Basic attenuator question - diyAudio
Go Back   Home > Forums > Source & Line > Analog Line Level

Analog Line Level Preamplifiers , Passive Pre-amps, Crossovers, etc.

Please consider donating to help us continue to serve you.

Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving
Reply
 
Thread Tools Search this Thread
Old 21st May 2011, 09:17 PM   #1
diyAudio Member
 
Join Date: Apr 2008
Default Basic attenuator question

Hi,

I searched and searched but still could not clear up my question.

For example: the poweramp has an input impedance 50 kohms. I bought a matching stepped log attenuator with 50 kohms and measured the values in the steps.

Full turned up the attenuator I measured 0 ohms. O.K. Full turned down I read 50 kohms. At 80 % of full turned on position about 40 kohms (because logarithmic). No problem.

What I could not understand was that people often recommend attenuators with a wide range of specifications. For example 10 to 50 kohms. So letīs compare a 10 and 50 kohms attenuator.

The 10 kohm has a value of 10 kohm in position turned down full. But this would pass 4 times more current than the 50 kohms turned on 80 % of full position (40 kohms). So imho it would be a fatal choice to take a 10 kohms instead of the 50 kohms (same of the input impedance of power amp)

Could anyone put in a statement ? Thanks in advance
  Reply With Quote
Old 21st May 2011, 09:32 PM   #2
tomchr is offline tomchr  United States
diyAudio Member
 
tomchr's Avatar
 
Join Date: Feb 2009
Location: Greater Seattle Area
I sense two parts in your question:

1) Power amp loading of the attenuator
2) Attenuator loading of the signal source

For a "passive preamp" i.e. an attenuator/potentiometer where the signal source is connected across the entire resistive element and the output is taken at the different taps (attenuator) or at the wiper (potentiometer), the power amp will load the attenuator. The loading will wary depending on the setting of the attenuator/potentiometer. This will cause the volume control to deviate slightly from the logarithmic curve. So you may get 0.5 dB per step at the lower volume settings, but will get more like 3 dB per step for the higher volume settings (assuming you're using a 50 kOhm attenuator with a 50 kOhm power amp input impedance). To combat this, you need to reduce the load on the attenuator. Commonly, this is done by inserting a buffer (near infinite input impedance, near zero output impedance, gain of 1 V/V (0 dB)) between the attenuator and the input to the power amp. This ensures that the attenuator "sees" almost no load and also ensures that the power amp does not present any significant load on the preamp output. Some people don't like the buffer and prefer a completely passive solution. In this case, you will have to ensure that the power amp does not present any significant load on the attenuator. For most electrical engineering "significant" means a factor of 10. I.e. for a 50 kOhm attenuator, the load should be 10x50 = 500 kOhm or greater to ensure that it does not present any significant load on the attenuator. Obviously, we can't easily change the input impedance of the power amp, but we can specify a different attenuator. For a 50 kOhm power amp input impedance, I'd choose an attenuator that's no higher than (50 kOhm)/10 = 5 kOhm.

Now a 5 kOhm attenuator will present quite a load on the signal source and it will need to be ensured that the source can drive the 5 kOhm impedance while still maintaining high performance.

Of course, you can aim for the happy medium and go with a 10 kOhm attenuator. Then the signal source will probably be able to drive it, and while the volume control will deviate slightly from the logarithmic curve due to the loading by the power amp, it'll probably be good enough. At least until any of the components (signal source, power amp) change...

Personally, I prefer to have my interfaces defined by the circuit designer (i.e. me). Hence, I'll have a buffer on the preamp input, then the volume pot, followed by a buffer on the output. That's my approach. Others are free to differ.

~Tom
  Reply With Quote
Old 21st May 2011, 11:14 PM   #3
diyAudio Member
 
Join Date: Apr 2008
Hi Tom, thanks much for your Input.
Buffer Yes/No I would choose a Yes because I read often about the sense of it's use in impedance mismatches between source and amp.

Honestly I have not understood completely what you tried to explain. I just considered for example to the output voltage of the CD Player. Let's take a value of 2 Volts at a specific load 200 ohms. In case of max. output of attenuator (zero ohms tap) the passing voltage would be 2 volts, or not ?

I measured at 80 % of full turned on position of my 50 kohm attenuator 40 kohms. The amp should be loud enough in this position - in any way louder than average listening levels. That means to me: 50 kohms position --> no music, and 40 kohms position ---> loud listening level ... ?!
This confused me a bit. 40 kohms is still a high value of resist.

Maybe I have to learn more basics regarding electronics in audio.
  Reply With Quote
Old 21st May 2011, 11:36 PM   #4
Banned
 
Join Date: Jan 2008
Blog Entries: 2
The thing is that across a range of values of as much as 2 decades there may not be much difference in performance between values. Input impedances from 10k to 1M are considered workable in a power amplifier. Output impedances of various stages can likewise vary a lot, so it's hard to be prescriptive. Sometimes the equipment at hand may not have exactly the distribution of impedances one would ideally like, so the decision as to what attenuation or where to employ it may be moot.

w
  Reply With Quote
Old 22nd May 2011, 12:14 AM   #5
tomchr is offline tomchr  United States
diyAudio Member
 
tomchr's Avatar
 
Join Date: Feb 2009
Location: Greater Seattle Area
Quote:
Originally Posted by TheDealer View Post
Honestly I have not understood completely what you tried to explain. I just considered for example to the output voltage of the CD Player. Let's take a value of 2 Volts at a specific load 200 ohms. In case of max. output of attenuator (zero ohms tap) the passing voltage would be 2 volts, or not ?
If the attenuator does not load the CD player significantly (read if the resistance connected to the CD player output is significantly greater than the output impedance of the CD player), the output voltage of the attenuator at max volume setting will be the same as the output voltage of the CD player. So 2 V in your example.

Quote:
Originally Posted by TheDealer View Post
I measured at 80 % of full turned on position of my 50 kohm attenuator 40 kohms. The amp should be loud enough in this position - in any way louder than average listening levels. That means to me: 50 kohms position --> no music, and 40 kohms position ---> loud listening level ... ?!
This confused me a bit. 40 kohms is still a high value of resist.
40 kOhm from where to where?

The attenuator should be a voltage divider and not a series resistor.

~Tom
  Reply With Quote
Old 22nd May 2011, 11:46 AM   #6
diyAudio Member
 
Join Date: Apr 2008
Quote:
Originally Posted by tomchr View Post

40 kOhm from where to where?

The attenuator should be a voltage divider and not a series resistor.

~Tom
I am sure that I am wrong because I obviously make a systematic mistake.
Let me explain it this way:
My attenuator is a 21 step. I was curious about the impedance values between input and output at different steps. With the multimeter I measured for the 50 kohms log pot.

- Step 1 (turned off full left direction): 50,9 kohms
- Step 2 : 50,6
- Step 3 : 50,2
..
- Step 17 : 40,0
- Step 18 : 35,0
- Step 19 : 30,0
- Step 20 : 20,2
- Step 21: 0,0 (turned into full right position)

So you see at step 17 a value of 40 kohms. This is a common big output Vol. position. I can particullary explain my mistake myself. The current is constant determined by the value of total Resistance value, or not ? In my case 50,9 kohms. So let's consider a constant voltage of 2 Volts coming from the CD Player -----> Current = 2 volts/50900 ohms = 0,00004 amperes.

In every attenuator step the current is same as above calculated, O.K. ? Only the divided voltages change depending on attenuator steps. Like the calculation: U1/R1=U2/R2=U3/R3= ... 0,00004 amperes. The bigger the Resistance the bigger must be the voltages to get same amperes. But why I get 50,9 kohms (Step 1, Vol. minumum !) which should mean that the voltage is here full input (2 Volts). I suspect that my attenuator has no resistors in series where the voltages are as different as in their different taken positions. I think that in my attenuator every step has it's own value based on different resistors. On the other hand I would get this time different currents at constant voltage and different resistor values. But the poweramp's output power is adjusted by different gains usually measured in volts and not amperes .... I added a pic of the attenuator - maybe there is another mistake.attenuator.JPG

Last edited by TheDealer; 22nd May 2011 at 11:56 AM. Reason: correction
  Reply With Quote
Old 22nd May 2011, 12:08 PM   #7
diyAudio Member
 
Join Date: Apr 2008
Quote:
Originally Posted by tomchr View Post

40 kOhm from where to where?



- Step 1 (turned off full left direction): 50,9 kohms
- Step 2 : 50,6
- Step 3 : 50,2
..
- Step 17 : 40,0
- Step 18 : 35,0
- Step 19 : 30,0
- Step 20 : 20,2
- Step 21: 0,0 (turned into full right position)

Hi, do you mean I should take the difference:
Voltage value of Step 17 = 50,9 kohms - 40 kohms = 10,9 kohms

U total/R total = 0,00004 amperes ----> U at position 17 = 10900 * 0,00004 = 0,436 Volts. That would make sense the 0,0 kohms value at position 21.

When I take the difference 50,9 kohm - 0,0 kohms = 50,9 kohms and the result is full input voltage (2 volts).
  Reply With Quote
Old 22nd May 2011, 01:38 PM   #8
DF96 is offline DF96  England
diyAudio Member
 
Join Date: May 2007
There are two impedances which matter in a volume pot or a step attenuator: input impedance, and output impedance. The impedance from input to output does not matter so there is no point in measuring it.

In most cases the higher the input impedance and the lower the output impedance, the better. Typically, the highest output impedance occurs at the -6dB position, where output voltage is half input voltage, and is equal to 1/4 of the total resistance. In your case this would be around steps 19 and 20, which have an output impedance around 12k.
  Reply With Quote
Old 22nd May 2011, 11:46 PM   #9
diyAudio Member
 
Join Date: Apr 2008
Quote:
Originally Posted by TheDealer View Post

Hi, do you mean I should take the difference:
Voltage value of Step 17 = 50,9 kohms - 40 kohms = 10,9 kohms

U total/R total = 0,00004 amperes ----> U at position 17 = 10900 * 0,00004 = 0,436 Volts.
As I expected, this was the solution. I measured with a 2,66 Volts supply and the voltage values confirm exactly my calculation above.
  Reply With Quote
Old 23rd May 2011, 03:16 PM   #10
diyAudio Member
 
Join Date: Dec 2010
Default resistor value matching

I would like to build a fixed ~10dB attenuator for connecting my DAC directly to the power amplifier.
For doing this I do need an advice regarding of how to match resistor values for a L-pad voltage divider.

My DAC has a output impedance of 100ohms
My amplifiers have a input impedance of 22000ohms, but used in bi-amped configuration the load will be 11000ohms.

So with 100ohms and 11kohms, what would be the optimal resistor matching?

Last edited by Wahlis00; 23rd May 2011 at 03:24 PM.
  Reply With Quote

Reply


Hide this!Advertise here!
Thread Tools Search this Thread
Search this Thread:

Advanced Search

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off


Similar Threads
Thread Thread Starter Forum Replies Last Post
Basic Electronics Question corrieb Solid State 3 26th March 2008 08:33 PM
Basic questions about 256 steps attenuator Mrpong Pass Labs 2 9th May 2007 06:46 PM
really basic question cuallito Tubes / Valves 6 15th April 2004 08:25 PM
Basic question Tommy Solid State 6 22nd February 2004 03:08 PM
really basic question MbayAQ Digital Source 3 7th January 2003 12:57 PM


New To Site? Need Help?

All times are GMT. The time now is 06:07 AM.


vBulletin Optimisation provided by vB Optimise (Pro) - vBulletin Mods & Addons Copyright © 2014 DragonByte Technologies Ltd.
Copyright Đ1999-2014 diyAudio

Content Relevant URLs by vBSEO 3.3.2