# TDA1308 amp circuit



## teikjoon

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 :
   
   
   
   

   
  (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... 
	

	
	
		
		

		
			




   
  Thanks in advance for an help


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## Avro_Arrow

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.


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## cobaltmute

teikjoon said:


> 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....


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## Avro_Arrow

+2 for the data sheet example


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## teikjoon

This one?
   
   

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


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## cobaltmute

Pin 8 is the voltage input.


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## teikjoon

Oh, thanks 
	

	
	
		
		

		
		
	


	



   
  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...


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## wakibaki

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
   
  w


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## teikjoon

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 
	

	
	
		
		

		
		
	


	



   
   
  Quote: 





wakibaki said:


> 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.
> 
> ...


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## Avro_Arrow

Why do you need such a high gain?
  Usually 5 or less is good enough...
   
  Quote:


teikjoon said:


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


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## teikjoon

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* /img/forum/go_quote.gif
> 
> Why do you need such a high gain?
> Usually 5 or less is good enough...


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## Avro_Arrow

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).


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## teikjoon

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

	
	
		
		

		
		
	


	



  Quote: 





avro_arrow said:


> 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).


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## lordearl

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)?


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## Avro_Arrow

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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## lordearl

Thanks Avro
  I'm using 5v supply, limiting value is listed as 6v - I'll try increasing it and see if it makes any difference (prior to changing resistors)


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## Yodeb

I feel this thread up again, but someone can tell me that this output impedance of the amplifier?


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## nikongod

yodeb said:


> I feel this thread up again, but someone can tell me that this output impedance of the amplifier?


 
  
 http://www.nxp.com/documents/data_sheet/TDA1308.pdf


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## Yodeb

nikongod said:


> http://www.nxp.com/documents/data_sheet/TDA1308.pdf


 
 Sorry, but can not find. If you see, tell me please. Thank you.


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## nikongod

0.25ohm.


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## Yodeb

nikongod said:


> 0.25ohm.


 
 Thanks you very much


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