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-   -   "rule of thumb" used to determine xxVA for a power supply's transformer? (http://www.diyaudio.com/forums/chip-amps/126658-rule-thumb-used-determine-xxva-power-supplys-transformer.html)

CopyofAudiˇfilo 20th July 2008 03:26 PM

"rule of thumb" used to determine xxVA for a power supply's transformer?
 
I've been reading and exploring not all, but most of the already posted threads, and every time a power supply is suggested for an amplifier it seems that there isn't any specific current (Ampere) limitation, just someone saying.... "you are using xxxxx chip, then a transformer with xxxxVA should be enough". I wonder if someone knows how to determine the maximum current an amplifier of xxW RMS needs, given the supply voltage (specified by chip's manufacturer)?

This is my first post and I have been reading for hours to search for an answer, but I haven't found it yet. If this subject is already posted and I have missed it, I'm truly sorry. Thanks.

AndrewT 20th July 2008 04:40 PM

yes,
many have a rule of thumb for guesstimating both the transformer VA and the output peak current.

The rule I use
Ipk~<=[Vsupply-Vloss] / Rload / 0.35
This peak current comes from the smoothing caps and decoupling caps, not from the transformer/rectifier.

VA>= 1 times maximum total output power to two times maximum total output power and >=100VA.
I avoid using very small transformer of <100VA for power amps due to the very high regulation these transformers have compared to toroids of 160VA and above.

woody 20th July 2008 04:45 PM

This isn't the real answer you asked for but.

The Dyna st400 amp a 200w per chanel amp used a 1.2k va transformer and the halfler 220 a 100w per chanel amp
uses a 460va transformer this is from memory and I may be
off a little off on those figures. Now I saw one poster on the
powersupply forum say ~1.6 x the va of your amp so for a
50w stereo amp at 8 ohms you would need a 160va transformer.
But supose you do this then deside to test this amp into 4 ohm
loads you may need a new transformer. Besides a larger transformer gives you better regulation.

CopyofAudiˇfilo 20th July 2008 11:59 PM

Results...
 
Thanks for replying so soon...

I have analyzed both equations. In order to compare both under the same conditions, I used an old car stereo with a HA13158 power amplifier. It has a 10A fuse and Vsupply = 13.8~13Volts. Using woody's equation, 20 W (13.2Volts) x 1.6 = 32VA ===> I=2.4A. As it is a 4 channel P.Amp., Itotal=Ix4=~9.6A (I suppose the rest is for internal electronics and lighting). So far, woody's approach fits quite well.

Andrew's equation it's a bit more complicated as I don┤t know wich parameters I have to use. I supposed that vsupply-Vloss = voltage applied to P.Amp.'s voltage supply terminals (after rectifiers and smoothing caps). Using HA13158 specs, Ipeak=4A and Output Power Max=34W, for 18V Psupply. So, for a 4Ohm load, Ipk=18/4/0.35=12.85A and ItotalPK=51.4A (4 channels). I'm sure I'm doing something wrong, so Andrew, If you read this again, please specify your equation's parameters and if it's possible, use the HA13158 specs (just search for the datasheet, it will be one of the 10 first results, if you can't find it, I can attach it next time)

Thanks a lot for your replies.Greetings.

AndrewT 21st July 2008 07:37 AM

Re: Results...
 
Quote:

Originally posted by CopyofAudiˇfilo
.... that vsupply-Vloss = voltage applied to P.Amp.'s voltage supply terminals (after rectifiers and smoothing caps). Using HA13158 specs, Ipeak=4A and Output Power Max=34W, for 18V Psupply. So, for a 4Ohm load, Ipk=18/4/0.35=12.85A and ItotalPK=51.4A (4 channels)
a supply rail of 18Vdc can never deliver 18Vpk to the load. Guess that 5V is lost through the wires and semiconductors and emitter resistors, .i.e.Vloss~5V
[18-5]/4/0.35~=9.3Apk
IF ALL 4 channels require the same peak at the same moment then the maximum instantaneous peak current drawn from the supply caps can be around 37Apk.

You can make the chipamps job much easier by using an 8ohm speaker. This halves the current to 4.6Apk. note this is very close to the chipamp specification limit.

or,
is this a single polarity supply?
Is this chip internally wired to operate bridged?

CopyofAudiˇfilo 21st July 2008 01:24 PM

2 Attachment(s)
Ha13158 Specs: The HA13158 is four-channel BTL amplifier IC designed for car audio, featuring high output and low distortion, and applicable to digital audio equipment. It provides 34 W output per channel,with a 13.7V power supply and at Max distortion.

-Is this a single polarity supply?===> It has Vcc and ground for power supply, not VCC and -VCC.

-Is this chip internally wired to operate bridged? ===> Do you mean BTL = Bridge-Tied Load?

I have attached Ha13158 datasheet. Greetings.

AndrewT 21st July 2008 03:35 PM

Hi,
no wonder Car Audio gets such a bad reputation.
You need to learn to read the spec sheet.
34W Pomax @ unspecified distortion. I suspect this is square wave output power.
20W max power @ 10% distortion.
~15W max power @ 0.1% distortion, into 4r0 from 13.7Vdc.
expect about 8W into 8r0 per channel.
Now look at the dissipation capabilities of the chipamp.
Pd=83W when fitted to an infinite heatsink. But only Pd=25W when fitted to that example heatsink, if you can keep the chip case temp down to 25degC.
Now look at the dissipation for one channel when delivering 10W output. How many channels could run simultaneously when fitted to your chosen sink at your expected operational temperatures?

8ohm is by far the better choice for speakers and you then have 4 by 8W into 8ohms if you can keep it cool.
Increasing the supply to may give as much as 15W into 8r0 per channel, but you have the dissipation problem to overcome.

CopyofAudiˇfilo 21st July 2008 08:11 PM

Thanks Andrew for those observations. I've never paid attention to that specification before. I am trying to figure out what that figure (Power Dissipation vs. Output Power) means but I think the ay to read it is this:

For 1 channel, for PT~=12W, the IC can deliver 10W of Power to a 4Ohm load. If I want more power delivered, PT decreases, but THD increases. Besides, If I use all 4 channels, PT is multiplied by 4 (assuming all 4 are the same drivers,linear addition for power?). So, if I want a decent operation I need to use that Power Amp. to deliver at its maximum 10W of power to only 2 channels for 4 ohm loads because its PT=26W (ideal conditions)? Is that assumption right?

I wonder where can I find a good book or application note expalining how to read specs.Can you recommend me some good reading material or it is just a matter of paying really good attention to datasheets?

Thanks for everything Andrew. I think that the past two days opened my eyes.I'm just trying to understand as much as possible before facing a real project an that car stereo showed me that comercial stuff sucks! I am now thinking that Tamb inside a car, with the stereo put inside a plastic box next to the drivers seat with an outside temperature of about 35 to 40 ║C and after some time playing music at its near maximum power, would be of about 50 ║C and its PT ~ <20W...so Po per channel is much lower than 10W if I'm using all 4 channels using a heatsink with the specified characteristics...Is that ok? I have trouble understanding why the PT vs Po curve decreases after 10W of Po. I assume is the THD that affects after those 10W (of Po), or is it something else? Thanks Andrew.Greetings.

e_fortier 22nd July 2008 12:46 AM

TNT Audio
 
Hi,

Have a look at TNT Audio and do not forget to read all 3 sections.

Enjoy :)

Crash course on power supply

CopyofAudiˇfilo 22nd July 2008 02:11 AM

Thanks e_fortier for the link....I'm going to read everything tomorrow, today is too late...It seems to be what I was looking for... Greetings.


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