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TDA1308 amp circuit

post #1 of 21
Thread Starter 

Hello forum,

 

Was looking around for a low-voltage headamp circuit I could build, which I could power with the USB battery pack I already carry around (an XtremeMac InCharge Portable) when I came across this TDA1308 circuit :

 

 

 

 

1000

 

(original source : http://www.dz3w.com/diycn/music/1119.html)

 

From what I can gather from the Google Transla-tion, this is powered by two 3.6V lithium batteries, but I had a few questions, and was wondering if someone could help with, such as :

 

a) Is JK3 a charging/mains power jack?

b) What is the purpose of the row of capacitors (C2, C3, C4, C5, C6)? I think the translation says it is a "RC damper network", but I'm not sure what it does.

c) Is there a virtual ground circuit? I can't tell... confused.gif

 

Thanks in advance for an help biggrin.gif

post #2 of 21

Yes, JK3 is the charging jack.

 

There is no virtual ground.

It uses a "Real" ground formed by the split rail battery pack.

One battery for the + rail and one for the - rail.

post #3 of 21
Quote:
Originally Posted by teikjoon View Post

Hello forum,

Was looking around for a low-voltage headamp circuit I could build, which I could power with the USB battery pack I already carry around (an XtremeMac InCharge Portable) when I came across this TDA1308 circuit :

Would it not be easier just to implement the circuit on the datasheet? It is single supply already....
post #4 of 21

+2 for the data sheet example

post #5 of 21
Thread Starter 

This one?

 

 

1000

 

Might be a silly question, but where on the diagram do I hook up the battery?

post #6 of 21
Pin 8 is the voltage input.
post #7 of 21
Thread Starter 

Oh, thanks biggrin.gif

 

Finally...am assuming R(L) is the gain resistor, and the datasheet has this at 32 ohms...but I am not sure what formula or values to calculate the gain.

 

Looked at Tangent's cmoy tutorial, and I'm not sure if his formula is applicable... confused.gif

post #8 of 21

No, RL is the load, or to put it another way, the headphones.

 

32 ohms is often used as a typical value for headphone impedance.

 

The gain is set by the 3k9 resistors (in the datasheet) (there being 2 of these connected to each input, one connected to the source, and one to the output), giving -3900/3900 or -1 (the amplifier is inverting). The modern trend is to make audio amplifiers preserve the signal phase as is the practice in mixers (non-inverting). See e.g. here:- http://www.radio-electronics.com/info/circuits/opamp_basics/operational-amplifier-gain.php

 

post #9 of 21
Thread Starter 

Oh...alright...

 

So I will need to change the ratio of R4:R3 (and R6:R5) to determine the gain?

 

Therefore to get gain of 10, I will need R4 = 39K ohm and R3 = 3.9K ohm...

 

Does it matter what values I use for these resistors, or merely their ratio? I think the cmoy uses 10K ohm / 1K ohm to achieve gain of 11.

 

Thanks for everyone's help so far biggrin.gif

 

 

Quote:
Originally Posted by wakibaki View Post

No, RL is the load, or to put it another way, the headphones.

 

32 ohms is often used as a typical value for headphone impedance.

 

The gain is set by the 3k9 resistors (in the datasheet) (there being 2 of these connected to each input, one connected to the source, and one to the output), giving -3900/3900 or -1 (the amplifier is inverting). The modern trend is to make audio amplifiers preserve the signal phase as is the practice in mixers (non-inverting). See e.g. here:- http://www.radio-electronics.com/info/circuits/opamp_basics/operational-amplifier-gain.php

 


Edited by teikjoon - 7/23/12 at 9:30pm
post #10 of 21

Why do you need such a high gain?

Usually 5 or less is good enough...

 

Quote:

Originally Posted by teikjoon View Post

Oh...alright...

 

So I will need to change the ratio of R4:R3 (and R6:R5) to determine the gain?

 

Therefore to get gain of 10, I will need R4 = 39K ohm and R3 = 3.9K ohm...

 

Does it matter what values I use for these resistors, or merely their ratio? I think the cmoy uses 10K ohm / 1K ohm to achieve gain of 11.

 

Thanks for everyone's help so far biggrin.gif

 

 

post #11 of 21
Thread Starter 

Hrmn...perhaps I worded it poorly, but I was just using those value to verify the formula.

 

Like you say, a gain of 5 would probably be best..so R4 = 19.5K and R3 = 3.9K?

 

Also..does it matter what the values of R4 and R3 are, or can I change their values as long as their ratio R4:R3 are consistent (i.e. 19.5K/3.9K = gain of 5, 5K/1K = gain of 5).

 

 

Quote:

Originally Posted by Avro_Arrow View Post

 

Why do you need such a high gain?

Usually 5 or less is good enough...

post #12 of 21

You are correct, the ratio determines the gain.

Higher values of R use less current, but cause more noise.

Lower values of R use up more of your current, but cause

less noise. Values of 1K and 5K would be reasonable values.

Maybe even as low as 500 and 2k5 (2k5 means the same as 2.5k).

post #13 of 21
Thread Starter 

Cool thanks, I'll get this breadboarded and try it out L3000.gif

Quote:
Originally Posted by Avro_Arrow View Post

You are correct, the ratio determines the gain.

Higher values of R use less current, but cause more noise.

Lower values of R use up more of your current, but cause

less noise. Values of 1K and 5K would be reasonable values.

Maybe even as low as 500 and 2k5 (2k5 means the same as 2.5k).

post #14 of 21

teikjoon - how did you go with this project?

 

I've built a TDA1308 amp based on the circuit proposed in the datasheet, sounds great with my KRK KNS8400 phones @ 32 ohms impedance, however it certainly doesn't have the juice to power my 600 ohm Beyer DT880s.  Did you try playing around with the gain resistors?

 

Any suggestions as to how I can customise the gain resistors for DT800 (600 ohm)?

post #15 of 21

Try R3/R5 = 1k and R4/R6 = 5k

How much supply voltage are you using?

You may need to increase the supply voltage as well.

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