Hello

I'm trying to build a rectifier to power our IoT device which would be deployed in the field where there is no DC voltage available. I only have a 120VAC power supply.

I'm using bridge MDB6S to rectifier the AC voltage. Please refer to the attached picture for the circuit diagram. Then I'm using voltage divider to get 15V across the load resistor. The voltage divider is between 30K and 4.7K. The smoothing capacitor is connected across the 4.7K resistor.

https://www.digikey.com/product-detail/en/on-semiconductor/MDB6S/MDB6SFSCT-ND/3137112

When I don't connect anything to the load resistor (4.7K), I measure 15V by a multimeter.

But when I connect our device across the load resistor, it only reads 7.8V. The device pulls 0.3 to 0.4 A of current. Keeping this in mind, I now connected a 470uF capacitor across the load resistor (4.7K). It still does the same.

Any suggestions?

Mod eidt: Direct to mains / transformerless power supply schematic deleted - JohnInTX

 

Playing on the power line is dangerous.
You should get a low voltage power supply.
A voltage divider only works when there is no load. Applying "our device across the load resistor" changes the resistor ratio.
I don't know what you are powering but you should be using a "wall wort" or "phone charger" type power supply.

 

But when I connect our device across the load resistor, it only reads 7.8V. The device pulls 0.3 to 0.4 A of current

Ohm law:
I = V/R
I = 110V/(30k + 4.7k) = 0.00317A
You need to lower your resistors value to supply 0.3-0.4A
Resistors value of 300 and 47 will source 0.317A
This method is very inefficient and dangerous, use a 'wall wart' instead

 

Hello

Thank you so much for your reply. That was the initial plan. But out device has an ADC on the line that power's it. Let me explain a little.

I'm trying to build a sensor that can measure AC voltage. Our device has ADC's ports which can read DC voltage. My idea was to convert AC voltage using a bridge rectifier and use voltage rectifier to bring the measurable voltage below 40V (Max ADC rating) then measure it using ADC"s.

We normally power the device via a 12V car battery. The device power line is also an ADC. This way we are constantly monitoring the power to which the device is connected to. That's why I'm using this sensor to power the device. This way the device gets power and also senses the AC voltage it's connected to.

Is there any way that I can still power the device this way?

Also, I now understand that the voltage divider is changed when I connected the device across the 4.7K resistor. Will it be safe if I take the device's resistance into account and modifier the circuit?

 

It's dangerous to deal with unisolated main's voltage.
I suggest you use an isolation transformer to isolate the line, otherwise your chances of damaging you equipment and/or electrocuting yourself are quite high..

 

Thank you all for the replies.

One follow up question:

If I use a standard AC adapter and power the device. Can the above-mentioned circuit use to sense the AC voltage?

TO be clear, Can I connect the positive terminal of the smoothing capacitor to the ADC line of our device and ADC ground to the negative terminal? This way the device tells me whats the AC voltage it's connected to?

Is this safe?

 

For a few US dollars you may want to just buy a few of these. 2x AC Voltage Sensor Module Voltage Transformer Module ZMPT101B 250V. AGoogle of AC voltage sensors should bring up a dozen hits, additionally a Google of AC Voltage Transducers will yield even more hits. Do just want to know if AC line voltage is present or do you want to measure it getting an accurate number?

Ron

 

Find a wall wart that has an AC output. For example, one rated at 120 VAC input and 12 volts 1000 ma output.
The 12 VAC output of the wall wart will be directly proportional to the AC input voltage, and you can rectify and filter the 12 VAC to get about 16 VDC.

 

Direct to mains, transformerless power supplies are forbidden by the TOS/UA for safety reasons. I've deleted the original schematic but will leave the thread open for now. Please continue beginning with a proper power or isolation transformer.
Thanks.

 

Thank you all for the replies.

One follow up question:

If I use a standard AC adapter and power the device. Can the above-mentioned circuit use to sense the AC voltage?

TO be clear, Can I connect the positive terminal of the smoothing capacitor to the ADC line of our device and ADC ground to the negative terminal? This way the device tells me whats the AC voltage it's connected to?

Is this safe?

Is it safe? Yes, usually.

Will it work? Maybe, depends on the kind of adapter you have.

I favor adapters with safety marks, genuine safety marks if possible

That means that I don't have to worry much about safety and can just get on with the interesting part of the project.

The adapters I have around here can be classified (very broadly) as:

• Transformes that plug into the wall and provide isolated lower voltage AC out.

• Transformer power supplies that plug into the wall and provide isolated and unregulated DC out. <== you probably want this one.

• Regulated, isolated power supplies with DC outputs.

If you use either of the first two types you can sense the AC line voltage. Keep in mind that there will be voltage measurement errors because the input-to-output voltage ratio is not absolutely constant, for the most part varying as a function of load, though unless you are doing something that requires high accuracy this is probably not going to be a problem.

(You can stop reading here but I just want to include a standard ground leakage safety test)
Reference: https://www.chromausa.com/applications/leakage-current-test/


The test is performed with the unit under test ungrounded. The bottom of the 1500 Ω resistor and 0.15 uF capacitor go to earth and an AC voltmeter is connected across the combination so you can read the voltage drop across the impedance. I am pretty sure this capacitor is the right one for 60 Hz AC for 50 Hz, you might use .18 uF or something close to it.

The probe is put in contact with the unit being tested, in this case, the output leads of the power supply or transformer, and the voltage drop across the resistor and capacitor noted.

As I recall, in the United States, Underwriters Laboratories specifies a limit of 500 microamps leakage to ground for most consumer products.

Personally, I just grab a 1k resistor and measure the voltage drop across it. If it the voltage is over 500 mv I don't use the device I am testing. Toss it or scrap it for wire and parts, but don't use it.

It is a good idea that in operation your circuit is well earthed and fused (most wall warts have a fuse somewhere).

 

Hello All,

Thank you so much for your responses. To answer one of your questions.

I want to monitor the voltage of a heater module. The heater is powered by plugging into a standard wall mount connection. The heater displays the voltage to which it's connected to on an LCD screen but I want to access this data remotely. Hence this setup. I want to know the voltage value. It doesn't have to be accurate. +5 or -5 voltages is fine too.

After reading all your comments, I have altered the schematic. Please refer to the attached doc.

The step-down transformer that I plan on using is https://www.mouser.com/ProductDetail/Bel-Signal-Transformer/CL2-25-24?qs=/ha2pyFaduhcNXvlILAyimwF1kKa1dl7RneqqZtIUkuogd36lQeeqw==

The bridge is https://www.digikey.com/product-detail/en/on-semiconductor/MDB6S/MDB6SFSCT-ND/3137112

Please let me know if this is safe. I'm sorry for asking stupid questions. I worked more on DC circuits and my primary work is writing firmware. I'm new to AC circuits(course work knowledge only). Please bear with me.



Note: The 30K and 4.7K resistors on the ADC circuit have a power rating of 2W. They won't burn at 120V.

Also, thank you so much.

Varun R

 

When you form a voltage divider using resistors you get a good reference voltage. In your case you're referencing 15 volts DC. But keep in mind what happens when you put two resistors in parallel. Suppose you have two 100Ω resistors in parallel. The overall resistance changes to 50Ω. So your voltage divider will change your output voltage. Your 4.7KΩ resistor is being placed in parallel with a load of some unknown resistance value. Assume just for the sake of discussion your load is also 4.7KΩ. When you form the new resistance you end up with 2.35KΩ. You can see how that will change things.

Also consider the voltage drop across a diode. Typically 0.6 to 0.7 volts. In a bridge rectifier 1.2 to 1.4 volts is dropped.

If you need 15 VDC then start with a transformer rated to give you about 12 volts. Why 12 volts? Because that's AC voltage. When rectified and filtered the value will go up by 1.414 times the rated voltage. RMS voltage is 12 volts. Peak voltage is 16.97 volts when filtered. BUT WAIT! I forgot to downgrade the voltage from the rectifiers. 12 - 1.4 = 10.6 volts. So 10.6 X 1.414 = 14.9884 volts. That's pretty darn close to 15 volts. In fact, I'd happily call it 15 volts DC.

As others have already said, working with mains is dangerous (and a forbidden topic). Use of an isolation transformer (a "one to one" transformer) will isolate you from the mains voltage, but you're still trying to get rid of a whole lot of un necessary voltage. Best to start off with a 12 volt transformer. From there you have a low voltage to work with. On top of that - you get as many amps as the transformer is capable of. And placing a load across the final rectified and filtered voltage won't cause you the headaches you're getting from trying to use a voltage divider. They DO have their purposes, but not as a power supply.

 

Hello Tony,

Thank you for your reply. I'll address each paragraph of your reply separately.

I realize it now. Hence I changed my circuit. I have attached the new schematic to my reply above and to this as well. I got rid of the voltage divider and replace it with a stepdown transformer with a turn ratio of 4.7:1. I connected the device in series with a protective resistor. There is no voltage divider anymore. This should be safe right? I used the isolation step down transformer and got rid of the voltage divider. I'm only dealing with 24VAC from the transformer then 120 before.

I need any voltage between 12V to 32V DC for this device. Since the minimum voltage available is 110V AC here, I tunned it to give 15V for that voltage. If we have to connect it to 220 VAC then we get 30VDC which should still power the device. That was the idea.

Yes, I agree with you. After reading the above posts, I realized the mistakes I have made. Therefore I changed the circuit. I have attached this message.

Please let me know if the changes I made are good enough for my setup.

Thank you

Varun R

 

Where is your ground? (To what are the voltages going to be referenced?)

 

There is no voltage divider anymore. This should be safe right?

Yes, that's a lot safer AND much more reliable.

As for commenting on your diagram, sorry, I don't quite understand it. I'll do what I can but when I don't know - I'll proudly state that I don't know the answer. There's no shame in not knowing something. It's how we learn.

 

The fuse should be between the transformer and the BR (Bridge Rectifier). The capacitor should be across the + & - of the BR. Not sure why you want or need the 32Ω resistor though.

 

If you put the fuse in the transformer's secondary circuit your would have to rely on a fuse wire inside the transformer in case there is a problem with the transformer. Granted it is a Signal transformer and they are very good but even with Signal transformers I used to put the fuse in the primary before the power switch.

@Tonyr1084 Good catch about the 100 uf cap.

 

What exactly are you trying to do? If all you want to do is look for the presense of AC line voltage be it 120 VAC or 240 VAC they make opto-coupler chips expressly for this purpose. The H11AA1 opto-coupler comes to mind but there are likely a dozen others. Actually the chip I linked to is I believe obsolete but again there are a dozen others which do the same thing. Using one gets you live isolation and a simple output right to whatever your IOT device is. These chips all cost about $1.00 USD. Now if you just want to power this IOT device and only have 120 VAC available in the field why not just use a simple wall wart or common inexpensive DC power supply costing a few USD? You mention 12 VDC but do not mention the power needed as in 12 VDC at ?? Amps? 12 VDC line powered power supplies are inexpensive, efficient and safe.

Ron

 

Where is your ground? (To what are the voltages going to be referenced?)

The reference ground for the DC part of the circuit is the negative of the bridge rectifier. I know its a floating ground but I don't know how to fix this. Normally, When I work with DC circuits, I use pull up or pull-down resistors to fix this issue but I can't apply the same logic here. Any ideas?

Tony,
True. I will switch the fuse to the secondary circuit of the transformer. I initially put the capacitor across the resistor because of the voltage divider circuit. The capacitor that I have has a max voltage rating of 35V. At that time, the voltage across the bridge was 120 VDC. But It's just 24 VDC now so I'll switch the capacitor. Thank you

Ther output current of the transformer is 1.1A and the bridge can rectify 1A continuously. But I want to be a little cautious, Hence I'm using a 32ohm resistor to bring the current down to 0.75A. I know the device has unknown resistance, Hence I'm gonna test this circuit out in an hour so see if this works. If it doesn't then I'll remove the resistor. It's basically a protective resistor.

 

Hello Ron,

An Optocoupler will tell me if a line is active or not. I also need to know the voltage on the line. Not Very accurate but close to + or -5 V is fine as well. Hence II part of the circuit. I intend to connect it to the ADC and read the voltage. Then I'll work it backward and calculate the AC voltage present on the line. It will be a little off but that fine.

I can power it via wall wart but the problem here is that the power line of the device is also an ADC. If I use a Wall wart, then I won't be sensing the voltage to which the heater module is connected to, rather I will get a constant voltage which makes the ADC on the powerline worthless. Hence I'm going through this trouble to sense the AC voltage and power the device at the same time. Therefore the I part of the circuit.

I did mention it before. The device when connected to DC power supply pulls around 0.3A at 12V, that's around 4W. Sometimes, when the device doesn't have good reception, it pulls more because of the GPS and cellular modem on it. To be safe, I'm going for 6W power. Without load, there should be 0.75A of current in the circuit. The bridge rectifier and transformer (Secondary coil) can handle up to 1A of current. Nothing should burn up.

Please refer to the schematic attached to this message.

Thank you

Varun R