Hi guys,

What's the quickest and simplest way of running an 8 ohm earpiece to a
32 ohm earpiece output socket without suffering too much power loss?

Thanks,

p.
--

"What is now proved was once only imagin'd." - William Blake, 1793.

What's the quickest and simplest way of running an 8 ohm earpiece to a
32 ohm earpiece output socket without suffering too much power loss?
======
Any small transformer with a winding ratio 2 : 1 . Check with low
voltage -50 Hz , 1 V in ......0.5 V out. Ratio is not all that critical .
You can also use a tranformer's single winding with a centre tap .

Frank GM0CSZ / KN6WH

On Tue, 13 Jul 2004 23:48:53 GMT, "Highland Ham"
wrote:


What's the quickest and simplest way of running an 8 ohm earpiece to a
32 ohm earpiece output socket without suffering too much power loss?
======
Any small transformer with a winding ratio 2 : 1 . Check with low
voltage -50 Hz , 1 V in ......0.5 V out. Ratio is not all that critical .
You can also use a tranformer's single winding with a centre tap .

Thanks, Frank. Turns out on further investigation the circuit
specifies * two* 32 ohm insets in series for 64 ohms altogether. I've
still only got an 8 ohm earpiece, though. So that's an 8:1
transformation which adds up to.... a tap a quarter way along one
winding? Or is it three-quarters. Or of course a full, 4:1
transformer, I suppose. I always get confused with transformations,
for some reason. :-/
--

"What is now proved was once only imagin'd." - William Blake, 1793.

Paul Burridge wrote:

On Tue, 13 Jul 2004 23:48:53 GMT, "Highland Ham"
wrote:


What's the quickest and simplest way of running an 8 ohm earpiece to a
32 ohm earpiece output socket without suffering too much power loss?

======
Any small transformer with a winding ratio 2 : 1 . Check with low
voltage -50 Hz , 1 V in ......0.5 V out. Ratio is not all that critical .
You can also use a tranformer's single winding with a centre tap .


Thanks, Frank. Turns out on further investigation the circuit
specifies * two* 32 ohm insets in series for 64 ohms altogether. I've
still only got an 8 ohm earpiece, though. So that's an 8:1
transformation which adds up to.... a tap a quarter way along one
winding? Or is it three-quarters. Or of course a full, 4:1
transformer, I suppose. I always get confused with transformations,
for some reason. :-/

The winding ratio goes as the square root of the impedance ratio,
because you're transforming the current UP at the same time your
transforming the voltage DOWN. So you need a root-8:1, or about 3:1.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Turns out on further investigation the circuit
specifies * two* 32 ohm insets in series for 64 ohms altogether. I've
still only got an 8 ohm earpiece, though. So that's an 8:1
transformation which adds up to.... a tap a quarter way along one
winding? Or is it three-quarters. Or of course a full, 4:1
transformer, I suppose. I always get confused with transformations,
for some reason. :-/
============================
For a (lossless) transformer V*V/Z = constant , hence the input to output
impedance ratio is proportional to the square of the voltage ratio.
For your 64 to 8 Ohms application ,hence 8 : 1 impedance ratio you need a
small transformer with a sqrt (8:1) = 2.8 :1 voltage ratio (equals winding
ratio). You possibly have a small transformer in your junkbox from a
wallwart , those low voltage DC power supplies you plug into a wall socket.
They often have a switch enabling different output voltages 3-5-6-9-12 V
The transformer inside has a single secondary winding ,which could be used
as an auto transformer for your application.
Perhaps you can also wind your own transformer on a toroid from an old
switch mode PSU or the like, say with 50 and 140 turns or a single winding
of 140 turns with a tap at 50 turns. The actual winding ratio is probably
not critical.
The impedance of your earpiece is only around 8 Ohms for a limited audio
freq range ; it is usually specified for 1000 Hz

Good luck with you endeavours

Frank GM0CSZ / KN6WH

On Wed, 14 Jul 2004 11:00:09 GMT, "Highland Ham"
wrote:

Turns out on further investigation the circuit
specifies * two* 32 ohm insets in series for 64 ohms altogether. I've
still only got an 8 ohm earpiece, though. So that's an 8:1
transformation which adds up to.... a tap a quarter way along one
winding? Or is it three-quarters. Or of course a full, 4:1
transformer, I suppose. I always get confused with transformations,
for some reason. :-/
============================
For a (lossless) transformer V*V/Z = constant , hence the input to output
impedance ratio is proportional to the square of the voltage ratio.
For your 64 to 8 Ohms application ,hence 8 : 1 impedance ratio you need a
small transformer with a sqrt (8:1) = 2.8 :1 voltage ratio (equals winding
ratio). You possibly have a small transformer in your junkbox from a
wallwart , those low voltage DC power supplies you plug into a wall socket.
They often have a switch enabling different output voltages 3-5-6-9-12 V
The transformer inside has a single secondary winding ,which could be used
as an auto transformer for your application.
Perhaps you can also wind your own transformer on a toroid from an old
switch mode PSU or the like, say with 50 and 140 turns or a single winding
of 140 turns with a tap at 50 turns. The actual winding ratio is probably
not critical.
The impedance of your earpiece is only around 8 Ohms for a limited audio
freq range ; it is usually specified for 1000 Hz

Good luck with you endeavours

Okay, many thanks, Frank (& Tim). That should be plenty to get me up
and running. I've got some toroids that should do the trick. I assume
since you say it's okay to use ones from PSUs that loss isn't a
problem here. I've got some low-loss powdered iron RF types, but
they're probably too small for the number of turns required. Still, a
bit of experimenting is what it's all about!
Thanks again,

p.
--

"What is now proved was once only imagin'd." - William Blake, 1793.

"Paul Burridge" wrote in message
...
On Wed, 14 Jul 2004 11:00:09 GMT, "Highland Ham"
wrote:

I always get confused with transformations,
for some reason. :-/

Paul, maybe others. I don't know if this would help but I had to explain
this "square root of" thing to some students who had some basic knowledge of
DC and knew enough about AC electricity to understand a transformer stepping
voltage up or down. And it seemed to help?

It went like this.

Draw two boxes on a sheet of paper to represent transformers.

The first transformer has an input of 20 volts at one amp; so the impedance
(Volts /amps = 20/1 = 20 ohms.)
This first transformer has an output of 20 volts at one amp (forget about
transformer losses and all that stuff) so the secondary impedance is also
20/1 = 20 ohms.
In other words it has a one to one ratio both for voltage and impedance
matching.

The second transformer also has an input of 20 volts at one amp. 20 ohms
again.
This second transformer has an out put of 10 volts at two amps. So 10/2 = 5
ohms. Notice that this is not HALF of the input impedance but one quarter;
which start to give one a clue to what is happening?
The voltage ratio of this second transformer is two to one. That is 20/10 =
2
The impedance ratio however is 20/5 = 4 to one. Think about it!
Voltage is 2 to one, but impedance is 4 to one.

You can try it again with say a ten to one transformer.
Twenty volts at one amp input to, say, 2 volts out at 10 amps. OK?
Voltage ratio = 20/2 = 10
And the secondary impedance will be 2 volts divided by 10 amps = 0.2 ohms!
So the impedance ratio of this ten to one trans former will be 20 divide by
0.2 or 200/2 which is 100 to one!

So in all the examples the impedance transformation ratio of the transformer
depends on the 'Square of the voltage ratio. Or;
Doing it the other way round the transformation ratio is the 'Square root of
.......' etc.

Last example (Pause for breath!)
A 6V6 output tube has, say, a recommended out put load impedance of 5000
ohms?
To match that to a 4 ohm speaker the IMPEDANCE ratio is 5000/4 = 1250
So I need a suitable transformer with a TURNS ratio which is the 'square
root' of 1250.
The transformer should have a ratio of 'around' 36 to 1. (36 x 36 =
1296).
And 1296 x 4 = 5184 which is very close.
Since this is an audio 'output transformer' voltage is not usually
mentioned, except the ability of the transformer's insulation to withstand
radio DC B+ voltages of 200 to 350 volts above chassis.

AIUI in practical radio work the ratio is 'not that critical'; since we are
dealing with a range of frequencies and typically one or a couple of
speakers in a wood or plastic box which may represent different impedance
loads at different frequencies anyway; so in practice anything from 30 or 40
to one would at least be likely to work OK. Certainly to get a set working
with a non triode type output tube working somewhere in the right region in
regard to matching the speaker?

My first experience, at a tender age, was connecting some relatively high
impedance, war surplus, headphones to the four ohm speaker output of a 1930s
HMV and wondering why the sound was all "Wuffly" and distorted! I was
rescued by my mentor (Elmer) my favourite uncle who set me on the right
road. Ultimately I used a scrapped 'front door bell transformer' which had a
ratio of roughly 240 volts to 20, (turns ratio 12 to one) backwards, to
transform the approx. four ohm radio set output to something closer to my
headphones!
If I can remember; volt/turns ratio was 240/20 = 20.
Impedance transform 20 x 20 = 400.
And 400 x 4 = 1600! And I think the head phones were 2000 ohms?
Anyway it sounded much better!
Sorry for the long 'rant'. Terry.