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?
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?
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
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
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!
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 - 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!
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!
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