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-   -   TDA7052A - power, & (real world) heat levels ...Sanity Check sought! (http://www.diyaudio.com/forums/chip-amps/148776-tda7052a-power-real-world-heat-levels-sanity-check-sought.html)

 HankMcSpank 9th August 2009 09:32 AM

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?

 MaxS 9th August 2009 09:41 AM

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

 HankMcSpank 9th August 2009 11:58 AM

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!

 mjf 9th August 2009 04:19 PM

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 9th August 2009 05:08 PM

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

 HankMcSpank 9th August 2009 05:23 PM

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!

 HankMcSpank 9th August 2009 06:00 PM

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.

 wwenze 11th August 2009 01:46 AM

This thing is rated for 85 degrees, which would be pretty hot if you touch it. Best to check with a thermometer. :smash:

And if you're paranoid, just heatsink it. Though I doubt it'd be necessary.

 AndrewT 11th August 2009 12:17 PM

go and read some literature on power amplifier driving an inductive load.