Converting 8VAC to 5VDC

Other Hardware General Electronics
1

For a smart doorbell project, I'm considering using a uninterrupted 8VAC supply to provide power to a SoC module and some relays.

A quick google came up with this:

https://www.ebay.co.uk/itm/L7805-AC-DC-DC-Step-down-Regulator-Power-Supply-Module-7V-36VDC-6-27VAC-to-5VDC/232022503091

Which seems to do the trick, but uses an L7805. Is this the best way to achieve what I need? I ask because I read that it's pretty inefficient. As an alternative I also spotted this:

https://www.ebay.co.uk/itm/AC-DC-Buck-Step-down-Voltage-Converter-3-3V-5V-6V-9V-12V-Rectifier-Filter-Module/253333543363

But partly as a learning exercise, I'm wondering if I can do better by building my own, based on this:

However I'm having some difficulty working out what I'll need.

At 8VAC, the resulting DC will be 8 x 1.414 ~ 11.31 - 1.4 ~10v. I was hoping to just use a buck after this point like this:

But do I still need the various capacitors? If so, how can I work out which ones?

Also can I use 1N4148 for the rectifier bridge?

2

A simple bridge rectifier, storage capacitor and linear regulator would work OK.

If you want higher efficiency a buck convertor is better, but a 'lashup' on a breadboard will almost certainly not work - the devices involve very fast high current switching and need a carefully designed PCB.

You can buy such premade devices capable of up to an amp or so at 5v very cheaply (~£1 ) on eg eBay, and I suggest you go that way.

You'd still need a rectifier and storage capacitor, of course

Allan

3

Based on your other thread all you want to power is an Arduino so you don't need much current. This 1A buck converter seems to be the ideal solution: plug and play.

If you want to experiment:
Bridge rectifier, gives you a 100 Hz half-sine voltage. Add a big cap (that module you linked to uses a 1000 uF cap) on the output to smooth it, then a linear regulator like indeed the good old 7805 with its own caps (330nF ceramic both sides iirc - check data sheet). That should also give you a rather clean supply.

Such a linear regulator may need a small heat sink; 200 mA is typical for an Uno so that's 0.2 * 5 = 1W of power dissipated.

4

Not sure what you mean by an "SoC module", but if it's anything like the typical Arduino [this is an Arduino forum, after all], then it probably has a built in, voltage regulator [could be Linear and it could be a Switch Mode Buck Converter].

So, you may be able to get away with rectifying and filtering that 8VAC with a Bridge Rectifier arrangement, and filter capacitor, like in that first link of yours [sans 7805], and just use the Arduino's regulator. Can't advise you further, because you didn't divulge what device you're using.

BUT, it depends on how much current that 8VAC transformer can put out, and what the over all current requirement will be for everything you want to drive [e.g. relays and such]. Also, transformers are notorious for having a higher open circuit voltage than the rated voltage [i.e. the *rated voltage* is at the *rated current*]. So, your 8VAC transformer might run at 8VAC when it's ringing the bell, but when the bell isn't ringing, that voltage could be more like 11VA, or more!

Also, if your application doesn't require much current, then why even be concerned about efficiency? I mean if we're talking about a different of fractions of a penny [assuming your using US currency wherever you are ;)], then what's the fuss?

You might get away with using 1N4148s, but being a beginner, I suggest not adding another failure mode variable and use 1N4004s [or 1N4005 or 1N4006 or 1N4007].

Well, it's late...so more tomorrow...unless someone else wants to take over :smiley:

5

Thank you for all the replies. To answer some of the questions:

I have a habit of wanting to "do things right", but also learn along the way. Although this has gotten me deep into many rabbit holes, I suppose part of my question is why I would want to use a regulator over a buck or vice versa. Hence although it might not make a difference in practical terms, having something that's as efficient as can be would make me feel warm and tingly inside.

I said "SoC module" because I'm still not sure what I'll be using - I'll probably settle on a Wemos D32, but it might be a D1 Mini Pro instead. I actually had a look at the schematics and I think they all use a ME6211 which operates on a voltage range of 2-6v which sounds too narrow for a rectified 8vac? Also looking around I can expect these to draw around 800mA peak since they'll be using the wifi a bit.

For context here's the current bell:

current bell.png
current bell.png964×266 11.7 KB

I've pulled out the multimeter and have found the following for the transformer:

  • The transformer idles at 9V
  • It's a DAT03A which is rated at 2A
  • When the button is pressed the voltage drops to 8.7V.

And for the chimes:

  • The label of the chime says AC 8V ~ 1A
  • The resistance across the AC chime terminals (I tested both also while button was pressed which I assumed shouldn't matter but did it anyway) is around 4 ohm
  • The voltage at across one of the chimes when the button is pressed is around 6.4V
  • I can't seem to measure the current with my multimeter (I later realised it only works with DC), but according to the above it draws around 1.6A (6.4v/4ohm), which sounds pretty high.

So unless my schoolboy testing is in error, it sounds like the transformer shouldn't be able to provide enough current for the two chimes let alone the wemos/relays. Am I missing anything?

current bell.png
current bell.png964×266 11.7 KB

6

Yes.

The supply is AC, so the impedance it sees in the chime depends both on it's dc resistance and it's inductance. I'll assume the AC is a pure sinewave.

  • ie R and 2pi F * L.

Where R is the dc resistance in ohms , F is the frequency of the supply, and L is the inductance in Henries.

I'll call this second ezpression ZL. The complex impedance of the inductive part.Still measured in ohms.... but.

You can't just add these quantities as you would with series resistors , since the current in the inductor has a phase advance of 90 degrees .

The actual expression for the addition is:

Z ( the complex impedance of the R and L in series ) = R + j ZL where j is SQRT(-1).

So you have to add them as two orthogonal vectors - ie

|Z| ( the magnitude of the complex impedance) = SQRT( R^2 + ZL^2))

You can then apply ohms law : I = V/|Z|

The phase angle of the current flowing compared to the applied ac voltage is given by

arctan ( ZL / R )

Allan

7

The bell is operated by electromagnet, basically a big inductor, current is dominated by the inductance, not
the DC resistance. Apply 8Vdc to the thing and it will carry much larger current and then overheat
and fry itself.

8

Thank you both for the responses. Interestingly I can connect a 9V battery to it but I suppose something else is going on there.

So if our conclusions are correct, does that mean I'm on a fool's errand trying to tap into the 8VAC? My circuit will sit next to the transformer so I will have access to 240v too and so could use a USB wall wart for my 5v.

It just feels like I'm giving up too easily :wink:

9

sshaikh:
... I suppose part of my question is why I would want to use a regulator over a buck or vice versa. Hence although it might not make a difference in practical terms, having something that's as efficient as can be would make me feel warm and tingly inside.

Well, there is what you might want vs what might be an industry practice, or the practice of most of the rest of us :wink:

If you want warm-and-fuzzy efficiency, then by all means, use a Buck Converter. But, generally, and this is somewhat subjective, if the current demand is less than 200 mA or so, then a Linear solution probably makes more sense.

BUT, if the current demand is, as you say, in the 800mA range, then a Switch Mode solution makes more sense.

sshaikh:
I said "SoC module" because I'm still not sure what I'll be using - I'll probably settle on a Wemos D32, but it might be a D1 Mini Pro instead.

Well, this is an Arduino forum, so I'm sure you'll forgive the fact that I have no experience with any of the platforms you mentioned.

sshaikh:
I actually had a look at the schematics and I think they all use a ME6211 which operates on a voltage range of 2-6v which sounds too narrow for a rectified 8vac?

Not at all. 8VAC will easily accommodate that voltage range, as long as the 8VAC source can supply the required current. But, of course, a regulator will be required.

sshaikh:
The resistance across the AC chime terminals (I tested both also while button was pressed which I assumed shouldn't matter but did it anyway) is around 4 ohm

What you tested is the DC resistance. AC resistance, or reactance is the quantity in play. That 100Ω on the label is, likely, the AC resistance at 50 or 60 Hz [depending on your locale]. 8V/100Ω = 800mA which is in the vicinity of 1A. AND, actually, the actual resistive quantity in question, here, is Impedance, which is a Vector summation of both pure resistance and inductive reactance. And, one would expect that 100Ω assessment to be the impedance.

My knowledge of AC is a bit rusty and my knowledge of electromechanical math and physics is just plain lacking, so I have no explanation for the 200mA disparity between the 1A rating and the 800mA math result.

sshaikh:
The voltage at across one of the chimes when the button is pressed is around 6.4V

Sounds like a lot of voltage is being dropped in the wires. A smaller gauge [fatter wire] will fix that.

sshaikh:
...but according to the above it draws around 1.6A (6.4v/4ohm), which sounds pretty high.

Yup -- 'cuz you're dealing with AC, not DC.

So, each bell has a 1A rating on the label [perhaps that's the maximum recommended current], an apparent actual 800mA current demand at 8VAC, and you're running two bells off a transformer rated at 2A max. That doesn't leave much room for an SoC module demanding peaks of 800mA. Though you can get a bit of relief using a Switch Mode DC-to-DC converter [SMC]:

If the SoC runs at 3.3V [most with WiFi, do], then if, say, the SMC has an efficiency of 85% [conservative], then:

ViIi = eVoIo where: Vi & Ii are the input voltage and current, Vo & Io are the output voltage and current, and e is the efficiency.

Vi = VAC√2 - 1.4 where: VAC is the the AC voltage of the bell transformer and 1.4 is the bridge rectifier forward voltage drop
thus
Ii = eVoIo / (VAC√2 - 1.4)= .85(3.3)800mA / [8.7(1.414) - 1.4] = 206mA

So, 0.80A*2 + 0.2A = 1.8A or 200mA shy of the max of 2A

This is the magic of power transfer!

Note: these calculations are based on a nominal current of 800mA, but with your present setup, that doesn't seem to be the reality. If there is only around 6.4V at the bell, then that's more like a current of 640mA. You do the math :wink:

10

ReverseEMF:
What you tested is the DC resistance. AC resistance, or reactance is the quantity in play. That 100Ω on the label is, likely, the AC resistance at 50 or 60 Hz [depending on your locale].

Eeek. That 100Ω is an artifact of the circuit drawing app I used - it was the default and has no basis in reality and my focus at the time was to draw up the circuit. So my lesson learned here is to never be flippant when providing these diagrams. That said, I don't think it changes the conclusions too much so thank you.

Not at all. 8VAC will easily accommodate that voltage range, as long as the 8VAC source can supply the required current. But, of course, a regulator will be required.

Can you elaborate please? My understanding is that after rectifying I'll end up with 10vdc (8 x 1.414 ~ 11.31 - 1.4 ~10v). Isn't that what would be fed into the ME6211?

Thank you for the rest of the explanation. Since I have no real information about the AC resistance of the chimes or the current they draw, I'm going to be a little conservative and assume the chimes /are/ drawing 1A, which won't leave much out of the transformer. As another engineering friend told me KISS is very important when getting stuff to actually work reliably so as excited as this thread has made me I think I'm going to wuss out and just use a USB wall wart with 240vac.

#sad

11

Probably wise if you just want to get something going. But better to choose a higher voltage wallwart ( 8-15v)
and feed it into the Vin of the arduino - it's internal high quality linear regulator will take care of the rest....

Allan

12

sshaikh:
Thank you both for the responses. Interestingly I can connect a 9V battery to it but I suppose something else is going on there.

That 9V battery - I assume a small block battery - can supply no more than 200 mA or so on a good day. It wont fry your bell or so.

I suppose what you have is the kind of bell where the attraction of the hammer to the chime also breaks the contact, causing the oscillation of the hammer and the ringing sound. That will work on both AC and DC indeed, and in AC the additional inductance effect limits the current. With your 9V battery it's the internal resistance of the battery.

That also means that you will have a very dirty power supply. Peaks every time the coil makes and breaks; almost certainly no flyback diodes either, so any power supply you connect to it has to deal with at least the voltage variations (those flyback diodes you may add yourself, don't know how and if it will affect the ringing).

As you have mains available, a USB power supply is the quick and cheap solution here.

ReverseEMF:
so I have no explanation for the 200mA disparity between the 1A rating and the 800mA math result.

When it comes to electrical appliances I see that rated currents are routinely overstated: if you use that amperage to calculate wattage, you often end up 20-50% higher. Or maybe that's simply peak current rating (when e.g. a fan motor just switches on) vs. average power rating.

13

allanhurst:
Probably wise if you just want to get something going. But better to choose a higher voltage wallwart ( 8-15v)
and feed it into the Vin of the arduino - it's internal high quality linear regulator will take care of the rest....

I'm still a little confused about this.

The Ardiuno Uno seems to use this lp2985-33dbvr which takes in 16V which sounds quite handy.

However the ESP32 board I'm looking at has a ME6211 (https://wiki.wemos.cc/_media/products:d32:sch_d32_v1.0.0.pdf) which seems to take in 6v max, which I presume means the 8-15v is out?

Why would I go for a higher voltage anyway? For more options later? If it is handy to have then I can externally step down the input to 3.3v for the ESP32 (using what? A lp2985-33dbvr?).

I also need 5v to feed the relays, though, so if I stuck to the 5v looking at the schematic or board photo (https://wiki.wemos.cc/products:d32:d32) I can't tell how I would provide that 5v to the board itself - there doesn't seem to be a 5v in pin?

14

Just use a USB adapter then. 5V for the relays & the ESP32 (it'll be stepped down to 3.3V).

15

sshaikh:
For a smart doorbell project, I'm considering using a uninterrupted 8VAC supply to provide power to a SoC module and some relays.

Should be OK.

sshaikh:
A quick google came up with this:

Which seems to do the trick, but uses an L7805. Is this the best way to achieve what I need? I ask because I read that it's pretty inefficient.

Ten volts in, five out. Wastes as much power as what it feeds, uses. Better not to get things any hotter than you can manage to avoid.

sshaikh:
As an alternative I also spotted this:

Includes bridge rectifier, and interestingly, the cheapest of those you cited. Looks good. Can't see the capacitor rating but probably adequate as claimed.

sshaikh:
But partly as a learning exercise, I'm wondering if I can do better by building my own, based on this:
How to convert 6V AC to 5V DC - Electrical Engineering Stack Exchange

Bad idea. Very bad. If you are referring to the switchmode design shown, such regulators are non-trivial to construct and operate stably - and you do not want it to misbehave in any fashion. Common discussion here, best leave it to proven designs, not to mention that it is cheaper to do so.

sshaikh:
At 8VAC, the resulting DC will be 8 x 1.414 ~ 11.31 - 1.4 ~10v. I was hoping to just use a buck after this point like this:

Would work, but easier to use the one with the bridge and capacitors built in. Note that the rectified voltage will peak at 10 V, but drop between cycles as the capacitor discharges. However this design should have relatively low "dropout" voltage (and better than the 7805), so apart from thermal efficiency, this is an advantage of the switchmode.

sshaikh:
But do I still need the various capacitors? If so, how can I work out which ones?

Use the complete module.

sshaikh:
Also can I use 1N4148 for the rectifier bridge?

Absolutely not.

sshaikh:
I said "SoC module" because I'm still not sure what I'll be using - I'll probably settle on a Wemos D32, but it might be a D1 Mini Pro instead. I actually had a look at the schematics and I think they all use a ME6211 which operates on a voltage range of 2-6v which sounds too narrow for a rectified 8vac

That regulator is intended to regulate your 5 V supply (like USB) down to 3.3 V. It is never intended to be supplied from 10 V.

sshaikh:
Also looking around I can expect these to draw around 800mA peak since they'll be using the WiFi a bit.

800 mA sounds excessive, even as peak. 300 mA more likely as I recall for the ESP8266.

sshaikh:

  • The transformer idles at 9V
  • It's a DAT03A which is rated at 2A
  • When the button is pressed the voltage drops to 8.7V.

Door chime transformers are generally rated for intermittent use only, but the design with separate primary and secondary tends to be a sort of "choke transformer" limiting its output current.

sshaikh:
And for the chimes:
The label of the chime says AC 8V ~ 1A

About right - for intermittent use only.

sshaikh:
The resistance across the AC chime terminals (I tested both also while button was pressed which I assumed shouldn't matter but did it anyway) is around 4 ohm

And as others have explained, the inductance in addition, will limit the current to 1 A at 8 V.

sshaikh:
The voltage at across one of the chimes when the button is pressed is around 6.4V[/li][/list]

Given a few more ohms in the wire.

sshaikh:
I can't seem to measure the current with my multimeter (I later realised it only works with DC), but according to the above it draws around 1.6A (6.4v/4ohm), which sounds pretty high.

No, add the inductance, less than 1 A.

sshaikh:
So unless my schoolboy testing is in error, it sounds like the transformer shouldn't be able to provide enough current for the two chimes let alone the WeMOS/ relays. Am I missing anything?

Yep. 1 A each chime, two of them is 2 A, the WeMOS will not draw too much.

sshaikh:
Interestingly I can connect a 9V battery to it but I suppose something else is going on there.

What sort of 9 V battery? Did it do anything at all?

sshaikh:
I'm still a little confused about this.

Mmmm.

sshaikh:
The Ardiuno Uno seems to use this lp2985-33dbvr which takes in 16V which sounds quite handy.

It may be rated at 16 V, but without a proper heatsink, will be useless at anything above 8 V.

sshaikh:
However the ESP32 board I'm looking at has a ME6211 (https://wiki.wemos.cc/_media/products:d32:sch_d32_v1.0.0.pdf) which seems to take in 6v max, which I presume means the 8-15v is out?

You provide this with 5 V - five Volts - from a regulator, such as the second, switchmode, module you cited. It is designed for five Volts.

sshaikh:
Why would I go for a higher voltage anyway? For more options later? If it is handy to have then I can externally step down the input to 3.3v for the ESP32 (using what? A lp2985-33dbvr?).

Not sure what you mean.

sshaikh:
I also need 5v to feed the relays, though, so if I stuck to the 5v looking at the schematic or board photo (https://wiki.wemos.cc/products:d32:d32) I can't tell how I would provide that 5v to the board itself - there doesn't seem to be a 5v in pin?

It is labelled "USB".

So, you use this regulator module to develop stable 5 V from your 8 V transformer.

Note that now, all the 8 V AC devices must be connected separate to the 5 V system, controlled by relay contacts (or SSRs with optocouplers or such).

16

@Paul__B thank you for the comprehensive and deep response. I've since settled on using a "5v USB charger" solution.

I'm considering using a micro USB breakout board (like this) to get the 5v out, but I'm a bit puzzled as to how to feed this into the board without having to go back through its usb port.

Schematic is here: https://wiki.wemos.cc/_media/products:d32:sch_d32_v1.0.0.pdf

I could use a microusb plug (like this) or (maybe inadvisable?) stick a wire on the USB socket on the board (like this?), but I was wondering what the the VIN label on the Power schematic was for (I can't see it anywhere else).

On the other hand, I notice from the same schematic that VBUS terminates at the pin 15 (USB) - would it be safe to power the rest of the system (essentially two relays) from this pin instead?

17

How about just sticking it into the USB socket of your WeMOS? That's what that socket is for.

18

I need 5v for the circuit too so was asking how to access it before/after the wemos socket.

Unless I'm missing something obvious? A USB splitter or something?

19

Doesn't it have a 5V pin connected directly to the USB? May have a different name such as Vin.

20

I did notice that the vbus label terminates at pin 15, but this is the first time I've looked at these type of schematics.