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!

Thanks for that Marley, very informative. I have read the article on rectifiers and understand it, the only thing I don't quite understand is the part where you say: "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".
Could you explain how the current can be continuous and not continuous? My understanding is the capacitors turn the peaks into a continuous voltage ,albeit with ripple. Also, why is the current during the peak many times that of the continuous current?

 

"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".

I changed a few words and I think it would make more sense:

"the AC current in the transformer is not continuous (sine wave), It will have a peak every cycle. The current during this peak will be many times the rectified continuous current to make up for the rest of the cycle"

We can think of AC voltage and current in a number of ways:
1) p-p or peak to peak
2) In terms of v(t) and i(t)
3) In terms of RMS

The instantaneous current is time varying; just like the voltage is: e.g. V(t)=169*sin(wt); I could be I(t) = 2*sin(wt)

an RMS voltage is the same as what DC voltage would cause ohms law to be valid for the AC waveform. Say R=V/I ; For R-1and I=1, the RMS voltage has to be 1 for the AC waveform and the DC voltage of the same value.

Many meters assume a sine wave input, precision rectify and filter, and multiply this value by a fudge factor so the answer of rms is only valid for a sine wave between 2 frequency limits. TRMS usually have the ability of DC+AC or just the AC (time varying) component.
By definition, the TRMS value of 10 VDC is 10 V and the TRMS of 10 V RMS is 10 V RMS.
Your probably better reading the DC+AC as Direct Coupled and AC coupled.

 

why is the current during the peak many times that of the continuous current?

The average value of the peak current has to equal the average value of the DC output current.
This means the peak current is many times the average current, since the peak current occurs for only a short period of time at the peaks of each cycle.

 

the only thing I don't quite understand is the part where you say: "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".

I have tried to illustrate this with a simple diagram. This shows a half wave rectifier. There are better drawings of this online and in textbooks. The red line is the capacitor voltage (the output voltage). Between cycles the capacitor voltage drops because it is discharging into the load (providing current into the load). This is why there is a "ripple voltage".

The rectifier diode only conducts when the AC voltage is higher than the capacitor voltage. As you can see, the rectifier diode current only happens for a short time (during the shaded part). When the rectifier is not conducting, there is no transformer secondary current (because the transformer secondary and the diode are in series). The slope of the red line is proportional to the capacitor charging (or discharging) current. Steeper the line - more current. So the short charging current is much larger than the longer discharging current.

Another way of looking at this is that the energy supplied to the capacitor must, on average, be the same as the energy removed from the capacitor if its average voltage (over many cycles) remains constant. All the energy must be supplied during the shaded time. The energy is removed all the time. Because the time is much shorter, the current must be much higher.

The current coming out of the capacitor (flowing into the load) is continuous - although not constant, the current supplied by the transformer stops when the diode is not conducting so is not continuous.

The capacitor is like a tank of water with a small leak (a small current) that runs continuously. You have to maintain the water level by tipping a bucket of water (a big current) in every so often. Hope this helps!

 

I have tried to illustrate this with a simple diagram. This shows a half wave rectifier. There are better drawings of this online and in textbooks. The red line is the capacitor voltage (the output voltage). Between cycles the capacitor voltage drops because it is discharging into the load (providing current into the load). This is why there is a "ripple voltage".

The rectifier diode only conducts when the AC voltage is higher than the capacitor voltage. As you can see, the rectifier diode current only happens for a short time (during the shaded part). When the rectifier is not conducting, there is no transformer secondary current (because the transformer secondary and the diode are in series). The slope of the red line is proportional to the capacitor charging (or discharging) current. Steeper the line - more current. So the short charging current is much larger than the longer discharging current.

Another way of looking at this is that the energy supplied to the capacitor must, on average, be the same as the energy removed from the capacitor if its average voltage (over many cycles) remains constant. All the energy must be supplied during the shaded time. The energy is removed all the time. Because the time is much shorter, the current must be much higher.

The current coming out of the capacitor (flowing into the load) is continuous - although not constant, the current supplied by the transformer stops when the diode is not conducting so is not continuous.

The capacitor is like a tank of water with a small leak (a small current) that runs continuously. You have to maintain the water level by tipping a bucket of water (a big current) in every so often. Hope this helps!

Thanks for that, I understand it completely now. Although, I've always understood it as the diode would conduct the whole time the AC signal was in the positive portion and the capacitor filled in for the negative.

 

The average value of the peak current has to equal the average value of the DC output current.
This means the peak current is many times the average current, since the peak current occurs for only a short period of time at the peaks of each cycle.

Thanks for the reply. I never realised the diodes only conduct for a sort time, Ive always understood it to be conducting the entire positive half and then stop conducting when it goes past the zero crossing point into the negative cycle.

 

I changed a few words and I think it would make more sense:

"the AC current in the transformer is not continuous (sine wave), It will have a peak every cycle. The current during this peak will be many times the rectified continuous current to make up for the rest of the cycle"

We can think of AC voltage and current in a number of ways:
1) p-p or peak to peak
2) In terms of v(t) and i(t)
3) In terms of RMS

The instantaneous current is time varying; just like the voltage is: e.g. V(t)=169*sin(wt); I could be I(t) = 2*sin(wt)

an RMS voltage is the same as what DC voltage would cause ohms law to be valid for the AC waveform. Say R=V/I ; For R-1and I=1, the RMS voltage has to be 1 for the AC waveform and the DC voltage of the same value.

Many meters assume a sine wave input, precision rectify and filter, and multiply this value by a fudge factor so the answer of rms is only valid for a sine wave between 2 frequency limits. TRMS usually have the ability of DC+AC or just the AC (time varying) component.
By definition, the TRMS value of 10 VDC is 10 V and the TRMS of 10 V RMS is 10 V RMS.
Your probably better reading the DC+AC as Direct Coupled and AC coupled.

Thanks for all the help. Could you explain this, "The instantaneous current is time varying; just like the voltage is: e.g. V(t)=169*sin(wt); I could be I(t) = 2*sin(wt)".
I'm familiar with ohms law but, I don't understand where 169 came from? and what is (wt)?
I also don't understand the last paragraph. Is TRMS the average voltage/current over time?

 

Thanks for the reply. I never realised the diodes only conduct for a sort time, Ive always understood it to be conducting the entire positive half and then stop conducting when it goes past the zero crossing point into the negative cycle.

That's true for a resistive load.
But for a capacitor load, the diode stops conducting as soon as the sinewave voltage becomes less than capacitor voltage.
Since the capacitor charges to near the peak of the sinewave voltage then the diode will only conduct for the sort period when the sinewave peak voltage exceeds the capacitor voltage.

 

One thing at a time:

and what is (wt)

I cheated. It's actually ωt or omega * t which is called the radian frequency. ω=2πf or 2 * 3.14159... * f
t is time and in the units of seconds. In circuit analysis, it's just a lot easer to use ωt.

The sin() function varies from +-1, so 169 * +1 is 169, so 169 * sin(ωt) is a waveform that varies from -169 to +169 and repeats forever. I just picked 169 as that waveform would have an RMS value of about 120 V.

See: http://www.wolframalpha.com/input/?i=root+mean+square+169*sin(t)

Wikipedia defines RMS as: The RMS value of a set of values (or a continuous-time waveform) is the square root of the arithmetic mean of the squares of the values, or the square of the function that defines the continuous waveform

https://en.wikipedia.org/wiki/Root_mean_square

You have to integrate over a full cycle. The 1/(T2-T1) * integral from t1 to t2 of f(t) is just the average value of the function f(t)

for a single number is √(mean(n)^2), So, for 2; the value is 2 and for -2, the value is 2.
The square root of the square, basically makes the negative portion of the sine wave become positive.

The average of a sine wave is zero, because it contains an equal number of positive values and negative values. With RMS, we mathematically flip the negative portion and averages the resulting all positive waveform.

 

One thing at a time:



I cheated. It's actually ωt or omega * t which is called the radian frequency. ω=2πf or 2 * 3.14159... * f
t is time and in the units of seconds. In circuit analysis, it's just a lot easer to use ωt.

The sin() function varies from +-1, so 169 * +1 is 169, so 169 * sin(ωt) is a waveform that varies from -169 to +169 and repeats forever. I just picked 169 as that waveform would have an RMS value of about 120 V.
Thanks for clearing some of that up. I tend to get lost with all the mathematical terminology so I'm going to have to brush up on that. I will definitely have a read over the links you provided. Thanks you.
See: http://www.wolframalpha.com/input/?i=root+mean+square+169*sin(t)

Wikipedia defines RMS as: The RMS value of a set of values (or a continuous-time waveform) is the square root of the arithmetic mean of the squares of the values, or the square of the function that defines the continuous waveform

https://en.wikipedia.org/wiki/Root_mean_square

You have to integrate over a full cycle. The 1/(T2-T1) * integral from t1 to t2 of f(t) is just the average value of the function f(t)

for a single number is √(mean(n)^2), So, for 2; the value is 2 and for -2, the value is 2.
The square root of the square, basically makes the negative portion of the sine wave become positive.

The average of a sine wave is zero, because it contains an equal number of positive values and negative values. With RMS, we mathematically flip the negative portion and averages the resulting all positive waveform.

 

That's true for a resistive load.
But for a capacitor load, the diode stops conducting as soon as the sinewave voltage becomes less than capacitor voltage.
Since the capacitor charges to near the peak of the sinewave voltage then the diode will only conduct for the sort period when the sinewave peak voltage exceeds the capacitor voltage.

Ah right ok that makes a lot more sense now, thanks. The only other question I have is, wouldn't all of these power supplies have a capacitive load? otherwise you couldn't smooth out the ripple.

 

Some types of load don't care if there is ripple. For example simple battery charging or running a DC motor.

True, the capacitor helps to smooth out the ripple. That's why they are often called smoothing capacitors. For more sensitive electronic loads this is not good enough. Further regulation of the voltage is required. Using IC regulator as in your original post is one way of doing this.

 

Ah right ok that makes a lot more sense now, thanks. The only other question I have is, wouldn't all of these power supplies have a capacitive load? otherwise you couldn't smooth out the ripple.

You can also have an L-R filter. These are used primarily in tube/valve circuits or no filter as Marley states. Switching regulators may include some L-R filtering. Using a switching regulator for audio generally means to keep the switching frequencies higher than what you can hear and the L-R filter may be better at filtering ripple in say in the MHz range.

 

You can also have an L-R filter. These are used primarily in tube/valve circuits or no filter as Marley states. Switching regulators may include some L-R filtering. Using a switching regulator for audio generally means to keep the switching frequencies higher than what you can hear and the L-R filter may be better at filtering ripple in say in the MHz range.

Great, thanks for clearing that up. I now have a much better understanding of these power supplies.

 

Some types of load don't care if there is ripple. For example simple battery charging or running a DC motor.

True, the capacitor helps to smooth out the ripple. That's why they are often called smoothing capacitors. For more sensitive electronic loads this is not good enough. Further regulation of the voltage is required. Using IC regulator as in your original post is one way of doing this.

Thanks for all the help Marley, I really appreciate it. I have a much clearer understanding now

 

Well, I finally built the regulated power supply, but, it's not working. I powered it up and the circuit was getting power but wasn't working properly and the regulators got absolutely boiling after 10 secs of trying it out, so I'm a bit nervous to plug it back in now in case it destroys the synth. I have been having a look on the net to see what the problem could be and I came across someone mentioning that most wall warts these days don't have a ground connection (centre tap), could this be the problem? I connected what is labeled as "neg" on the plug to the ground of the circuit. Could this be the problem? Shall I disconnect that one and connect it to the negative rectifier diode instead and have no ground going to the regulator circuit?

Any help would be appreciated as this is the last part of the circuit to be done and i'm getting tired of connecting croc clips to 2 9 volt batteries.

 

This is the power supply I bought. Its seems the output only has Live and Neutral outputs, (page 2). PDF below.

 

This is the power supply I bought. Its seems the output only has Live and Neutral outputs, (page 2). PDF below.

You need two of those if you want ±15Vdc.

 

You need two of those if you want ±15Vdc.

Ah no, really? What am I missing then? Ray says on the MFOS website to get a AC-AC Wall wart. It seems I've missed a vital feature of the AC wall wart, but what is it? ....A ground connection?

 

What am I missing then? Ray says on the MFOS website to get a AC-AC Wall wart. It seems I've missed a vital feature of the AC wall wart, but what is it? ....A ground connection?

Yes.
You cannot get separate ±15Vdc outputs from one two-wire AC source.
To get plus 15Vdc and minus 15Vdc (to ground) you need a separate 15VAC source to ground for each voltage.
This can be done with two separate AC sources, or a 30VAC center-tapped output transformer with the center tap going to ground.