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Need help modeling IF transformer in LTspice

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joeh100

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I need help modeling the following IF transformer; see data sheet: http://www.mouser.com/ds/2/449/XC-600131-205883.pdf
It is the model no. 42IF123-RC 25K in, 4K out, cap color green.

I notice it's resonating at around 8 MHz instead of 10.7 MHz. Can anyone offer any feedback and/or advice on my circuit model? Do I have the Inductance values correct based on the data sheet? Do I have to use a more complex spice directive?

I have attached my spice circuit for LTspice using a simplified spice directive for the transformer.

Thanks,

Joe
 

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JimB

Super Moderator
Most Helpful Member
I cannot comment on the simulator, but a simple calculation shows that the resonant frequency of 52pF and 4.5uH is 10.45MHz.

JimB
 

joeh100

Member
Thanks Jim but it has something to do with the transformer and resonance which I don't quite understand. My circuit should be resonating around 10.9 MHz based on a primary inductance of 4.5uH and 47pF. If I comment out the spice directive for the transformer it works as expected.

I need someone that understands RF and or transformers to help.
 

ronsimpson

Well-Known Member
Most Helpful Member
Take L2 out of the K1 expression. The leakage inductance is not coupled.
-----edited-----
Maybe I don't understand what you are doing with L2. It looks like leakage inductance.
----edited again-------
L1=9 turns
L2=5 turns
L3=2 turns
Then why L4?
-----edited-----
The problem is in coupling. Give me time to think about it.
-----edited-----
The problem is that 5 turns +9 turns makes 14 turns.
14T=4.5uH
There is a turns squared in the formula.
L1+L2 does not equal 4.4uH it is L1 squared + L2 squared = 4.5uH.

Think about a center tapped transformer. It the total = 4uH then each half = 1uH. If 2 turns=4uH then 1 turn=1uH
-------------
Please check my work.
1.86uH and 0.574uH for the primary.
92nH or the secondary. L3 only, L4 removed.
Now 10.9mhz
 
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joeh100

Member
Thanks!

I actually came up with those figures earlier but changed my math to force L1+L2=4.5uH, because I "knew" that inductors in series have to add up, lol. Well, obviously not for mutual inductance I see.

I split the secondary to experiment with. My first calculation was again right at 92nH but I divided it in half in error and used 2 46 nH inductors for the secondary.

Again, thanks a lot. That really helped.
 

RCinFLA

Well-Known Member
The impedance ratio is the square of turns ratio between pins 1 & 2 primary and 4 & 6 secondary. N_ratio^2 = impedance ratio.

The inductance is 1-3 primary.

It looks like over coil pocket turning core which has a lot of tuning range. The xx pF +ext xx pF means the internal capacitance plus external contribution of ext. xx pF. The external capacitance contribution looks like typical number for collector of transistor, assuming low feedback Ccb transistor. 1/y22 for bipolar or FET is very high real part, so load will be basically what is transformed from secondary load on transformer (through the impedance ratio transformation) to high impedance primary side of transformer.

Q is unload Q of primary tank. Q being Rp_coil / (2*PI*10.7 MHz * L primary pin 1-3)
 

Kaizer

New Member
I wonder how did you guys calculate the secondary which is 92 nH. I followed and understood how the primary inductances (1.86 uH and 0.57 uH) were calculated.
 

joeh100

Member
Tp/Ts = sqrt(Lp/Ls)

Tp (primary turns) = 14
Ts (secondary turns) = 2
Lp (primary inductance) = 4.5uH

Solving for Ls = Lp/(Tp/Ts)^2
4.5e-6/49 = 91.837nH
 
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Kaizer

New Member
Thanks, Joe. I really appreciate it. I was modeling two IF Transformers for AM detection and AGC. This thread helped me tremendously. I got it to work last night.
 

Tsif

New Member
Hi Kaizer, I want to modeling IF Transformers 455KHz color yellow, can you help me. I'm new in LTspice. Please can you give me your model and what to chango for 455KHz IF transformer. Thanks.:)
 

ronsimpson

Well-Known Member
Most Helpful Member
IF Transformers 455KHz color yellow
There is not much information given.
Can you measure the inductance?
Some transformers have capacitors inside.
We need to know the turn ratio. How many taps.
 
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Tsif

New Member
Thanks for your answer ronsimpson, This is the instruction in datasheet datasheet.png , I have not an inductance meter for mesuring the real inductance :(.
 

ronsimpson

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Inductance = turns squared. Impedance = turns squared.
Inductance from 1 to 3 = 680uH.
Inductance from 1-2 = 135uH. Inductance from 2-3 = 208uH. Inductance from 4-6 = 1.35uH
In LTSpice there is no "turns" or "60k ohms". Those are just comments. It is all about inductance.
Inside the transformer there is 180pF. They assume you will add 5pF with a transistor.
upload_2017-3-1_13-11-45.png
L1,L2,L3=three inductors. "k l1 l2 l3 1" makes the three a transformer. "1" connects them together very well. 0.98 might be better.
Note I added a Rser to V1. I also added R1. for a test.
I did not think about "Q". That could be fixed later.
upload_2017-3-1_13-17-48.png
Here is a frequency sweep of 400khz to 500khz. Vertical is in db not volts.
 

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alec_t

Well-Known Member
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Inductance from 1-2 = 135uH. Inductance from 2-3 = 208uH.
Doesn't that give a total primary inductance of 343uH? The datasheet says 680uH.
 

ronsimpson

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Doesn't that give a total primary inductance of 343uH? The datasheet says 680uH.
No.
Inductance in a coupled transformer do no add up. They to the "turns squared" thing.
example: If you had two separate inductors, of 100uH they would add up if in series. (200uH)
BUT
In a transformer they will be 400uH. (2x the turns is 2 squared or 4x the inductance)
Inductance = turns squared. Impedance = turns squared.
 

alec_t

Well-Known Member
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I get the squared thing, but how is the 680uH figure arrived at?
 

ronsimpson

Well-Known Member
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how is the 680uH figure arrived at?
upload_2017-3-2_17-59-40.png
upload_2017-3-2_18-0-52.png
From the data sheet.
There is a cap of 180pF from pin 3 to 1. (they made this cap 5pF small to make up for loading on the outside)
The transformer should resonate at 455khz.
Inductance is 680uH. (What pin????)
455khz=2 pi squart(180pf * 680uH)
By reading between the lines; I assume 680uH must be from pin 3-1 because it must be across the 180pF.
------------------------------
More reading between the lines:
Pin 1-2 = 70Turns, Pins 4-6 = 7Turns. That is a 10:1 turns and 100:1 impedance.
Primary = 60k ohms. Secondary = 600 ohms. So secondary must be pins 4-6. Primary must be pins 1-2.
The "resonant" part must be pins 1-3.
------------------------------
I know I guested on some of this. Reading a data sheet is an art-form.
I have written data sheets. Usually only 50% of the information in the engineer's head makes it to paper. The other 50%, he assumes you know. Every one in the factory knows the primary is pins 1-2. So it did not get written down.
------------------------------
Another example of 50% of the information.
Romans had "Century", a group of 100 men. Every one knows this.
During the 2nd Punic wars, they had problems getting enough men so Century=60 men. A historian wrote that they went out with 60. BUT We know there are 4 men that don't fight. (trumpet and flag, etc) So did they have 60 men that included the 4 non-fighters or did they have 60 fighters +4 =64. Those who know directly are all dead. Those who wrote about this assumed that every one knows. OR Those who wrote about this did not really know. OR Those who wrote, did not realize that there would be a question.
 

Tsif

New Member
Pardon c'est la première fois que je vais utiliser les transformateurs FI. Est-ce que je peut inverser le primaire et le secondaire lors que j'utilise le transformateur? et encore merci à ronsimpson pour ses explication. :)
 

alec_t

Well-Known Member
Most Helpful Member
Depending on your circuit, reversing the primary and secondary could still work, Tsif. (Anglais, s'il vous plait).
455khz=2 pi squart(180pf * 680uH)
Not quite. 455kHz = 1/(2*pi*sqrt(180pF*680uH)).
Intuitively I still think the inductance sum of the two primary sections (i.e. from pin 1 to 3) should be 680uH, with the sections having inductances in the ratio (87/70)^2, i.e 1.54. This would give values of 680uH*1/(1+1.54) and 680uH*1.54/(1+1.54), i.e. 268uH and 412uH respectively. Correct me if I'm wrong.
 
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