Turns ratio on a 70V 10W Line matching transformer??

[royalmp2001]

I have come across a project that uses an old Radio Shack 70V 10W Line
transformer.

The primary has taps for "C", 10W, 5W, 2.5W, 1.25W and 0.62W
The secondary has taps for "C", 4ohms, 8Ohms and 16Ohms.

What is the turns ratio if the 5W and 0.62W taps only are used on the
primary and the "C" and 16Ohms taps are used on the secondary?
How is it calculated??
THanks

 

[Phil Allison]

"royalmp2001"

I have come across a project that uses an old Radio Shack 70V 10W Line
transformer.

The primary has taps for "C", 10W, 5W, 2.5W, 1.25W and 0.62W
The secondary has taps for "C", 4ohms, 8Ohms and 16Ohms.

What is the turns ratio if the 5W and 0.62W taps only are used on the
primary and the "C" and 16Ohms taps are used on the secondary?
How is it calculated??

** The whole primary takes 0.62 watts at 70 volts rms when the secondary is
loaded as indicated.

P = V squared / R

so R = V squared / P

= 70 x 70 / 0.62

= 7903 ohms

The turns ratio relative to 16 ohms is

sq.rt 7903 /16 = 22.2:1


For the 5 watt primary, the load impedance is

70 x 70 / 5 = 980 ohms.


The ratio relative to 16 ohms is:

sq rt 980 /16 = 7.82:1



Finally, the ratio for the connection you gave is

22.2 - 7.8 = 14.4:1

The partial primary impedance is:

14.4 squared x 16 = 3318 ohms.



BTW:

No account of copper losses is included.

Actual ratios will be about 15% smaller if you include them.




........ Phil

 

[John Fields]

I have come across a project that uses an old Radio Shack 70V 10W Line
transformer.

The primary has taps for "C", 10W, 5W, 2.5W, 1.25W and 0.62W
The secondary has taps for "C", 4ohms, 8Ohms and 16Ohms.

What is the turns ratio if the 5W and 0.62W taps only are used on the
primary and the "C" and 16Ohms taps are used on the secondary?
How is it calculated??

---
Normally, the source is connected to the primary between 'C'
(common) and the tap delivering the required power into the load,
and the load is connected to the secondary between 'C' and the tap
corresponding to the impedance of the load.

Your request implies that the source will be connected between the
'5w' and '0.62W' taps. Is that really what you want to do?

If it is, would you tell us why?

 

[John Fields]

---
Normally, the source is connected to the primary between 'C'
(common) and the tap delivering the required power into the load,
and the load is connected to the secondary between 'C' and the tap
corresponding to the impedance of the load.

Your request implies that the source will be connected between the
'5w' and '0.62W' taps. Is that really what you want to do?

If it is, would you tell us why?

---
In any case, since for 5 watts:


Vp = 70.7V

and

Vs = sqrt (5W * 16R) ~ 8.9V,

the turns ratio will be:


Np Vp 70.7V
n1 = ---- = ---- = ------- ~ 8
Ns Vs 8.9V


For 0.6 watts it'll be:

70.7V
n2 = ------- ~ 22
3.2V

Then, since the winding between 'C' and '5W' won't be connected,


n3 = n2 - n1 ~ 14,


in pretty close agreement with Phil's 14.4.

 

[BobG]

n3 = n2 - n1 ~ 14,
in pretty close agreement with Phil's 14.4.

===========================================
I bet I can make Phil cuss John.... watch this......
John, I think your number is correct.

 

[John Fields]

===========================================
I bet I can make Phil cuss John.... watch this......
John, I think your number is correct.

---
Actually, they're both correct.


Phil did his with n = sqrt (z1/z2)

Vp
I did mine with n = ----
Vs

The difference in our results is because I rounded more.

 

[Phil Allison]

"John Fields"

The difference in our results is because I rounded more.

** But aren't you Texans famous for " rounding up " rather than rounding
down ??





........ Phil

 

[John Fields]

"John Fields"




** But aren't you Texans famous for " rounding up " rather than rounding
down ??

 

[royalmp2001]

---
Normally, the source is connected to the primary between 'C'
(common) and the tap delivering the required power into the load,
and the load is connected to the secondary between 'C' and the tap
corresponding to the impedance of the load.

Your request implies that the source will be connected between the
'5w' and '0.62W' taps. Is that really what you want to do?

If it is, would you tell us why?

 

[royalmp2001]

Normally, the source is connected to the primary between 'C'
(common) and the tap delivering the required power into the load,
and the load is connected to the secondary between 'C' and the tap
corresponding to the impedance of the load.

Your request implies that the source will be connected between the
'5w' and '0.62W' taps. Is that really what you want to do?

If it is, would you tell us why?

Good question, John.

I got this from an article published in the 80s to step-up the output
of a BK 3011 function generator by sending the square wave output to
the C and 16Ohms secondary taps, and taking the output from the 5W and
0.62W taps on the primary. According to this article this arrangement
best matches the impedance of the generator.

 

[jasen]

I have come across a project that uses an old Radio Shack 70V 10W Line
transformer.

The primary has taps for "C", 10W, 5W, 2.5W, 1.25W and 0.62W
The secondary has taps for "C", 4ohms, 8Ohms and 16Ohms.


What is the turns ratio if the 5W and 0.62W taps only are used on the
primary and the "C" and 16Ohms taps are used on the secondary?
How is it calculated??

using V=sqrt(P*R)

turns ratio for 5W to 16 ohms is 7.82:1

for 0.62W is 22.23:1

between 5 and 0.62 will be the difference 14.41:1 (inpututput)

Bye.
Jasen