Power Supply for Chip Amp

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I've been reading quite a bit about chip amps lately and am very interested in putting one together. My electrical know-how is pretty minimal, but I feel fairly confident about my ability to put together an amp kit from instructions. However, I'm not entirely comfortable working with wall voltage for the power supply. Is there a way to power one of these from a wall wart or other enclosed external adapter? In addition to my lack of technical expertise, I find this solution attractive because I want to keep the amp enclosure as small as possible.

If this isn't a viable solution, could anyone recommend simple step-by-step instructions or a good power supply kit?

Thank you kindly in advance.
 
It's always a bit of a risk encouraging anybody to 'play' with mains voltages but having said that, IF you are careful and sensible, it isn't a great problem to build a PSU!

Have a read of this and the other Gainclone pages and get a good idea of what is entailed before you make a decision one way or the other! ;)
 
This might be a dumb question, but can you tell me what would be appropriate to use for a ground point in a wooden case?

Assuming that you are putting the PSU and amp in the same case, you will (should) have your power and signal star grounds as usual in the amp section. If you then want to connect the power star ground to mains ground, that would be done using a 'lifter' consisting of a cap and resistor, and that could be connected to the ground terminal of your IEC input socket, or the earth wire of a captive mains lead.

Most people do not seem to connect the power star ground to mains earth when they use a non metallic case.
 
One of those will only provide 18 VDC, barely enough for chip amp operation. The amp would need to be configured for single supply operation, which means an output coupling cap would be needed, unless you build a bridge amp. You could use two of the supplies for a +/- 18 VDC, closer to what you’d want. The 2 amp output current limit may not work well on musical peaks with some speakers though, so you’d probably need to put some large capacitors on the output similar to in the diagram that I posted. Probably better to take the plunge and build something. The wall wart I had in mind for the above diagram was Mouser P/N 507-XT1840, if you decide to go that way.
 
Probably better to take the plunge and build something.

I am leaning in that direction. I'm torn between my enthusiasm at having discovered this forum and my desire not to overburden everyone with really basic questions. Toward that end, I've purchased an introductory electronics book and have been making frequent use of the archives. Thanks so much for all the help so far.
 
kaos said:
I’ve seen some wall warts rated at 18 VAC at over 2 amps. If you half-wave rectify it you could get a split +/- 24 VDC supply, adequate for many chip amp applications. Just connect one lead to the gnd point and the other to the diodes.

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Um, am I missing something? How would you get +/- 24 VDC from 18 VAC with a basic half-wave rectifier? Wouldn't it be +/- 8.3 VDC max?
 
It’s a basic voltage doubler. One side of the trans winding is connected to gnd, the other through the rectifiers to their associated capacitors. Each half cycle alternately charges the positive and negative rails respectively. 18 V x 1.414 yields a bit over 25V/rail. Subtract a diode drop and it's just over 24 volts/rail.
 
kaos said:
It’s a basic voltage doubler. One side of the trans winding is connected to gnd, the other through the rectifiers to their associated capacitors. Each half cycle alternately charges the positive and negative rails respectively. 18 V x 1.414 yields a bit over 25V/rail. Subtract a diode drop and it's just over 24 volts/rail.


I'm still confused. I can see +/- 17.3 VDC, but where does the 1.414 come from? And I don't understand the .01 uF cap.

Don't mean to hijack...
 
The 1.414 is simply the relationship between the RMS voltage and the peak voltage for an AC sine wave. Transformers are always specified at their RMS value. To find the DC value with a diode/cap combo multiply the transformer voltage rating by the 1.414. This ignores factors like transformer load regulation, so the number is a simplified approximation of the real world value for a given situation.

The .01 uF cap helps to quench switching arcs that can induce a clicking sound in the amp when it's switched on or off.
 
kaos said:
The 1.414 is simply the relationship between the RMS voltage and the peak voltage for an AC sine wave. Transformers are always specified at their RMS value. To find the DC value with a diode/cap combo multiply the transformer voltage rating by the 1.414. This ignores factors like transformer load regulation, so the number is a simplified approximation of the real world value for a given situation.

The .01 uF cap helps to quench switching arcs that can induce a clicking sound in the amp when it's switched on or off.


Oh, ok. Thanks kaos. So, any ac wallwart would actually be rated for RMS?
 
Hi,
if you can source an 18Vac 2A plug top mounted transformer it can only be a 36VA transformer inside.
This limits the available maximum power to between 24W and 36W. If you power two channels from this 36VA, do not expect more than 20W/channel.
To achieve this you need +-20Vdc to +-22Vdc when outputting 20W into 8r0. The unloaded rails will be about 4 to 5Volts higher than this, so design for +-25Vdc. This is just about right for an 18Vac transformer. The half wave rectification will be less efficient than a dual secondary with bridge rectifier. This will show as more ripple on the supply lines and more droop (sag) in supply voltage as the power output rises.
It will work, but choose fairly efficient speakers to go with this set-up.
 
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