TDA7052A - power, & (real world) heat levels ...Sanity Check sought!
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 Chip Amps Amplifiers based on integrated circuits

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 9th August 2009, 09:32 AM #1 HankMcSpank   diyAudio Member   Join Date: May 2008 TDA7052A - power, & (real world) heat levels ...Sanity Check sought! I'm using a TDA7052A to drive a coil that's 9.2 Ohms (the end device isn't important here!) OK, so I put a 330Hz sinewave into the chip & crank the input level until the signal is showing 3V peak to peak output across the coil - which is about the level I need it at. But at this level, the TDA7052A gets very hot (ie I can only press my finger firmly on the chip's top surface for a second or two). So, are these following figures correct... 3V peak to peak = 1.062V RMS (ie multiply by 0.354 ) Current flowing through coil = V/R, therefore 1.062/9.2 Ohms = 118mA Total power = VI, therefore 1.062 x 0.118 = 125mW So with a 3V peak to peak sine wave across the coil, this TDA7052A is only running at about 125mW, the datasheet says it can output 1W (which is about 8X the level where I already think it is getting too hot!) Have I stuffed up with my calculations...or am I just underestimating what this tiny little package can disappate safely?
 9th August 2009, 09:41 AM #2 MaxS   diyAudio Member   Join Date: Mar 2006 Location: France Coil use : welder ? Hello, If I'm right : you are driving a coil in sinusoidal current. A coil has complex impedance : Z = R + j L 2 Pi 330. But have a look, a coil is charged by current and gives back some voltage in the chip. I do not know how fine is your electronic knowledge but I want let you know it is like get a punch in your head. We do not like that ! Regards, MaxS __________________ Ca marche pas ! Tu ne l' as pas branché non plus ....
 9th August 2009, 11:58 AM #3 HankMcSpank   diyAudio Member   Join Date: May 2008 Ok, I've solved my problem - doh, of course, it's a BTL chip & I was only scoping the positive leg - therefore the output signal is not 3V peak to peak, but 6V peak to peak. therefore (if my figures are correct)... 6V sine signal peak to peak = 4.24V RMS (ie peak to peak voltage x 0.707) Coil DC resistance is 9.2 ohms and its inductance is 1.7mH , but at 330Hz, overall resistance (impedance) is 9.98 ohms. (http://www.cvs1.uklinux.net/calculators/ ) Therefore current = Voltage/resistance, 4.24V/9.98 ohms = 425mA Power = VI, 4.24V x 0.424mA = 1.8W No wonder it's getting hot ...I need a more powerful chip!
 9th August 2009, 04:19 PM #4 mjf   diyAudio Member   Join Date: Jan 2009 Location: Austria hello. what power supply voltage do you use with this opamp? have a look at the datasheet..........with 6v supply and 8 ohm you will get around 1w.........and with 10ohm probably 0,9w . greetings........
ttako
diyAudio Member

Join Date: Dec 2002
Location: Hungary/ Budapest
Quote:
 Originally posted by HankMcSpank Ok, I've solved my problem - doh, of course, it's a BTL chip & I was only scoping the positive leg - therefore the output signal is not 3V peak to peak, but 6V peak to peak. therefore (if my figures are correct)... 6V sine signal peak to peak = 4.24V RMS (ie peak to peak voltage x 0.707) Coil DC resistance is 9.2 ohms and its inductance is 1.7mH , but at 330Hz, overall resistance (impedance) is 9.98 ohms. (http://www.cvs1.uklinux.net/calculators/ ) Therefore current = Voltage/resistance, 4.24V/9.98 ohms = 425mA Power = VI, 4.24V x 0.424mA = 1.8W No wonder it's getting hot ...I need a more powerful chip!

Hi,

6V p-p is 3Vpeak, which is 2.12Vrms
With this voltage you get ca 0.45Wrms...

Tamas

 9th August 2009, 05:23 PM #6 HankMcSpank   diyAudio Member   Join Date: May 2008 mjf - I was using 9V to the TDA7052A as supply voltage. ttako - yes, I think I made a mistake in my previous calculations (I'm rusty which is why I'm posting here for the sanity check!). Let me walk through this with you.... I concur with the 2.12Vrms, but to work out the RMS power, we need at least one more bit of data - either current or resistance. The coil is 9.2 ohms, but at 330Hz the actual total impedance is 9.98 Ohms. So we have a 2.12V RMS signal across an impedance of 9.98 ohms, therefore the current through the coil must be 0.212mA. Power = V x I therefore power through the coil is 450mW (which I guess concurs with your 0.45W!) Have I got it now then? If so, then perhaps I don't need a different (beefier) audio power chip, but at 0.45W it seems very hot - I can't imagine that being twice this hot (1W) is going to be good for it!
 9th August 2009, 06:00 PM #7 HankMcSpank   diyAudio Member   Join Date: May 2008 Just a quick typo in that last post (& the edit button seems to have gone?), it should have read... So we have a 2.12V RMS signal across an impedance of 9.98 ohms, therefore the current through the coil must be 212mA.
 11th August 2009, 01:46 AM #8 wwenze   diyAudio Member   Join Date: Mar 2008 This thing is rated for 85 degrees, which would be pretty hot if you touch it. Best to check with a thermometer. And if you're paranoid, just heatsink it. Though I doubt it'd be necessary.
 11th August 2009, 12:17 PM #9 AndrewT   diyAudio Member   Join Date: Jul 2004 Location: Scottish Borders go and read some literature on power amplifier driving an inductive load. Start with ESP site. He shows just how bad the amplifier dissipation can become when the reactance starts to exceed 30degree phase. __________________ regards Andrew T.

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