Transformer Impedance calculation, with shunt?

[rhythminmind]

I understand how to use this formula Zp/Zs = (Np/Ns)^2 to show out what impedance values reflect at the in/output of a transformer.
If you add a load resister shunt to the secondary, how does that fit into the formula?
Zp/Zs = (Np/Ns)^2 + Shunt maybe?

Say you have a 10000:600 with 1k across secondary, how will that reflect impedance values?

 

[CJ]

ok z ratio  is 10000/600 = 16.666, spooky eh?

so the turns ratio is the sq root of the z ratio,

so 16.66 root is about 4,

so to find the reflected load, simply divide 1 k by 4,

that would be 250 ohms reflected if the resistor was on the 10000 side,

if it was on the 600 side then multiply by 4 which would be 4 k load on the 10000 side,
somebody check my math, i just got up and did the wake and bake, 

 

[Jean Clochet]

somebody check my math, i just got up and did the wake and bake, 

;D  ;D

CJ, I think you need to bake a bit more 
I know you know this stuff.

Reflected impedance to the primary with a 1K resistor on a 10K:600 transformer secondary is 16.666K.  Ignoring the additional terms from primary and secondary DCR of course as we haven't been told what these are.

 

[rhythminmind]

Thanks for the replies.
Thats just the reflected value of the 1k shunt correct? Still unclear how you factor in the values with the source, shunt, & load all connected? I don't know how the shunt changes the original Zp/Zs = (Np/Ns)^2 formula.

Example -
50ohm into a 10k:600, with a 1k shunt across the 600 secondary, feeding a 1m input. ( 50 > 10k:600 > 1k across Ns > 1m )

 

[Jean Clochet]

Example -
20k into a 10k:600, with a 1k shunt across the 600 secondary, feeding a 1m input. ( 20k > 10k:600 > 1k across Ns > 1m )

I don't really understand what you mean by 20k into a 10K:600 transformer.  What's the 20K?
As far as adding a shunt to the secondary which is already connected to a 1M ohm load, the 1KR and 1MR are in parallel so, again ignoring DCR's, the parallel load on the secondary is 999R which would reflect a primary impedance of 16.65K. 

The 10K:600 transformer doesn't really present any impedance by itself.  All that spec indicates is the ratio and the typical impedances that it is designed to work between.  There is some reflection from pri. to sec. and sec. to pri. due to the DCR of the windings though. 

 

[rhythminmind]

Yeah my bad (copy/paste mishap) should be 50ohm not 20k. Thanks for that explanation of the load in parallel, that helped me out.
If the 999r is on the secondary that makes Zp/999 = (10000/600)^2  =  277500 reflected to the primary. That dosen't seem right.

 

[Jean Clochet]

If the 999r is on the secondary that makes Zp/999 = (10000/600)^2  =  277500 reflected to the primary.  That's dosen't seem right.

Reflected impedance is just the same as 10K/600  so multiply the secondary load by 16.666 
Turns ratio is the same as voltage ratio and is the square root of impedance ratio which, in this case is 4.082
Which is a voltage loss (or gain if you're going 600:10K)  of 12.22dB

If you really needed to be exact with impedance calculations (for noise considerations maybe), you'd also factor in the DCR of the windings as well. 


Going the other way from your 50r source impedance, and being more precise, you'd add the source 50r to the DCR of the primary.  Divide this by the impedance ratio of 16.666.  Add to that the DCR of the secondary and then figure the parallel of this sum with your 999r load.  This would be the source impedance presented to the load destination.

 

[rhythminmind]

Ahhh I was getting it wrong with the turn ratio.
Ok for archives sake, let me check this.

Example -
50ohm into a 10k:600, with a 1k shunt across the 600 secondary, feeding a 1m input. ( 50 > 10k:600 > 1k across Ns > 1m )

If I want to know the impedance reflected to the primary

The turn ratio of the 10k:600 = 4:1 , The 1KR + 1MR parallel resistance = 999R

Using the Zp/Zs = (Np/Ns)^2 to find the reflected impedance's.

Zp/999 = (4:1)^2 makes Zp=15984R

The impedance reflected  for the secondary would then be

50/Zs = (4/1)^2 making Zs = 3.125R + the 999R load = 1002.125

Correct? (ignoring DCR that is)

 

[Jean Clochet]

Correct?

Yes 

 

[rhythminmind]

Thanks for the patience/help.

 

[CJ]

transformer dcr might be at a 90 degree angle to the resistor, so it will not do as much damage as you think,

you can look at this from a power transformer point of view,

a 1:4 transformer will make 1 volt into 4 volts,

if you have a 4 ohm resistor on the secondary, then 1 amp will flow.

this means you have a 4 watt system on the secondary.

now the current needed on the primary to make 1 amp on the secondary will be

volts times amps pri = volts times amps sec

so 1 volt times X amps = 4 volts times 1 amp

so 1 volt times X amps = 4, so X = 4 amps pri current needed

or use the watts in = watts out and say we have a 4 watt system on the pri also, this means that 1 volt needs 4 amps to make 4 watts, a little easier way of calculating,

so 1 volt at 4 amps, use ohms law on that, to figure out the load reflected  to the pri

1/4 = 0.25 ohms,

so yes, the reflected load is by way of the Z ratio, not the turns ratio,

since 0.25 ohms is 4 ohms/0.25 = 16 which is the z ratio.

use this power approach as a brute force way to correct yourself, 

the main thing to see is that since power is I^2R, the R must follow this formula, which is quadratic in I, not linear, like the volts times amps formula.
so the Z ratio formula is really derived from the Watts formula,


sec: 1 amp ^2  times 4 ohms = 4 watts

on the pri side, 4 amps ^2 = 16, so if I^2R = 4 watts , then 16 times R = 4, R = 0.25.