impedance of transformer mathed to pot

[neazoi]

Hello I have a small audio transformer (8R:1K) at the output of a radio, having a secondary (secondary used as output to high impedance headphones) with 1k impedance.
The original circuit used a 5K potentiometer, connected as a voltage divider to the secondary, as an audio volume.
I was wondering if a 1K potentiometer should be better used instead of a 5k, to better match the impedance of the transformer?

 

[neazoi]

Here is the schematic.
Should R4 better be 1K pot instead of 5K pot, since the secondary of T3 is 1K?

 

[Venkadesh_M]

What is the circuit next to POT ?

The Transformer impedance = input impedance of the next circuit. which includes POT and the LOAD connected to POT.

 

[FvM]

You'll use whatever gives better sound quality and higher volume. I guess 5k will be better.

 

[neazoi]

What is the circuit next to POT ?

The Transformer impedance = input impedance of the next circuit. which includes POT and the LOAD connected to POT.

Next to pot there are phones connected.
One set of phones (available to me) is 1.2K magnetic types.
The other set of phones (available to me as well) are 20Mega ohms crystal types.

 

[Venkadesh_M]

I think the output is already matched for 1.2 k type phones. So what is your problem with that ? Dont you get maximum power transfer ?

 

[Borber]

If you talk about matching first requirement is that audio source must have 8ohms output impedance and second is that transformer must be terminated with 1k. What is your audio source output impedance?

 

[FvM]

Audio output circuits are usually not required with impedance matching.

If you load the transformer with 1k, the amplifier is already loaded with it's rated impedance, when connecting an earphone, the total load falls short of 8 ohm.

 

[neazoi]

Audio output circuits are usually not required with impedance matching.

If you load the transformer with 1k, the amplifier is already loaded with it's rated impedance, when connecting an earphone, the total load falls short of 8 ohm.

As explained, a set of 1.2K earphones and another set of 20M headphones will be connected to the output (interchanged to see which one is best)
The pot is there to provide volume adjustment.

I thought that since the output of the transformer is 1K, then a 1K pot would be ideal for minimum loss.
However if the 1.2K phones are directly connected to the transformer, then they are already about ok matched to it.

I basically need to determine if 1K pot will be ok, since I have only this available, and if it will be actually better (match)

 

[FvM]

As said, I would choose a 5k potentiometer, but 1 k will work too.

 

[neazoi]

As said, I would choose a 5k potentiometer, but 1 k will work too.

Will a 5k be better because of the lighter loading of the transformer? Is this the thing that defines audio quality?
To be honest since only single tones would appear at the output (morse code), audio quality is not of so much concern, but sensitivity is more important.

 

[Audioguru]

Most amplifiers (except vacuum tube amplifiers) work perfectly with no load or with a high impedance load. A modern amplifier NEVER has its output impedance matched because its output impedance is extremely low at 0.04 ohms or less but it has a load of 4 or 8 ohms.

Headphones are very close to your ears so the output of an amplifier must be attenuated to drive them.
Your transformer is doing the opposite of attenuation, it is boosting the level 11 times which might destroy the headphones or your hearing.

Turn down the volume of the radio then connect the headphones directly to the speaker output. If it is too loud then turn down the volume more or make an attenuator with two series resistors.

Cheap old fashioned 1200 ohms magnetic or crystal headphones produce poor quality sound.

A radio is supposed to have automatic gain control so that a weak distant station sounds as loud as a strong local station.