i have tried every combination i can think of to create a trace (stub) from between the caps to the audio ground and it cant be done. to create the trace (stub) i have to reduce the size of the GNDA to isolate the trace (stub) and when i do the outer reg caps wont connect. even changing the name of the GNDA to GND does not help. there is just not enough room on the board. if i had tree layers to play with then yes no problem but i'm using the free version of Eagle.
the only option i can think of is to use two wire pads and take a piece of insulated wire between them. Eagle will throw a wobbly but i will know what to do when the times comes to solder the boards up. messy but it will work.
the only option i can think of is to use two wire pads and take a piece of insulated wire between them. Eagle will throw a wobbly but i will know what to do when the times comes to solder the boards up. messy but it will work.
heheheeee, its free and if i can use it it must be easy
looks like lots of coffee time as i'm going to remove the V+/- rails from the reg/rec section and tidy things up.
A real gound plane is actually a return route for currents flowing in another layer.
The return route at DC is via the shortest low resistance piece of plane irrespective of the Flow route of the trace.
But as frequecy has become AC then impedance rules. The return route will take the lowest impedance part of the plane. As frequencies get higher the lowest impedance becomes dominated by the inductance between the flow and return routes. The lowest impedance is when the return route is very close coupled to the flow route of the trace, i.e. the return route follows the trace in the other plane.
For this reason AC flow currents must NEVER pass across a gap/break in a return plane. That gap forces the current to find a different route than under/over the Flow trace. That different route will have a higher inductance and will also have become an EMI problem. It emits radiation and receives radiation that is minimised when the upper and lower traces are only separated by 8thou of insualtion.
The return route at DC is via the shortest low resistance piece of plane irrespective of the Flow route of the trace.
But as frequecy has become AC then impedance rules. The return route will take the lowest impedance part of the plane. As frequencies get higher the lowest impedance becomes dominated by the inductance between the flow and return routes. The lowest impedance is when the return route is very close coupled to the flow route of the trace, i.e. the return route follows the trace in the other plane.
For this reason AC flow currents must NEVER pass across a gap/break in a return plane. That gap forces the current to find a different route than under/over the Flow trace. That different route will have a higher inductance and will also have become an EMI problem. It emits radiation and receives radiation that is minimised when the upper and lower traces are only separated by 8thou of insualtion.
I think the regulator's ground plane should connect directly to the audio ground plane,
and not to the stub. Then both planes connect to the power net with the wire.
my day off so i have been out and about.
Andrew, i think i get what you mean. the cold audio signal will try and migrate to the hot via the pads as trace pads even though isolated will migrate. how does this work with off board power/regulation supplies as my brain is saying that what i have done is exactly the same as an off board reg/rec supply its just that its on the same substrate.
all this mixing of AC/DC really confuses me especially how connecting the +/- of the spare rectifier terminals together creates a GND. i'm used to working with vehicle electrics where everything is live (when needed) apart from when it gets back to the negative terminal due to it being flow and return just like switching on a light bulb. its the ground in electronics that i have a hard time understanding as my brain says that its not a ground its a return/com.
rayma, if i connect both reg and audio ground planes then i wont need a wire and use a stub from the middle of the cap T?
Andrew, i think i get what you mean. the cold audio signal will try and migrate to the hot via the pads as trace pads even though isolated will migrate. how does this work with off board power/regulation supplies as my brain is saying that what i have done is exactly the same as an off board reg/rec supply its just that its on the same substrate.
all this mixing of AC/DC really confuses me especially how connecting the +/- of the spare rectifier terminals together creates a GND. i'm used to working with vehicle electrics where everything is live (when needed) apart from when it gets back to the negative terminal due to it being flow and return just like switching on a light bulb. its the ground in electronics that i have a hard time understanding as my brain says that its not a ground its a return/com.
rayma, if i connect both reg and audio ground planes then i wont need a wire and use a stub from the middle of the cap T?
I think that there is an error in the PSU part of the schematic/layout. The positive bridge rectifier connects to the negative side of the capacitor.
The current flow of the input signal goes:
L/R+IN -> Pot -> C1-L/R -> R2-L/R -> GND -> PGND -> R -> AGND -> L/R-IN.
Could/should R2L/R and R3L/R not be connected to the input ground? Each channel could have its own input ground and resistor connecting the input grounds to the main ground (not PGND). This should also help reducing hum caused by ground loops.
The current flow of the input signal goes:
L/R+IN -> Pot -> C1-L/R -> R2-L/R -> GND -> PGND -> R -> AGND -> L/R-IN.
Could/should R2L/R and R3L/R not be connected to the input ground? Each channel could have its own input ground and resistor connecting the input grounds to the main ground (not PGND). This should also help reducing hum caused by ground loops.
Yes, the regulator and audio ground planes can be made into one plane.
Just keep the rectifiers/filter capacitors out of it, except via one direct connection.
The junction of the two filter capacitors is indeed a common, but many call it a ground.
well spotted, the whole regulation is upside down
R2/3-L/R are just drains to ground.
R2-L/R (and C1-L/R) form a high pass filter on the input (hot) to restrict at 20hz. with the amp being for headphones there is absolutely no point in going any lower than this as A,you cant hear it and B,you cant feel it. hooking up a CD/Digital file player will be ok but hooking up say a Phono stage could result in too much extension of the headphone diaphragm due to wow and flutter/rumble, ok so it wont stop WF/R at higher than 20hz but most dangerous are the ones below as they are the big 'cone flappers'.
R3-L/R is the drain for the feedback network.
i'll do that.
i still dont understand fully where the COM voltages go once they get to the rectifiers. ooohhh wait i think i have just sussed it out, the two REG caps are the equivalent to a battery.
looks like a i have a lot of work ahead. getting there though
Electronic circuits don't have a drain. Think loops, supplies and returns. The source, connected to the amp, has a supply and a return (GND). The signal current flows in a loop from the source though the interconnect, potentiometer, capacitor, resistors and returns back though the interconnect. The same goes for the amp feedback and the outputs. Positive/negative rails (capacitors) -> opamp -> Load and feedback -> Ground (capacitors).
The trick is to organise the grounds so that you don't mix the different currents though a ground plane. And, keep the supply and return loops small.
Could you please post a new schematic?
The trick is to organise the grounds so that you don't mix the different currents though a ground plane. And, keep the supply and return loops small.
Could you please post a new schematic?
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Yes exactly, and you could connect batteries where the capacitors are, instead.
Except there is noisy current flowing through the capacitors.
i think i have just had a eureka moment. is this why the supply voltage goes up as effectively i'm connecting two battery's in series and the rectification network side is acting as the cells of the battery's, then the regulation is there to reduce and smooth the voltage?
The rated transformer AC secondary voltage is an "rms", a kind of average. The secondary sine voltage has a
peak of 1.414 times that average. That peak voltage is used to charge the filter capacitors through the rectifiers.
http://www.circuitsgallery.com/2012/11/diode-bridge-rectifier-circuits.html
http://www.electronics-tutorials.ws/diode/diode_6.html
Or, you could think of this as the raw capacitors providing energy for the regulator to generate a high quality
clean DC voltage, a more general case. Some power supply circuits can make the output voltage higher than
the raw input voltage. The input power is the actual constraint on what can be done.
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i understand that thanks
does the specified voltage of a capacitor have anything to do with how quick it will charge given a certain
voltage input e.g. will say a 100v rated cap charge just as quick as a 30v rated cap given 25v supply?
The rated voltage is a limit before there could be damage, but has no direct affect on circuit operation.
The capacitance C is the key rating. This, and external circuits with resistance, etc. will determine
the behavior of the voltage across the capacitor. The basic "law" of a capacitor is C=Q/V, where C
is defined as the capacitance, V is the voltage across it, and Q is the charge on one of its plates.
The Q is proportional to the number of excess electrons on one of the capacitor's plates.
The other plate has an equal "deficiency" of electrons.
Foe example, a battery with voltage V, connected to a resistor R and a capacitor C, all in series.
If the capacitor is initially discharged (with 0V), and the connection is made, current will flow in the
series circuit loop "through" the capacitor until it is charged enough to equal the source voltage V.
Then current stops flowing in the circuit loop, since the voltage across the resistor is 0V, and I=V/R.
Current does not actually flow through C, but the charge being stored on its plates makes it appear so.
The rate of charge accumulation (current) in the capacitor varies as it stores more charge on its plates.
The equation for the voltage across the capacitor is: V(capacitor) = V (1 - exp (-t/RC) ).
So, the voltage across the capacitor starts at zero, and rises up to V, at first quickly, and then more slowly.
The loop current is: (V/R) (exp(-t/RC) ). It starts at (V/R), and decays to zero, at first quickly, then more slowly.
http://www.electronics-tutorials.ws/rc/rc_1.html
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