Voltage divider circuit is giving wrong value - Page 1
Forum section: electronics / beginners

author: rahulb
author_rank: Contributor
timestamp: March 31, 2025, 04:47:10 pm
content: Hi Everyone,I have made a voltage divider circuit. Problem is when I calculated the value that I should get, it was 0.8vBut, when I am reading the value, it is coming 2.5VI don't know what is happening.I am attaching the schematic, please let me know what the problem is with the circuit.thanks

author: grumpydoc
author_rank: Super Contributor
timestamp: March 31, 2025, 05:45:55 pm
content: Post a photo of your setup - the diagram looks fine but I find myself wondering if you have built it correctly and measuring the right voltage (as 2.5V is about the voltage you would expect across the LED).

author: lakis70
author_rank: Contributor
timestamp: March 31, 2025, 07:08:26 pm
content: or the resistors are both 10k

author: mzzj
author_rank: Super Contributor
timestamp: March 31, 2025, 07:28:24 pm
content: or cheapo dvm on AC range

author: coromonadalix
author_rank: Super Contributor
timestamp: March 31, 2025, 07:36:33 pm
content: you want a divide by x value ? the output of the divider goes to what ? you wanted a divide by 100because there is a difference between no load or load ... take that into accounta 10k and 1k is okay for divide by 100, but where it goes ? to an voltmeter, an mcu, an adc

author: Zero999
author_rank: Super Contributor
timestamp: March 31, 2025, 07:45:31 pm
content: Quote from: lakis70 on March 31, 2025, 07:08:26 pmThat still wouldn't be right. It would give 4.5V from a 9V battery.Perhaps the OP is mistakenly measuring the voltage across the LED, not the resistor?

author: TimFox
author_rank: Super Contributor
timestamp: March 31, 2025, 07:45:56 pm
content: Quote from: coromonadalix on March 31, 2025, 07:36:33 pm10k into 1k is divide by 11, where did you get 100?He was trying to get 9/11 V from 9 V.

author: grumpydoc
author_rank: Super Contributor
timestamp: March 31, 2025, 08:14:23 pm
content: Quote from: Zero999 on March 31, 2025, 07:45:31 pmYes, that was my thought also.

author: BeBuLamar
author_rank: Super Contributor
timestamp: March 31, 2025, 08:22:51 pm
content: Or if the terminal #2 of the LED and terminal #2 of the 1K resistor are shorted together.

author: golden_labels
author_rank: Super Contributor
timestamp: March 31, 2025, 08:45:23 pm
content: Also note that a fully charged battery with no significant load is 9.6 V, while a deplated one under load goes below 7 V.I concur with grumpydoc, please show the actual setup, and how you measure it. Preferably take a photo and add it as an attachment to the post.

author: grumpydoc
author_rank: Super Contributor
timestamp: March 31, 2025, 09:01:14 pm
content: Quote from: BeBuLamar on March 31, 2025, 08:22:51 pmAlso a distinct possibility, hence the request for a photo.

author: vk6zgo
author_rank: Super Contributor
timestamp: April 01, 2025, 12:29:26 am
content: Quote from: rahulb on March 31, 2025, 04:47:10 pmFirst thing to do is to measure the actual resistances of R2 & R3 individually, as it is not unknown for resistors to be incorrectly marked.If they look correct, remove the LED & associated resistor, then check that the 9v supply is definitely present across the series combination of R2 & R3.If your "9v" is a battery, it could be flat & producing around 5v at the terminals.If you then inadvertently used two 10k (or two 1k) resistors for R2, R3, there would be 2.5 v across R3, just as you measured.Yet another thing, some cheap multimeters start to give wildly inaccurate readings when their battery is becoming flat, well before they display the "Low battery" warning.We had early model UNI-T DMMs at one job, & on one occasion, the UNI-T measured a 27k resistor as 14k, which would have been well within spec, but the (just built) device wasn't working. My beat up old Fluke 77 identified it for what it was & after a change in battery, so did the UNI-T.

author: gbaddeley
author_rank: Regular Contributor
timestamp: April 01, 2025, 02:18:45 am
content: OP did not mention using a battery. Either R2 or R3 are not the stated values or the measurement is incorrect. Even with a high impedance dc voltmeter, it could not measure as high as 2.5v, so that’s not the issue.OP, pls check the resister values, and the measurement again. Is this a sim or a built circuit?

author: CaptDon
author_rank: Super Contributor
timestamp: April 01, 2025, 01:19:05 pm
content: No reply from O.P. Why???

author: squadchannel
author_rank: Frequent Contributor
timestamp: April 01, 2025, 01:30:58 pm
content: Quote from: CaptDon on April 01, 2025, 01:19:05 pmwhere's the gratitude?don't often get them, though.

author: vk6zgo
author_rank: Super Contributor
timestamp: April 01, 2025, 02:14:35 pm
content: Quote from: gbaddeley on April 01, 2025, 02:18:45 amIndeed, the OP did not mention using a battery, but people often do, assuming it is a true 9v.If the DMM had a massive change of accuracy for the reason I suggested, or some other one, there can be a gross error, as there was in my example.

author: BeBuLamar
author_rank: Super Contributor
timestamp: April 01, 2025, 04:31:40 pm
content: Quote from: CaptDon on April 01, 2025, 01:19:05 pmI found in many forums that the OP can ask question but can not be asked.

author: lakis70
author_rank: Contributor
timestamp: April 01, 2025, 07:28:22 pm
content: Quote from: BeBuLamar on April 01, 2025, 04:31:40 pmsuddenitis the modern disease...

author: rahulb
author_rank: Contributor
timestamp: April 15, 2025, 08:25:58 am
content: hi,sorry for late reply.It seems the PCB design I made had an error.the voltage I was getting was according to the parallel resistance value of R1 and R2On breadboard it was working fine but not on PCBI will try to make the design again.I want to thank everyone for help