Transformer winding in LTspice gives various output

[gauthamtechie]

Using Various Inductances in a transformer circuit in LTspice give me various current outputs, even out of a constant current source with a 1:1 ratio. Although when I change the winding to a higher value, the transformer works as it should in a 1:1 ratio.

Why is this?

 

[alec_t]

What do you mean by 'constant current source'? That term normally refers to a DC current source, and transformers don't work with DC.

 

[gauthamtechie]

I mean a sinusoidal current source with a constant RMS value at any instant; sorry for the ambiguity

 

[alec_t]

Using Various Inductances in a transformer circuit in LTspice give me various current outputs, even out of a constant current source with a 1:1 ratio.

I've just tried it using a 1:1 ratio and K=1, with inductance values from 1nH to 1H. In all cases the secondary voltage equalled the primary voltage and the primary and secondary currents were equal provided the load resistance was vanishingly small.

 

[ronsimpson]

You should send your circuit. A screen picture is fine.

A 1:1 transformer with a k=1 should have the same primary and secondary voltage.
Do you know how to make a 2:1 transformer?

 

[Jony130]

Do you know how to make a 2:1 transformer?

Well very simple task.
2:1 in turns ratio equals 4:1 in inductance ratio.

 

[ronsimpson]

I don't know what the problem is.

If the inductance is high the transformer will work fine like in you attached thumbnail.

If the inductance is low (4uH, 1uH) the input current will be high and the input/output voltage will be low. Maybe this is what you are questioning.
1) I did not get your V1 to come through so I made one. My signal source has 100 ohms if impedance. (by default) It can drive the 1k fine. It can drive the 4H because the impedance of 4H is much higher than 100 ohms. As you change the transformer impedance down, at some point the impedance of the inductance at 50hz is 100 ohms. At this point the internal 100 ohms in V1 and the 100 ohms from inductor makes a 2:1 divider and you loose 1/2 of your signal. At 10th that inductance the 100 ohms will eat up most of the signal. At 4uH most all the signal is across the internal 100 ohms and you get a very small voltage out. (2.3A is what happens when 230V is places across the 100 ohms inside the signal source) Current flows! Not much voltage will get out.

This is what happens with a real signal generator. There is 50 or 100 or 600 ohms of internal impedance inside the box. If you short out the leads no voltage is seen but current flows. At 50hz a 4uH coil is a short.

Does this help?

 

[audioguru]

LTspice input is the PEAK voltage, not the RMS voltage. Your input is 230V peak and the output is about 110V peak.

Your simulation shows two separate inductances that are not coupled together like in a transformer.

 

[ronsimpson]

Your simulation shows two separate inductances that are not coupled together like in a transformer.

L=L1 L2 1 This makes the two inductors coupled, 100%.

 

[audioguru]

L=L1 L2 1 This makes the two inductors coupled, 100%.

Thanks. I have never simulated a simple transformer. I calculate it, buy it and use it instead.

 

[Roff]

Did anyone else notice that the circuit posted was not from the OP (unless Jony130 has two accounts here)?

 

[Roff]

Using Various Inductances in a transformer circuit in LTspice give me various current outputs, even out of a constant current source with a 1:1 ratio. Although when I change the winding to a higher value, the transformer works as it should in a 1:1 ratio.

Why is this?

I suspect that you were using a sine wave input, with transient simulation. You have to make the input frequency substantially above the cutoff frequency, which is R/(2*pi*L), where R is the load resistance.
For example, if your inductance is 1mH, and your load is 1Ω, your high pass corner frequency is 159Hz. Make you sine wave input >1kHz if you want full output (see the green curve in the attached Bode plot).

 

[gauthamtechie]

Did anyone else notice that the circuit posted was not from the OP (unless Jony130 has two accounts here)?

You are right, it wasn't me. I didn't post that circuit

 

[Roff]

You are right, it wasn't me. I didn't post that circuit

OK. Did you see my reply in post #12?

 

[gauthamtechie]

OK. Did you see my reply in post #12?

Yes. I get the idea that the frequency must exceed a certain cut-off frequency in order for it to be able to reliably conform to the Turns ratio. I hope that is right.
And the Bode plot part I am yet to understand completely because the concepts of the plot itself, and how I need to interpret the gain is not known to me. I have to study these concepts further to comprehend this.

 

[MikeMl]

...the Bode plot part I am yet to understand completely because the concepts of the plot itself, and how I need to interpret the gain is not known to me. ...

Any node voltage or branch current plotted on an LTSpice Bode plot by default is plotted in db with respect to the main AC voltage source or AC current source used for the .AC analysis run.

In Roff's attachment, the independent variable of the simulation is the frequency of the I1 current source. The amplitude is 1A.

The dependent current through R1 shown as I(R1) Green trace on the plot is decibels relative to the 1A that I1 supplies. To the right of the corner frequency, I(R1) approaches the same value as I(I1)=1A, so the value approaches 0 db, which is a gain of 1 due to the turns ratio of the transformer. Remember that db of a current ratio is 20log(I2/I1).

At 1Hz, the green trace is down more than -42db, so the actual current through R1 would be

-42 = 20*log[I(R1)/1]
- 2.1 = log[I(R1)]
I(R1) = 10exp(-2.1) = 0.0079A = 7.9mA

(-40db is a factor of 1/100)

btw-you can turn off the db mode in an LTSpice Bode plot, and plot the ratio directly as a real and imaginary or magnitude and phase.

Roff will have to explain what the red and blue traces are?

 

[Roff]

Any node voltage or branch current plotted on an LTSpice Bode plot by default is plotted in db with respect to the main AC voltage source or AC current source used for the .AC analysis run.

In Roff's attachment, the independent variable of the simulation is the frequency of the I1 current source. The amplitude is 1A.

The dependent current through R1 shown as I(R1) Green trace on the plot is decibels relative to the 1A that I1 supplies. To the right of the corner frequency, I(R1) approaches the same value as I(I1)=1A, so the value approaches 0 db, which is a gain of 1 due to the turns ratio of the transformer. Remember that db of a current ratio is 20log(I2/I1).

At 1Hz, the green trace is down more than -42db, so the actual current through R1 would be

-42 = 20*log[I(R1)/1]
- 2.1 = log[I(R1)]
I(R1) = 10exp(-2.1) = 0.0079A = 7.9mA

(-40db is a factor of 1/100)

btw-you can turn off the db mode in an LTSpice Bode plot, and plot the ratio directly as a real and imaginary or magnitude and phase.

Roff will have to explain what the red and blue traces are?

Good explanation, Mike!

The green trace is when L=1mH. Blue is when L=30mH. Red is when L=1H. These are called out in
.step param L list 1m 30m 1.