A 1.5W mono amp with no capacitors in the signal path?

[peskywinnets]

Hiya,

I seek a mono 1W to 1.5W (power iC amp) with no capacitors whatsoever in the signal path (DC coupled?).

The power supply will be 5V-7V

My preamp is an opamp and already biased at 1/2 VCC, the input to the preamp will be a coil pickup.

I'm figuring if I connect one leg of the input pickup coil to 1/2 VCC & the other to the first opamp stage input pin , then that that bit (the input) is therefore covered off (ie no caps needed) - but most IC audio amps 'example circuits' have decoupling caps at their input pin ...if my preamp circuit is already operating at 1/2 VCC....I'm assuming I can simply ignore those caps?

This is exploratory territoy for me, so general advice warmly received along with an tips for suitable IC amp contenders!

 

[AndrewT]

Hi,
you can use a DC coupled amplifier when the signal input ground and the signal output ground are both at Zero Volts.

If you use a pre-amp that is biased to 1/2Vcc then the output of this amp should, if set up and working correctly, be at 1/2Vcc. It cannot be connected directly to a DC coupled amplifier.

 

[peskywinnets]

Not fully understanding your last sentence. If my input stage is biased at 1/2 VCC & the BTL output amp stage is biased at 1/2 VCC then there outght to be no need to use coupling caps?

The biggest unknown (to me) is what happens when not using a coupling cap into IC audio amps (I guess each one is different). For example....a TDA7052A (my preferred IC), with a supply of 8V, seems to have abbout 2.5V on its input pin - it really doesn't like having the coupling cap removed and connected to a previous stage whose output is biased at 1/2 VCC

 

[Frank Berry]

You may be able to use an LM4951. Not sure whether the input is biased at 1/2 VCC but it may be worth looking at.

 

[peskywinnets]

Thanks Franck that's a good lead (I wasn't familiar with that chip)

One I've been looking at is a TDA8941 ....if for no other reason, than they are being upfront with what's going on with the IC's input biasing (ie an internal schematic for the front end) - it looks to have the input pin biased @ 1/2 VCC through a 45K resistor.....

http://www.datasheetcatalog.org/datasheet2/2/05o8eudsuhecpf14chjy8gz5h7fy.pdf (page 13 has the input biasing circuit)

the supply voltage range is 6V to 18v (9V typical), so I'd be using it towards the lower end. Alas, it's only available in DIP08 (I'd rather have SOIC or some oter surface mount package)...but hey, I'll not be so choosy at this stage!

Not sure why they quote a 16 0hm load (i'll be using about 8 0hms), surely the output power is a function of RMS voltage across the load - and ultimately the package's heat disappation possibilities (so why do they even bother quoting 1.5W at 16 ohms?)

 

[AndrewT]

if your pre-amp is biased to 1/2Vcc ( about 4.5Vdc) then the pre-amp is feeding 4.5Vdc to the input of the amplifier.
The output of the amplifier/s will be at DCgain times the input voltage.
A BTL is some form of bridging where the gain of the two halves are +ve and -ve.
Let's assume the DC gain is 4times (+12dB). The outputs from the two halves will sit at [+4 * 4.5] and [-4 * 4.5] = +-18V
Is that higher than supply rail voltage?

 

[peskywinnets]

Humour me here!

if your pre-amp is biased to 1/2Vcc ( about 4.5Vdc) then the pre-amp is feeding 4.5Vdc to the input of the amplifier.

Agreed...but if the said amplifier's input is already biased to sit at 1/2 VCC internally (as in the TDA8941) - then there's no quiescent DC 'difference' ...so will there be any impact?

For simplicity lets assume a common supply voltage of 9V.

The BTL IC amps I've toyed with to date have all had 1/2 VCC sitting on both their output pins...therefore for a 9V supply, that'll be 4.5V DC. I guess what I'm not understanding (for example with the aformentioned TDA8941), is if it's input stage of the power amp is also quiescently biased to 4.5V, then will connecting a previous stage with 4.5v bias into the IC amps input pins have any impact? (other than any small DC offset between the two stages?)

 

[peskywinnets]

Ok, if I'd bothered to go into the shed to check my tray of 'plethora of audio amp ICs', I needn't have asked this question (I could have just breadboarded it - which is what I've just done).

I actually found a TDA8941 (wondered why it seemed familiar on my research today!)

Yes, it seems you can DC couple the previous stage into an audio amp IC - but any DC discrepancies between the (low impedance) previous opamp stage & the (high impedance) Audio IC amp input stage causes the dc at the input to be loaded slightlly...which gets amplified on the output pins (meaning one side of the 'balanced' output signal shows as clipped)

The solution is to place a series resistor into the input pin (ie between the opamp output & audio amp input pin) - this & allows the audio amp input pin drift to it's prefered bias level (set internally) - I used a 47k series resistor which has removed the clipping....

[IMGDEAD]http://img833.imageshack.us/img833/9297/scopetrace.jpg[/IMGDEAD]

(green trace is dc coupled input to the TDA8941, yellow trace is the TDA8941 output)


Here the same measurement, but with no series resistor (ie direct from the previous opamp stage into the TDA8941)...

[IMGDEAD]http://img215.imageshack.us/img215/7453/clipped.jpg[/IMGDEAD]