H Type Balanced Attenuator.

[GeeVee]

I am currently constructing and Aleph P1.7 preamplifier, and I have been investigating the most appropriate type of volume / attenuator control for balanced operation.

In discussion with a friend, he suggested that the most appropriate system is the H attenuator, as shown in the attached sketch.

Looking at it, I can understand that the additional resistor on the wiper of the pot will result in a linear pot behaving as a logarithmic type.

However I do not understand what the 330R resistors are for. To my way of thinking, I suspect that it is to keep the 2 sides balanced, regardless of attenuator position.

I would appreciate any comments regarding this arrangement.

Thanking you in advance.

Regards,
George.

[juma]

330 Ohm resistors are there to prevent shortcircuiting the outputs of the preamp and they can be bigger (1K - 2K2) in order to avoid overloading the preamp.
The other pair of 330R that connects to the input of the amp is not needed, there are always resistors in the input sections of amplifiers.

[wuffwaff]

Hi,

you need the two resistors connected to the preamp to make the attenuator work.
Together with the pot they form a voltage devider, without them the output voltage would stay more or less the same (when the preamp has a low output impedance) until you short it.

The two output resistors are not really needed and do raise the output impedance of the preamp.

I´m using something like this on my X-BOSOZ with two 1k2 resistors and a 24 step switch. Works very nice.

I tried it with a pot and here the difficulty is that for the lower volume settings the resistors are like 1 to 10 ohms and most pots don´t have the neccessary precision (left/right tracking). I tried a quite expensive laser trimmed one but couldn´t get it working on low volume settings.

Here´s a table with the values for the shunt resistor with 1k2 output resistors:

First Rshunt, then attenuation in dB, then output impedance:

5818 -3 1699
2411 -6 1203
1320 -9 852
805 -12 603
519 -15 427
346 -18 302
235 -21 214
162 -24 151
112 -27 107
78 -30 76
55 -33 54
39 -36 38
27 -39 27
19 -42 19
14 -45 13
9,6 -48 10
6,8 -51 7
4,8 -54 5
3,4 -57 3
2,4 -60 2
1,7 -63 2
1,2 -66 1
0,9 -69 1

as you can see the lower values are not that far apart.


William

[GeeVee]

Originally posted by juma
330 Ohm resistors are there to prevent shortcircuiting the outputs of the preamp and they can be bigger (1K - 2K2) in order to avoid overloading the preamp.
The other pair of 330R that connects to the input of the amp is not needed, there are always resistors in the input sections of amplifiers.

Hello Juma.

Thank you for your reply. I now have a better understanding. At first I was simply going to wire a 2 gang pot (1 bank in the - line and 1 bank in + line) in a configuration that would be simillar to an unbalanced set up.

I was then going to run 2 complete separate pots, so they would act as individual channel gain.

When I showed my idea to a friend, he said no, use the balanced approach (H Type)

Regards,
George.

[GeeVee]

Originally posted by wuffwaff
Hi,

you need the two resistors connected to the preamp to make the attenuator work.
Together with the pot they form a voltage devider, without them the output voltage would stay more or less the same (when the preamp has a low output impedance) until you short it.

The two output resistors are not really needed and do raise the output impedance of the preamp.

I´m using something like this on my X-BOSOZ with two 1k2 resistors and a 24 step switch. Works very nice.

I tried it with a pot and here the difficulty is that for the lower volume settings the resistors are like 1 to 10 ohms and most pots don´t have the neccessary precision (left/right tracking). I tried a quite expensive laser trimmed one but couldn´t get it working on low volume settings.



William

Hi William.

Thank you for your response.

Just to ensure that I understand it correctly, you have placed 1.2k
resistors on the preamp side, removed the resistors on the output / power amp side.

Looking at the value of your stepped atenuator, am I also correct in assumnig that the additional wiper resistor has been removed in your design?

Thanks in advance.

Regards,
George.

[juma]

Hi William !
I think that solution you presented is the most elegant and most viable when it comes to controling volume in balanced circuits (good quad pots are hard to find, quad switch attenuators are clumsy and expensive, relays, microprocessors and similar stuff can be even clumsier and pricier).
Volume control is a major obstacle for most people to start using balanced circuits in full differential mode.
Since you are great with spreadsheets (kudos for AXE), would it be a trouble to make a .xls for this (12 step, 24 step, choice of R values, column with -dB attenuation, schematics ...)?
I'm sure that community would be grateful.
Thanks

[wuffwaff]

Hi,

well I´ve got an excel sheet for this. Will put a few comments in it and post it this evening (or tomorrow....)

William

P.S. there´s no wiper because it´s not a pot but a 24 position switch.

[GeeVee]

Originally posted by wuffwaff
Hi,

well I´ve got an excel sheet for this. Will put a few comments in it and post it this evening (or tomorrow....)

William

P.S. there´s no wiper because it´s not a pot but a 24 position switch.

Hi William.
I am looking forward to viewing your excell worksheet.

With respect to the 'wiper"comment, sorry - poor choice of wording on my part.
Obvoiusly the stepped switch takes the place of the wiper so to speak.

However, are you still using a shunt resistor to the -ve line?

Regards,
George.

[wuffwaff]

Hi,

I think this resistor in combination with the pot makes the characteristic more logarithmic.

Since I use a switch where one resistor is used per step I don´t need this.

The switch is connected between +out and -out (after the series output resistors)

William

[wuffwaff]

Here it is,

it is kind of self explanatory......


William

P.S. since I found it as a Zip file it must be posted somewhere before.....