Contributor
timestamp: March 07, 2017, 12:42:38 am
content: Quote from: ataradov on March 07, 2017, 12:31:46 amThanks 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?

author: ataradov
author_rank: Super Contributor
timestamp: March 07, 2017, 12:48:50 am
content: Quote from: j57H8I4 on March 07, 2017, 12:42:38 amWhat 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.Quote from: j57H8I4 on March 07, 2017, 12:42:38 amCurrent 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.

author: j57H8I4
author_rank: Contributor
timestamp: March 07, 2017, 01:14:20 am
content: 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=ENand I bought it from there; but I think it is a Chinese one. I do not know if it is isolated or not.

author: ataradov
author_rank: Super Contributor
timestamp: March 07, 2017, 01:15:34 am
content: KD3005D is a decent power supply, don't worry about it.

author: j57H8I4
author_rank: Contributor
timestamp: March 07, 2017, 01:21:49 am
content: Quote from: hamster_nz on March 06, 2017, 11:59:50 pmThat is interesting, thanks. I did not know those 15-cell batteries, I am a beginner at electronics.

author: Brumby
author_rank: Unknown
timestamp: March 07, 2017, 04:06:22 am
content: Quote from: j57H8I4 on March 07, 2017, 01:21:49 amDon'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.

author: macboy
author_rank: Super Contributor
timestamp: March 07, 2017, 05:03:00 pm
content: Quote from: j57H8I4 on March 07, 2017, 01:14:20 amKD3005D 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 youcan'tjust 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.