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Originally Posted by Dennis Alwon
What screws...If you measure at the chime between the Transformer Terminal and the front door terminal then you will not get the true voltage as you will only have the true voltage when the button is depressed. I have no idea what you may be reading thru the pro doorbell controller
The screws where the wires are landed at the button.

The ring isn't installed yet, just the existing circuit. Yes at the chime, there is a spike in voltage when the button is depressed but for some reason it goes back to zero even when the button stays depressed.

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Originally Posted by gar
180503-1619 EDT

minesh21:

You list yourself as an electrical engineer. You should be able to answer your own question.

It is most likely your meter in AC mode provides an averaged value over a number of cycles calibrated in RMS for a sine wave whether it is a true RMS meter vs a full wave rectified type meter.

For a sine wave the RMS value is about 0.707 the peak value, and the full wave rectified average value is about 0.636 of the sine wave peak. So a Simpson 260 actually measures 63.6 V DC average from a 100 V peak sine wave. The scale on the meter is drawn 1.112 times greater than its actual DC value.

On a non sine wave signal the Simpson will usually provide an incorrect RMS reading.

Your doorbell transformer is probably designed with a high internal impedance and rated for output voltage at full rated load.

Your open circuit voltage is 22 V and the transformer is rated 16 V at 10 VA. This implies a high internal impedance.

First, is your input voltage equal to the rated input voltage. Assume it is, then you tell us what you calculate the transformer output impedance to be assuming resistive.

Second, based on that calculated internal impedance, then what current do you expect the doorbell is drawing? How does that compare with actual doorbell current?

Third, based on your calculated internal impedance, then what is a calculated short circuit current? What is the measured short circuit current?

These comparisons my not be real close. You can go back and try to determine the actual AC impedances, and redo calculations and experiments.

.
I can do without your judgements and attitude. I stopped reading after your opening statement. Try working on your people skills and maybe people will listen to what you have to say.

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Originally Posted by LarryFine
Because, a the button, you're measuring the voltage with the chime in the circuit. This is like thermostat wiring without a C wire.
The button wasn't depressed but I think it's the resistor that they placed in series at the button.

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Originally Posted by Dennis Alwon
The reading at the doorbell was measured with the button depressed?

I was finally able to add a photo! Had some trouble doing this on my phone. Not sure if you can see the resistor added there...I'm not sure why this was added.

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Originally Posted by minesh21

I was finally able to add a photo! Had some trouble doing this on my phone. Not sure if you can see the resistor added there...I'm not sure why this was added.
Not a resistor. Is a diode.

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Originally Posted by petersonra
Not a resistor. Is a diode.
Whoops. Thanks for the clarification. But that looks like it was added separately. Can you tell me what its use is for in this situation?

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Originally Posted by minesh21
Whoops. Thanks for the clarification. But that looks like it was added separately. Can you tell me what its use is for in this situation?
is that on the doorbell?

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Originally Posted by petersonra
is that on the doorbell?
That's the backside of the doorbell pushbutton in the picture.

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Originally Posted by minesh21
That's the backside of the doorbell pushbutton in the picture.
No idea what that is supposed to do.

However look at this youtube video

10. That diode is there to allow electronic door chimes to complete the programmed music after the button is released.

You should read zero volts across a doorbell button when it's being pushed.

Again, the voltage you're reading across the un-pushed button is in series with the door chime.

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