you raise a good point. I need to check that. I’m leaning towards ditching this doorbell transformer and getting an AC-AC adapter I think.
I think that was always the better option.
agreed. I just need to find a good one that’s available to me here in Canada
Here’s one from Amazon for $15.
Here’s the datasheet:
https://www.jameco.com/Jameco/Products/ProdDS/112336.pdf
Does that look ok?
Edit - added excerpt of electrical data from datasheet, to post. Moderator - BT
@Robert.Wall am I correct to say that particular transformer has ~17% regulation which would make it a good candidate?
I really should have searched first
If the transformer’s load is constant, your only concern is accurately knowing the output voltage with this load.
Regulation matters when the load changes and the resulting voltage change affects either this load or other apparatus fed by the transformer.
(Consider what would happen if your house was fed by a big version of your doorbell transformer: it would be 120 V when all your appliances and lights were on, and about 260 V when you had just one light on. Not a satisfactory state of affairs.)
Yes, the Jameco one looks OK - you’ll need to measure the output because the no-load voltage is still subject to the 5% manufacturing tolerance. It appears to be available from Amazon (Canada) for $17.
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Ordered one. Thanks!
@Robert.Wall random question… just poking around while I wait for that AC-AC adapter to arrive. I measured that doorbell transformer with a 10K load and the voltage remained at 20V… I had my daughter hold down the doorbell for 5 seconds and during that time the voltage dropped to around 14V. Is this just a crappy transformer all around? lol
I think it would be fair to say that it’s made down to a price and not up to a quality, in other words, only just good enough for the intended purpose.
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Sounds like a reasonable assessment 
So my AC-AC adapter arrived. I’m not getting the values I’m expecting so I’m goofing something up. I’m going to share a whole bunch of information and if you spot anything wrong, feel free to tell me I’m an idiot so I can fix it
Here’s the AC-AC adapter datasheet. I’ve extracted the specs I think are needed:
https://www.jameco.com/Jameco/Products/ProdDS/112336.pdf
I measured it with my scope:
With 117.4V input, it outputs 10.2 V RMS whether unloaded or connected to my measuring circuit. My measuring circuit looks like this:
link to circuit: http://tinyurl.com/yufs6emu
I’ve worked out my voltage calibration constant like so:
Voltage Divider:
Rd = (9960 + 997) ÷ 997 = 10.98996990972919,
VRef = 3.3 V,
CMax = -2048 to 2048 (12 bit ADS1015 that supports negative voltages and has a PGA. Voltage from ADC = ADC Value * (gain / 2047.0) - in this case the gain value is 4.096V
AC-AC Adapter:
Rt = 120 ÷ 10.2 = 11.76470588235294
voltage calibration constant = Rt × RD
10.98996990972919 x 11.76470588235294 = 129.2937636438728
Using 129.2937 for VCAL and 1.7 for PHASECAL I’m calling calcVI to measure 60 crossings with a 2000ms timeout.
typical output:
realPower: -736.009722, apparentPower: 731.292915, powerFactor: -1.006450, Vrms: 102.244487, Irms: 7.295094 samples: 1425, time: 570
I’m basically ignoring everything except for Vrms and Irms because (and correct me if I’m wrong) until I get those two values aligned with reality, there’s no point in looking at the other values. My Irms is very close to my test loads and I’m happy with it’s values. Vrms is low however. If I measure the voltage on the input side of the AC-AC adapter I see 117.4V but I only see ~102 reported in my code.
If I arbitrarily adjust the voltage calibration constant to 149.2 I see a voltage value more in line with what I see at the outlet, however I also see a power factor > 1 which I learned in this thread cannot be greater than 1.
realPower: 866.146270, apparentPower: 860.367767, powerFactor: 1.006716, Vrms: 117.071806, Irms: 7.349060 samples: 1402, time: 598
In an effort to confirm that I’m sampling properly, I decided to data log one of my calcVI calls. This is the data collected over 1/2 second of sampling.
and the raw data in CSV:
and here’s my datalog with the current readings charted as well if that’s useful info:
and a different scale for the current since it’s a 0.333V CT.
Thanks for taking your time to read my blog 
I appreciate it!
edit: Been wondering if PHASECAL may be messing with my values. I’m going to experiment with different values. Based on my timings, the time between reading the AC ADC and the CT ADC is 170us and each loop sample takes approx 360us.
ok so using VCAL of 149.2 (I took my calculated value of 129.2937 and slowly increased it until the Vrms looked close to my killawatt. Then I adjusted PHASECAL from 1.7 slowly down until I got to 1.2 and powerFactor hit 0.999 (my load is a purely resistive load)
reading from the software:
realPower: 861.102100, apparentPower: 861.531286, powerFactor: 0.999502, Vrms: 117.195045, Irms: 7.351260 samples: 1430, time: 508
My killawatt readings. Note that my clamp is on the 10X loop of my test probe so values should be 10X what the killawatt reads:
I’m fairly happy with the numbers, just a little unsure if I’ve arrived at them the proper way 
Those are the numbers you should have used in
but this always assumes your multimeter is 100% accurate. Agreeing with your Killawatt on one range doesn’t necessarily mean it’s right on every range - this goes for the resistance ranges too. But this (2.4%) doesn’t explain the 15% difference in the calculated Vcal values.
Is 997 Ω the value you measured for this one resistor? If so, it’s not the number to put in
because it has 2 × 10 kΩ in parallel with it (YES! - the 3.3 V supply is a short circuit to a.c.), so the value is nearer 831 Ω if you ignore the 10 µF, which makes quite a difference. And when you don’t, the 10 µF is 318 Ω at 50 Hz, 265 Ω at 60 Hz, so you have two steps of division there, the 10 kΩ and the 997 Ω, loaded by the 265 Ω of the capacitor and the 5 kΩ of the two 10 kΩ in parallel
Time to do the maths again. 
At the end of the day, if the power agrees with your supplier’s meter then as I’ve written before, it’s correct even though it might not be (because they never admit their meter is wrong unless it’s badly wrong).
It should never exceed 1.0 But the same voltage is not used for the two calculations, and the phasecal algorithm slightly distorts the voltage value, hence you can get a p.f. greater than 1.0 If you have loads none of which have a poor power factor, you’ll probably never notice the difference. If however you have a very low power factor load on its own, then you should adjust for the highest p.f. even though it is greater than 1. (If you really know what you’re doing, adjust for zero p.f. with only a capacitor as the load - but DON’T try it unless you are absolutely certain it will be safe, it’s inherently dangerous unless the capacitor is the right sort with the right voltage and current ratings.)
Thank you for spotting my error Robert. I’m getting there thanks to your help. I can now calculate the VCAL multiplier now and am getting much better Vrms values:
Voltage Divider:
Rd = (9960 + 997) ÷ 997 = 10.98996990972919,
VRef = 3.3 V,
CMax = -2048 to 2048
Parallel divider:
Rd = (9960 + 9960) ÷ 9960 = 2
Rd = 10.98996990972919 + 2 = 12.98996990972919
(this aligns with your value of 831Ω)
Rd = (9960 + 831) ÷ 831 = 12.985559566787
AC-AC Adapter:
Rt = 120 ÷ 10.2 = 11.76470588235294
voltage calibration constant = Rt × RD
12.98996990972919 x 11.76470588235294 = 152.8231754085787
Any advice on how I factor in the 265Ω of the capacitor into those calculations? I envy your EE math abilities. I’m just a lowly software engineer over here
So being a software engineer, I decided to see if ChatGPT could help me with an answer here. I’ll be curious to see if I ended up in the right ballpark.
Here is the advice ChatGPT gave me when I fed it my calculations and asked how to factor in the capacitor.
So I took my Rd value calculated above and the capacitor impedance and used the formula ChatGPT gave me:
Ztotal = 1 / ((1/12.985559566787) + (1/265))
Ztotal = 12.3789641827
VCAL = Rt * Ztotal
VCAL = 11.76470588235294 * 12.3789641827
VCAL = 145.634872738
which is pretty damn close to the value I ended up on with trial and error (149.2) … is that a fluke or did the machines guide me correctly?
This is what the equivalent circuit (drawn in LTSpice) looks like:
The ‘gotchas’ are you need to lump together R4 & R5 in parallel when you calculate the ratio of that stage of division, and you need to put the capacitor and those two in series (5265 Ω) in parallel with R2 when you calculate the ratio of that stage.
Don’t forget your 120 V wasn’t 120 V at the time you measured the output of the adapter.
thank you!
I think I’m pretty happy with how things are working. My numbers are aligning to my meters enough that I’m satisfied! Can I buy you a beer?
Because of component tolerances (probably 1% on the resistors, 10% on the capacitor) the calculated Vcal is really only there as a confidence check. If you have a calibrated voltmeter & ammeter, then those are usually what you’d want to calibrate against.
Don’t worry about that, I’m happy you’ve got a solution to suit your needs.
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My meter is an EEVBlog Bryman BM235 (True RMS) … It’s not a top of the line meter by any means but I think it’s good enough for my purposes here. I know I’ll never get dead on but I just wanted to try and be as accurate as I can be. I’ve learned a tremendous amount working on this project so far (and still going!)
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