Is the value of a capacitor placed in series with the secondary of a
transformer multiplied (as viewed from the primary) by the turns ratio
or the square of the turns ratio?

Thanks,

T

If the transformer has a step-up turns ratio (from primary to secondary)
then the resistance, inductive reactance and capacitive reactance connected
to the secondary (as measured at the primary) decrease as the square of the
turns ratio. Therefore effective capacitance measured at the primary
increases as the square of the turns ratio. For a step-down transformer the
effects are the opposite.

Bill W0IYH

TRABEM wrote in message ...
Is the value of a capacitor placed in series with the secondary of a
transformer multiplied (as viewed from the primary) by the turns ratio
or the square of the turns ratio?

Thanks,

T

On Sun, 18 Sep 2005 17:47:09 GMT, "William E. Sabin"
wrote:

If the transformer has a step-up turns ratio (from primary to secondary)
then the resistance, inductive reactance and capacitive reactance connected
to the secondary (as measured at the primary) decrease as the square of the
turns ratio. Therefore effective capacitance measured at the primary
increases as the square of the turns ratio. For a step-down transformer the
effects are the opposite.


OK Bill,

I have a series tuned loop that is resonant with 1000 pf and fed into
a 10 to one step up transformer.

If I remove the 1000 pF cap completely, I can resonate the loop at the
same frequency with 10 pF in the secondary?

It almost sounds too good to be true::

Regards,

T

Since the tranformer now becomes component of the tuned circuit, you
would seriously decrease the Q of the tuned circuit when doing so.

Harry C.

Your original question suggested an ideal transformer, which the loop
antenna undoubtedly is not, therefore the detailed answer is more
complicated. The loop impedance is an LCR network that would require a more
detailed study, which I am not able to offer right now.

If your 1000 pF gets the job done, that is probably a good solution that
tunes out the net effective inductive reactance.

Bill W0IYH

TRABEM wrote in message ...
On Sun, 18 Sep 2005 17:47:09 GMT, "William E. Sabin"
wrote:

If the transformer has a step-up turns ratio (from primary to secondary)
then the resistance, inductive reactance and capacitive reactance
connected
to the secondary (as measured at the primary) decrease as the square of
the
turns ratio. Therefore effective capacitance measured at the primary
increases as the square of the turns ratio. For a step-down transformer
the
effects are the opposite.


OK Bill,

I have a series tuned loop that is resonant with 1000 pf and fed into
a 10 to one step up transformer.

If I remove the 1000 pF cap completely, I can resonate the loop at the
same frequency with 10 pF in the secondary?

It almost sounds too good to be true::

Regards,

T

Thanks to both Bill and Harry.

I'm trying to vary the resonant frequency of a loop with an air
variable. The loop impedance is under 1 ohm, so it needs a step up
transformer anyway, and toroids are commonly used.

Since a 1000 pf varaible is much harder to get than a 10 pF, I was
pleased to learn that I could multiply the effective value by
multiplication of the actaul value times the square of the turns
ratio.

It makes the job much easier, even if the Q of the loop suffers some.

Thanks to you both,

T

TRABEM wrote in message ...
Is the value of a capacitor placed in series with the secondary of a
transformer multiplied (as viewed from the primary) by the turns ratio
or the square of the turns ratio?

Thanks,

T

Since I just started playing with the LMS impedance bridge featured in this
months QST (and the current QEX), I thought it would be interesting to try
to answer this one by experiment.

I have an old audio matching transformer labeled to transform 20 ohms to 8
ohms, a 2.5 ratio. I took a 10 uF capacitor and measured it at 9 uF, with Z
of 14.43/_ -85 degrees at a measurement frequency of 1225 Hz. Hooked it to
the low resistance side of the transformer and the meter to the high side.
The reading obtained was 1.99 uF or Z = 74.4/_ -62 degrees.

So I got a transformation ratio of 4.5 or 5 in terms of capacitance or
impedance, instead of the expected 2.5.

Now, how far is the transformer from ideal? Measuring its high side
inductance with the low side open, I get 10.1 mH. This is the magnetizing
inductance of the transformer, which appears in parallel with the
transformed capacitive reactance. In a transformer that approaches ideal,
the inductance is large enough to have a very large reactance compared to
the transformed load from the secondary. In this case it's only sort of
negligible. The transformed capacitance of 9/2.5 uF = 3.6 uF has a
reactance of -j36.1 ohms. The reactance of the transformer's magnetizing
inductance is j77.7 ohms. Combining the two in parallel gives a resultant
capacitive reactance of -j67.11. And this is equivalent to 1.94 uF at my
measurement frequency. Not too far from the measured value of 1.99 uF. I
did neglect leakage inductance, assuming that it would be pretty small for
this type of transformer.

Another interesting point to note is that the unadorned capacitor had a
phase angle of 85 degrees, but the transformed version showed 62 degrees.
The equivalent loss resistance of the transformer decreased the quality a
fair amount, as one would expect I guess.

So I guess the answer to your question is yes, it will be transformed just
like a resistive load would be, providing the transformed capacitive
reactance is fairly small with respect to the magnetizing reactance of the
transformer.

73--Nick, WA5BDU