[time-nuts] Zeeman Audio Source?

Brooke Clarke brooke at pacific.net
Thu Jan 13 00:37:54 UTC 2005


Hi Chusk:

Maybe I'm having a bad hair day.

Let's say we have a 4:1 step down transformer.
The 204B puts out 5 Volts open circuit.

When the transformer is installed between the 204B and the 50 Ohm Zeeman 
input we have:

Voltage at transformer secondary = 5 / 4 = 1.25
Source resistance of transformer secondary = 600 / (4*4) = 37.5 Ohms

The new voltage divider at the secondary is 37.5 Ohms over 50 Ohms or 50 
/(37.5 + 50) = .571
Output voltage is 0.771 * 1.25 = 0.714 Volts.

Another way to look at it is if a 600 Ohm load is added and if the 
internal impedance is 600 Ohms then the voltage will drip to 2.5.
power in the load = 2.5 * 2.5 / 600 = 10.4 mw vs the 20 mw needed for 
the Zeeman input.

73,

Brooke Clarke, N6GCE


Chuck Harris wrote:

> Hi Brooke,
>
> When I do the math, I get:
>
> P = (1V)^2 / 50 = 20mw
>
> and
>
> P = (5V)^2 / 600 = 42mw
>
> The power is there, so therefore, so is the voltage with the
> correct transformer.
>
> Or, another way, remembering that the ratio of the impedances
> is the square of the turns ratio:
>
> 600/50 = 12:1 impedance, therefore we need a transformer
> that has a SQRT(12) : 1 turns ratio, which is about 4:1
>
> -Chuck
>
>
> Brooke Clarke wrote:
>
>> Hi Chuck:
>>
>> My first thought was a transformer, but I did the math and don't see 
>> how it can work.
>> An amplifier sure will and that's probably the most inexpensive way.
>>
>> Have Fun,
>>
>> Brooke
>
>
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