Hi guys hope everyone is doing good

I have a question regarding an AC-AC, 15 VAC, 670mA power supply. I have put together the MFOS experimenter board synth and everything is working ok. The synth requires a dual supply range of +12/-12 Volt but can also be run at 9V+-. As Ray suggested on the website for people not used to working with mains power they should use 2 * 9v batteries connected in series with a center tap for ground, so that is what I have been using so far, however I would like to make the synth run off a mains powered supply. So, I have purchased an AC-AC, 15VAC power supply from rapid, https://www.rapidonline.com/ideal-power-77db-10-15m-15vac-670ma-plug-in-ac-ac-linear-psu-18-1499 , I have also purchased a L7915/L7815 regulated circuit board from ebay that has already been assembled to convert the AC signal into a dual +- supply rail. The question I would like some help with is:
The AC-AC power supply only has a positive and negative pin, where does ground go? The reason I ask is the L7915/L7815 board has 3 input pins which are labelled as AC, GND and AC. Am I right in thinking that the positive and negative connections of the supply would go to both the AC inputs, but, like I say, where is the ground connection then?

The other question I would also like to ask is; could I put a voltage divider (2 resistors in series), at the end of the regulated board to make it +- 12v instead of +- 15v?

P.S I am aware that the MFOS website has a section to build a wall wart power supply for the synth but I thought I could save some time and just buy the regulated supply part already assembled.

Any help would be greatly appreciated.

Thanks

 

Here's a drawing of how to get + and - out of a single AC winding.

The other question I would also like to ask is; could I put a voltage divider (2 resistors in series), at the end of the regulated board to make it +- 12v instead of +- 15v?

No, it won't be stable under load. If you want 12V, buy 12v regulator chips. Or get an LM317 and an LM337 and make them adjustable.

 

Here's a drawing of how to get + and - out of a single AC winding.

No, it won't be stable under load. If you want 12V, but 12v regulator chips. Or get an LM317 and an LM337 and make them adjustable.

Hi #12, thanks for the reply.
Im not sure how many windings are in the wall wart itself, but, I assume that there has to be 2 in order to take the voltage from 240v to 15v. As for the schematic you supplied, I'm having trouble understanding where this 3rd connection would come in. would AC be supplied on the positive line and the negative line goes to ground? If so, what is the other AC connection for on the regulated board?

I'm probably missing something really simple somewhere.

 

You're missing how to figure out the plug on the end of the wall wart wire.

 

Your probably missing a few concepts:

1) You can half wave or full wave rectify an AC signal
2) For audio applications you would probably want to use a full wave rectifier.
3) The DC RMS current will be much less than the AC rated current and different for full and half wave rectifiers.
4) Rule of thumb for linear rectifiers is at least 1000 uF/amp for the capacitor.
5) The DC voltage from an AC rectifier is √2 * the AC voltage or the peak of the AC voltage, so 15 VAC gives you 15 *1.4 VDC
6) You sometimes have to subtract one or 2 diode drops from the calculated output.
7) For +- outputs, you usually use a center tapped transformer or a dual secondary transformer with a full wave bridge rectifier.
8) So a 35-0-35 xformer gives you about +-50 VDC
9) The regulator needs at least 3 V more DC voltage at the input.
10) The input can't exceed about 40 V for the regulators your using.
11) The transformer regulates at about 10% at full load.
12) You can have a transformer rating greater than the current required for the circuit. Your house has say a 200 A service and the appliance draws what it needs.
13) Look here for rectifier use: https://www.google.com/url?sa=t&rct...k5oUMa3KwjgqL3-YA&sig2=dhe18WlAxVgJdJ35rxO7Ug
14) Your regulators probably need heat sinking.

So, for a full wave +-12 VDC output you need a center-tapped transformer that has an AC output greater than (12-1.2+3)/√2 or 9.85V, so a 12-0-12 transformer would work. So would a 2x12 V winding transformer. Now, I see how people get confused.

Your likely on the other side of the pond, but this http://www.jameco.com/z/JE215-Dual-...tml?CID=GOOG&gclid=COnJnbOeutECFcKEswodM6kNYA would work in the US - 120 V mains.

I know you can get wall warts with a DIN connector that have multiple DC outputs.

 

Hate to say it but you have probably bought the wrong power supply. Hard to generate +/- 12V DC with a single 15V AC input.

Also, the L7815 will give an output of +15V and the L7915 gives an output of -15V which is higher than you wanted. If you want 12V positive and 12V negative you should have used L7812 and L7912.

Then you could use a circuit similar to the attached to generate +/- 12VDC. Not ideal, because it only uses half-wave rectification but OK for small output currents.

Edit: Whoops! Electrolytic capacitors drawn wrong way up in circuit!!

 

You're missing how to figure out the plug on the end of the wall wart wire.

Thanks for that! I think I found the answer. Negative goes to ground and the positive must have to be split to go to both AC inputs.

 

Basically what I drew then but with a few more caps and diodes!

 

Hate to say it but you have probably bought the wrong power supply. Hard to generate +/- 12V DC with a single 15V AC input.

Also, the L7815 will give an output of +15V and the L7915 gives an output of -15V which is higher than you wanted. If you want 12V positive and 12V negative you should have used L7812 and L7912.

Then you could use a circuit similar to the attached to generate +/- 12VDC. Not ideal, because it only uses half-wave rectification but OK for small output currents.

Edit: Whoops! Electrolytic capacitors drawn wrong way up in circuit!!

Yeah I know, I had to buy a 15VAC because I couldn't find a 12VAC on rapid. I thought I might be able to reduce the voltage somewhere. Apparently, the circuit can run on a 15VAC, but, some biasing resistors will have to be changed which I'm not going to bother doing now, so I will just build one from scratch, just got to order more parts now

When I do build it with the L7812/L7912, can I still use the 15VAC wall wart? My understanding is the regulators need a higher voltage than the desired regulated voltage. Is that correct?

 

Basically what I drew then but with a few more caps and diodes!

Yes lol

 

Your probably missing a few concepts:

1) You can half wave or full wave rectify an AC signal
2) For audio applications you would probably want to use a full wave rectifier.
3) The DC RMS current will be much less than the AC rated current and different for full and half wave rectifiers.
4) Rule of thumb for linear rectifiers is at least 1000 uF/amp for the capacitor.
5) The DC voltage from an AC rectifier is √2 * the AC voltage or the peak of the AC voltage, so 15 VAC gives you 15 *1.4 VDC
6) You sometimes have to subtract one or 2 diode drops from the calculated output.
7) For +- outputs, you usually use a center tapped transformer or a dual secondary transformer with a full wave bridge rectifier.
8) So a 35-0-35 xformer gives you about +-50 VDC
9) The regulator needs at least 3 V more DC voltage at the input.
10) The input can't exceed about 40 V for the regulators your using.
11) The transformer regulates at about 10% at full load.
12) You can have a transformer rating greater than the current required for the circuit. Your house has say a 200 A service and the appliance draws what it needs.
13) Look here for rectifier use: https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0ahUKEwjR0urWm7rRAhWr24MKHfK9A4gQFggdMAA&url=http://www.hammondmfg.com/pdf/5c007.pdf&usg=AFQjCNFPNgWFEfkADk5oUMa3KwjgqL3-YA&sig2=dhe18WlAxVgJdJ35rxO7Ug
14) Your regulators probably need heat sinking.

So, for a full wave +-12 VDC output you need a center-tapped transformer that has an AC output greater than (12-1.2+3)/√2 or 9.85V, so a 12-0-12 transformer would work. So would a 2x12 V winding transformer. Now, I see how people get confused.

Your likely on the other side of the pond, but this http://www.jameco.com/z/JE215-Dual-...tml?CID=GOOG&gclid=COnJnbOeutECFcKEswodM6kNYA would work in the US - 120 V mains.

I know you can get wall warts with a DIN connector that have multiple DC outputs.

Thanks for such a detailed answer, much appreciated. Could you elaborate a bit more on point 2, point 4, specifically 1000uF/AMP, point 5, where did you get 1.4 from and what is VDC and a bit more info on point 11 would be good.

Thanks again

 

Don't think point 2 is valid really. For any application it is best to use a full wave rectifier. But not possible in this case.

Point 4: Because you are using half wave rectification, the smoothing capacitor only gets charged every 20ms. (once per cycle instead of twice per cycle). So the capacitor value needs to be big enough to maintain its voltage for long enough. How much current will you actually be needing to draw?
The circuit in #7 shows 3x 3000uF on each supply. I think this is overkill for your application. About 1000uF will be sufficient. The working voltage of these capacitors needs to be at least 25V to allow for the peak voltage of the 15V AC (15 X 1.414).

Point 11: This means that the 15VAC output of the power supply will drop about 10% at full load but as you are only wanting 12VDC on the output of your regulators, this will not be a problem here.

 

Don't think point 2 is valid really. For any application it is best to use a full wave rectifier. But not possible in this case.

Point 4: Because you are using half wave rectification, the smoothing capacitor only gets charged every 20ms. (once per cycle instead of twice per cycle). So the capacitor value needs to be big enough to maintain its voltage for long enough. How much current will you actually be needing to draw?
The circuit in #7 shows 3x 3000uF on each supply. I think this is overkill for your application. About 1000uF will be sufficient. The working voltage of these capacitors needs to be at least 25V to allow for the peak voltage of the 15V AC (15 X 1.414).

Point 11: This means that the 15VAC output of the power supply will drop about 10% at full load but as you are only wanting 12VDC on the output of your regulators, this will not be a problem here.

When you say not really possible in this situation, do you mean because I need a dual supply rail?
How did you work out that the Cap will only charge every 20ms? I am not quite sure how much current will be drawn from the supply. I have checked over all the descriptors for the project and I can't seem to find the answer. Apparently, most synth modules on the site will draw between 20 and 50 mA and I have 6 modules installed.

So would I need 2 * 1000uF caps for the supply, (one for positive and one for negative)? Also where did you get the 1.414 from?

 

Correct, because you need a dual supply rail.
With a half wave rectifier the capacitor only charges once per cycle. At 50Hz, that's one cycle every 20ms.
So the total current could be as much as 300mA. Actually a bit high. Your AC supply is only rated at 650mA. Still, see how it goes!
Yes you will need two 1000uF capacitors. One for each supply.
The 1.414 (or square root of 2) is the factor by which the peak value of an AC voltage is greater than its average (actually RMS) value.
The capacitors will charge up to the peak value (approximately).
For more on AC and rectifiers read An Introduction to Rectifier Circuits on this site.

 

Negative goes to ground and the positive must have to be split to go to both AC inputs.

I feel a need to polish up the grammar. When you have an AC supply and you nail one wire to ground, the other wire isn't positive, it's AC. The second wire contains both positive and negative and they take turns. Each of them gets its turn 60 times a second (in USA). That's why you can get both positive DC and negative DC out of a single wire. You just have to herd them into their own capacitors with a couple of diodes.

I really hope I'm making sense to your point of view.

 

Also where did you get the 1.414 from?

1.414 is approximately the √2; 1.4 was good enough.

it's the conversion between 0 to peak and RMS, so 120 Vac is (120 * √2) zero to Peak or about 170*2 or 340 V peak to peak (p-p).

120 Vac = 120*√2 (sin(2*π*f)); where f is frequency

 

Correct, because you need a dual supply rail.
With a half wave rectifier the capacitor only charges once per cycle. At 50Hz, that's one cycle every 20ms.
So the total current could be as much as 300mA. Actually a bit high. Your AC supply is only rated at 650mA. Still, see how it goes!
Yes you will need two 1000uF capacitors. One for each supply.
The 1.414 (or square root of 2) is the factor by which the peak value of an AC voltage is greater than its average (actually RMS) value.
The capacitors will charge up to the peak value (approximately).
For more on AC and rectifiers read An Introduction to Rectifier Circuits on this site.

Brilliant thanks for all the help, I really appreciate it. Just one more question; If the supply is rated at 650mA, why would 300mA be a problem?

 

I feel a need to polish up the grammar. When you have an AC supply and you nail one wire to ground, the other wire isn't positive, it's AC. The second wire contains both positive and negative and they take turns. Each of them gets its turn 60 times a second (in USA). That's why you can get both positive DC and negative DC out of a single wire. You just have to herd them into their own capacitors with a couple of diodes.

I really hope I'm making sense to your point of view.

Much appreciated, thank you for clearing that up and yes it makes complete sense. I was referring to them as positive and negative because that's how its labelled on the plug.

 

1.414 is approximately the √2; 1.4 was good enough.

it's the conversion between 0 to peak and RMS, so 120 Vac is (120 * √2) zero to Peak or about 170*2 or 340 V peak to peak (p-p).

120 Vac = 120*√2 (sin(2*π*f)); where f is frequency

Thanks for that Ill be honest i'm a little rusty on the maths side of things its been over 12 years since I did A level electronics. I really need to brush up on the maths.

 

Just one more question; If the supply is rated at 650mA, why would 300mA be a problem?

The wall supply is basically a transformer. The maximum current it can supply is set by two things:

• The thickness of the wire in the secondary of the transformer - it's current carrying capacity. Also the resistance of this wire.
  
• The magnetic field created by the current in the windings. The iron core of the transformer must be big enough to contain this field.

Now, although 300mA x 2 is only 600mA and your power supply is rated at 650mA so it seems to be good there is the rectifier to consider.
Due to the way the rectifier diodes work - you need to read that article on rectifiers - the current in the transformer is not continuous but will be a short peak every cycle. The current during this peak will be many times the continuous current to make up for the rest of the cycle when there is no current at all. So:

• The resistance of the wire will have a much greater effect - leading to voltage drop and heat generation
• The transformer core may saturate - have more magnetic field in it than it can handle - which will also reduce the output.

So it will not be able to produce as much power as if it states. But, as I said, better try it and see!