From 2-wire voltmeter to 3-wire? - Page 1
Forum section: electronics / beginners

author: ElectricPower
author_rank: Frequent Contributor
timestamp: July 25, 2022, 07:49:18 pm
content: Maybe a stupid question, but i want to make a led tester like in the vide below. (Please see link)https://youtu.be/cz2FptkGi9ABut i only have 2-wore version of that meter, but i can see an unused pin marked "IN" on the back. I can't see this pin work....Please see attached images of the meter.Is there any simple way i can convert this meter to an three pin (wire) meter?

author: RoGeorge
author_rank: Super Contributor
timestamp: July 25, 2022, 08:56:00 pm
content: https://www.reddit.com/r/AskElectronics/comments/cx5rk2/can_i_convert_2wire_selfpowered_voltmeters_to/

author: ElectricPower
author_rank: Frequent Contributor
timestamp: July 26, 2022, 01:49:02 am
content: Quote from: RoGeorge on July 25, 2022, 08:56:00 pmThanks. But the unit you linked to is not like the unit i have. I have no resistors or visible traves to cut etc....

author: ledtester
author_rank: Super Contributor
timestamp: July 26, 2022, 02:06:07 am
content: I think it is worth exploring if this hypothesis is true:I think IN and V are bonded together and V is the positive supply which must be greater than a voltage like 2.5V in order to power the MCU and display. So if you cut the connection between IN and V you might be able to use IN independently of V.I wouldn't be surprised if V goes off to a voltage regulator and IN goes to a voltage divider and then to an analog input of the MCU.

author: mindcrime
author_rank: Unknown
timestamp: July 26, 2022, 02:27:05 am
content: Quote from: ElectricPower on July 25, 2022, 07:49:18 pmQuote from: ledtester on July 26, 2022, 02:06:07 amUsername checks out!

author: Zero999
author_rank: Super Contributor
timestamp: July 26, 2022, 07:27:02 am
content: Quote from: ledtester on July 26, 2022, 02:06:07 amYes. The LED display probably has to be removed first, before that can be done.The correct solution is just to buy the right one to start with. I'm guessing it's because it's cheaper, or bought in error. I wouldn't bother just to save a small amount of money and as far as error is concerned, well live and learn.

author: mariush
author_rank: Super Contributor
timestamp: July 26, 2022, 07:33:21 am
content: The starting point would be desoldering the 7 segment display from the circuit board.There's a small microcontroller and a voltage regulator hidden under that display (maybe even some basic TL431 voltage reference but unlikely), so you can't really tell how the traces go on that other side without desoldering.Afterwards, it could be as simple as cutting the trace going to the voltage regulator input, or desoldering the voltage regulator to connect a wire directly where the output of the regulator was (if it outputs 3.3v or whatever).I'd suggest getting a boost led driver IC and configure it to boost up to 20-30v or whatever you think you're gonna test, and limit the current to something reasonable, for example 5mA , 10mA, 20 mAFor example, AP3032 works with 2.7v.... 9v and boosts it up to 27v (hard limit inside the chip) :https://www.digikey.com/en/products/detail/diodes-incorporated/AP3032KTR-G1/4470854Besides this chip, you'd need a suitable inductor (any small smd inductor, they recommend 6.8uF but considering the small test currents probably can also go up to 10uH) and a shottky diode and some small resistors and ceramic capacitors.See figure 31 in the datasheet (but note R1 and R2 can be omitted, they're optional) :https://www.diodes.com/assets/Datasheets/AP3032.pdfThe current limit is set with the sense resistor - ex R = 200mV / current desired .... you have an enable pin as well.So you could have 3-4 AA batteries for ~5v for led driver and for the meter chip. Or use a lithium battery and add a cheap 1$ battery charger ic to charge the battery when needed. But 4 AA batteries would last you for months.I'd add a on/off switch on the box so that the batteries won't be drained when it's not in use.if powered from 3-4 AA batteries, you can power the driver ic and the meter directly. If you use a 9v battery, i'd use a 5v regulator.Optionally you could have a RUN / TEST button which basically enables the led driver chip while the button is pressedThe meter can sit across the + - of the led driver output and measure the voltage produced by the led driver.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: July 26, 2022, 09:52:21 am
content: Thanks for all answersI have now removed the display from the board and i took a picture of the board with my microscope.I must take two pictures for you to see the entire board but the pictures is attached here.It is the three pins down to right that are the input pins. From right it is "G" "V+" and "IN" to the left.What should i do here for make this an three wire voltmeter?

author: mariush
author_rank: Super Contributor
timestamp: July 26, 2022, 11:13:51 am
content: Ok, so the regulator is 7833, a linear regulator that outputs 3.3v You can figure out the pinout using a multimeter ... but you can follow where V+ goes and that's your INPUT in the regulator, same for G which is ground ... it's either TAB + middle pin output voltage , or tab + middle pin is ground and the pin opposite to input voltage is ground.edit: if it's sot-89 HT7833 regulator -https://www.holtek.com.tw/documents/10179/82844a36-1633-498c-a6e3-4d99f2d0f6d7- then the tab and middle pin are actually input, left is ground and right is output voltage. Looks like there's a small ceramic capacitor on the output in your pictures, and the ground wire (right most wire) goes directly to the ground of the linear regulator.If you look on the pcb you can see two traces going across the whole width ... one trace is connected to Voltage, one is connected to where you'd have that 3rd wire (IN)So it looks like you could solder the third wire and then move that resistor with 303 on it, to the left, to the other trace that goes to input wireNow you input voltage is separated from input wire.The microcontroller can probably handle up to 5v, but if you want to stay safe keep the regulator. Without a regulator, I'd be comfortable powering this thing with 3 AAA/ AA rechargeable batteries (to get around 3.6v)The two resistors on the left form a voltage divider , reducing input voltage to a lower input voltage... though have to admit I'm a bit puzzled by "308" seems like a super high value to me. Would help if you can measure the two resistors with a multimeter.

author: pcprogrammer
author_rank: Super Contributor
timestamp: July 26, 2022, 11:24:10 am
content: First try to make a schematic of the circuit. You can use a black box symbol for the big chip, and trace the connections to the other parts.Maybe the two resistors on the left end give the possibility to change it to three wire mode, but that is hard to say.On your picture in the original post ground looks like to be next to a pin of the display, but in your latest pictures this is just a narrow wire going to these swap able resistors. The text on the little chip on the right is hard to decipher.mariush already answered this too.The resistor marking on the top is most likely 30Bhttps://kiloohm.info/eia96-resistor/30Band therefore 2Kohm. The other one is probably 30Kohm

author: SeanB
author_rank: Super Contributor
timestamp: July 26, 2022, 11:36:32 am
content: In board 2 you have 2 303 (30k) SMD resistors, with blank pads next to them. Move the one next to the "E", lower one, to the pads to the left, which will allow the in pad to be separate from the power, and allow it to work as 3 wire. Common will still be the negative rail.

author: Zero999
author_rank: Super Contributor
timestamp: July 26, 2022, 11:40:14 am
content: Quote from: ElectricPower on July 26, 2022, 09:52:21 amThe 303 resistor forms a potential divider with the 2k (30B) resistor and goes to an ADC input on the micro controller. At the moment, the 30k (303) resistor is connected to the middle pin V+ and IN is unconnected. You need to connect the 30kresistor to IN, by moving it left to the unpopulated pads.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: July 26, 2022, 01:11:09 pm
content: Quote from: Zero999 on July 26, 2022, 11:40:14 amThank you wery mutchI will try that soon. I will post the results

author: ElectricPower
author_rank: Frequent Contributor
timestamp: July 26, 2022, 01:59:50 pm
content: Quote from: Zero999 on July 26, 2022, 11:40:14 amThat worked fineI have moved the resistor. My only problem now is that the meassured voltage from power supply shows 0.1V less than actual voltage on the voltmeter. Som ideas how to fix that?See image below.

author: mariush
author_rank: Super Contributor
timestamp: July 26, 2022, 02:17:38 pm
content: You can tweak the resistors to change that ratio and get a more precise value..Basically what happens: the microcontroller has an ADC which converts a voltage to a number... if it's a 10bit ADC that means a value between 0 and 1023.If the microcontroller uses the input voltage 3.3v as voltage reference, that means 0...3.3v on that input pin gets converted to 0..1023.So for example, you have 9v into that resistor divider 30k / 2k results in 0.536v - the ADC reads this and converts into a number between 0..1023 ... 0.536 x 1024 / 3.3 = 166.32 or 166.The microcontroller then multiplies this number by a constant inside to get to 8.90 or whatever it shows on the 7 segment display.You can't change that constant inside the microcontroller, but what you can do is tweak the 30k resistor value a tiny bit so that instead of 0.536v the micro may get 0.55v instead, and then multiplied by that constant, you would get the proper 9v on the display.So you could add a small potentiometer in series with that 30k resistor and tweak it a bit and see what happens.

author: Zero999
author_rank: Super Contributor
timestamp: July 26, 2022, 02:58:03 pm
content: Quote from: mariush on July 26, 2022, 02:17:38 pmCorrection: a 30k and 2k potential divider will give an output voltage of1/16of the input, so 0.5625V out with 9V in.Quote from: ElectricPower on July 26, 2022, 01:59:50 pmIs that a constant 0.1V less i.e. an offset, or does it change with the input voltage i.e. it's a factor?Your reading is around 1% out, which is probably within the specification and is the sort of tolerance you should expect from a cheap panel meter.

author: mariush
author_rank: Super Contributor
timestamp: July 26, 2022, 04:22:08 pm
content: Thanks for the correction. I was too lazy and just googled "voltage divider online calculator" and entered the values... that's what it spit out.Besides the 1% precision of the resistors in the divider, it's worth pointing out that value could be wrong for other reasons.For example, the chip could have an internal voltage reference (ex 2.048v for a 10bit adc resulting in 4mV steps) or the chip could use the input voltage as a reference, hoping the linear regulator always outputs 3.3vWell, the output of the voltage regulator can drift with temperature of the chip and input-output voltage differential.If you power the thing with 9v, you'll have maybe 50mA or whatever it takes to light up all those segments so the chip will dissipate (9v - 3.3v) x 0.05A = 0.285 watts.The regulator getting warm may drift a bit from 3.3v.Also the microcontroller and the regulator will be packed between the pcb and the 7segment display so they'll be even warmer that way.It's possible, but unlikely, that they may have accounted for these temperatures, and calibrated the meter after a few minutes of operation, when the chips are warm(er) but I doubt it.I suspect it's just the low precision of the resistors, and the overall cheapness.

author: Zero999
author_rank: Super Contributor
timestamp: July 26, 2022, 06:39:14 pm
content: I wasn't just referring to the error of the potential divider, but the unit as a whole. It's built down to a price and the datasheet probably specifies that sort of accuracy. If it's changed since moving the resistor, it could be due to thermal drift from heating it up and it should settle back down after a few days or so.If this error is consistent and doesn't go on its own, then connecting a 3M resistor (go with 2M7 or 3M3, if 3M isn't available) in parallel with the 30k resistor will nudge its value down by about 1% and should correct the error. There appears to be enough room for another surface mount resistor to the left of the 30k, if there isn't space for it to be placed on top.