I am trying to install the ring pro doorbell. I have an existing doorbell and wiring in place. I measure the voltage at the doorbell and it was reading 14V. I measure voltsge at the transformer and it was 22V. Wierd thing is the transformer is rated at 16V, 10VA. So two questions:1. Why is the transformer outputting 22V? I dont know how to change the meter to peak or rms, but I'm pretty sure it's set to rms because the outlet at my house reads 122V.2. Why is there such a big voltage drop at the doorbell?
Doorbell wiring
- Thread starter minesh21
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- Location
- Chapel Hill, NC
- Occupation
- Retired Electrical Contractor
The reading at the doorbell was measured with the button depressed?
petersonra
Senior Member
- Location
- Northern illinois
- Occupation
- engineer
eta: ooops misread the post
petersonra
Senior Member
- Location
- Northern illinois
- Occupation
- engineer
I think the 16V is the loaded condition and what you are reading is the no load voltage.
Can you sketch up the circuit and post it?
I dont know how it was wired. Just observing externally. There is a button, a chime and transformer. Typical wiring is most likely the case.
- Location
- Chapel Hill, NC
- Occupation
- Retired Electrical Contractor
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
gar
Senior Member
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- Ann Arbor, Michigan
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- EE
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.
.
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.
.
LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
Because, a the button, you're measuring the voltage with the chime in the circuit. This is like thermostat wiring without a C wire.
I'm really concerned about the depressed button. I "googled" but found nothing about cheering a doorbell button 
(Every DB transformer I ever checked had a higher voltage than nameplate in a no load mode)
(Every DB transformer I ever checked had a higher voltage than nameplate in a no load mode)
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.
The button wasn't depressed but I think it's the resistor that they placed in series at the button.
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.
petersonra
Senior Member
- Location
- Northern illinois
- Occupation
- engineer
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?
petersonra
Senior Member
- Location
- Northern illinois
- Occupation
- engineer
is that on the doorbell?
That's the backside of the doorbell pushbutton in the picture.
petersonra
Senior Member
- Location
- Northern illinois
- Occupation
- engineer
No idea what that is supposed to do.
However look at this youtube video
https://www.youtube.com/watch?v=3vaGGiJrE2k
LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
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.
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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