EmonTx Shield + SCT013-030

h23934 · 2023-03-07T13:47:14.738Z

I bought the emontx-shield and soldered as needed (very hard for beginner I found). Using with SCT013-030
According to GitHub - openenergymonitor/emontx-shield: emonTx Arduino Shield I should remove the 33ohm resister for ct1, I could not remove it. Would soldering a path between 33ohm pins be enough to sidestep it and work?
Or using piece of wire soldered onto the holes on either side of the 33ohm resister, would that work?

h23934 · 2023-03-07T13:54:41.836Z

With it I’m trying to use this code:

github.com

openenergymonitor/emontx-shield/blob/master/firmware/Shield_CT1234_Voltage_SerialOnly/Shield_CT1234_Voltage_SerialOnly.ino

/*
  EmonTx CT123 Voltage Serial Only example
  
  Part of the openenergymonitor.org project
  Licence: GNU GPL V3
  
  Author: Trystan Lea
*/

#include "EmonLib.h"

// Create  instances for each CT channel
EnergyMonitor ct1,ct2,ct3, ct4;

// On-board emonTx LED
const int LEDpin = 9;                                                    

void setup() 
{
  Serial.begin(9600);

This file has been truncated. show original

Changing the voltage calib to:
ct1.voltage(0, 268.97, 1.7);
Seems to be getting me the right voltage

I cannot however get correct current (testing with 1800w kettle) using calib:
ct1.current(1, 30.0);
My multimeter shows 8A current, but the ct Irms reading gives me below:
0.18
0.19
0.15
0.17
0.18
0.19

Any tips or advice appreciated

Robert.Wall · 2023-03-07T13:54:42.187Z

h23934:

Would soldering a path between 33ohm pins be enough to sidestep it and work?

No, both these things would have the effect of making the reading zero under all circumstances.

You will need a medium-sized soldering iron so that you can heat up both ends of the resistor at the same time. When the solder melts, use the iron to wipe the resistor off its pads.

And don’t be afraid to let the iron get really hot before you try. Generally, you will do more damage to the circuit board and the components when the soldering iron is not hot enough - because you apply the heat for a much longer time.

h23934 · 2023-03-07T13:56:52.654Z

thank you. So I have to remove the 33ohm and then use any normal wire to bridge the ends where the resistor sat?

Robert.Wall · 2023-03-07T13:57:06.181Z

What you’ve done by not removing the 33 Ω burden is to give the c.t. two burdens, one internal and the 33 Ω. The effect will be to reduce the sensitivity significantly.

h23934 · 2023-03-07T13:58:56.557Z

Okay let me get that soldering iron out again. As a solder beginner, I struggled a lot to solder the shield together. But let me get that 33ohm out now that I know it cannot be sidestepped. Thanks a lot.

Robert.Wall · 2023-03-07T13:59:01.159Z

h23934:

So I have to remove the 33ohm and then use any normal wire to bridge the ends where the resistor sat?

No, remove the 33 Ω. The c.t. already has a burden inside it, you don’t need a second one. As I wrote above, a link will ensure you always read nothing - always zero.

h23934 · 2023-03-07T14:00:51.337Z

Noted, after removing the 33ohm resistor, do I need to bridge the points on the shield were the resister once sat, to ensure the path is connected with no resister (any plain wire?)

Robert.Wall · 2023-03-07T14:13:40.607Z

Go back and read what I’ve written all the way down this thread. Also, look at the ‘Learn’ section to see how the c.t and burden are connected in relation to the input to the ADC.

h23934 · 2023-03-07T14:33:23.010Z

You were spot on. I got the 33ohm off, and now the Irms values look great! Hovers around 0.2A no load, then 8.8A when the kettle comes on. When the kettle switched off, back to 0.2A

Thank you so much!