Transformer modeling in LTSpice

[Wond3rboy]

Hi, i am trying out the mutual indance directive to model a simple step down transformer in LTSpice and am using it in a bridge rectifier but for some reason, i cant get the expected output voltage from the secondary. I do understand that the square root of the turn ratio is is equal to the inductance ratios of the windings.

Turns-Ratio (TR) = Npri/Nsec

Inductance-Ratio (LR) = (Npri/Nsec)^2 = (TR)^2

So if i have LR=2, the TR=4, hence the output voltage at the secondary should be a quarter of the primary. I choose the primary inductance to be For a 20V 1Hz sine wave at the primary, the output should be 10V on the secondary. But due to inductive reactance, to prevent loading i make the resistance of the primary winding 1k to get a proper 20V signal on the primary. The questions i have are follows:

1. When i get 20v on the primary(with a coil resistance of 1k), the output of the secondary is in microvolts.

2. Is there anything wrong in the math?

thanks.

 

[MrAl]

Hi,

With a given inductance and a given frequency the inductance of the primary can not support the source voltage so what happens is the primary series resistance (source plus winding resistances) absorbs all the voltage. This means there is little left to transfer to the secondary. You'll also note that some of your specs are a little out of wack, such as 1k primary resistance.
A big factor is the connection of the secondary to the diodes is not correct to start with. The secondary
is supposed to drive the diodes not the primary.

Try these quick start fixies:
1. Increase the frequency of the source to 100Hz.
2. Decrease all the series resistances to 1 ohm (source, primary, secondary) so that is 3 resistances you have to decrease.
3. Connect the source to ground through a 10 meg resistor instead of directly to ground, then connect the secondary bridge output to ground directly. This allows you to view the output without using a differential voltage monitor.
4. Might have to increase the primary and secondary inductance.

For example, with 1 ohm for all the series resistances and 100Hz and 2H and 1H inductances, the output wound look fine.

One last thing:
To see the output wave properly you'll probably have to set the maximum time step to something like 10us instead of 0.1 seconds. This is so that the internal program routines take small enough steps in the solution so that they produce small enough steps in the output display so you can view it more as it would really look like in real life.
Might also have to decrease the output load to 10k or even 1k.

 

[Wond3rboy]

Hi, thanks MrAL.

The incorrect diode connections were embarassing

Made the changes that you recommended. I made the primary inductors resistance 1k so as to avoid loading, i got proper 20V at the primary so i guessed that i would get the converted voltage on the secondary. The problem that i get now is that the negative half cycle of the secondary voltage is nearly non existent. Is it because i have connected the -ve output to ground? I know i need to have a ground on the secondary loop of the circuit. Thanks again.

 

[MrAl]

Hi, thanks MrAL.

The incorrect diode connections were embarassing

Made the changes that you recommended. I made the primary inductors resistance 1k so as to avoid loading, i got proper 20V at the primary so i guessed that i would get the converted voltage on the secondary. The problem that i get now is that the negative half cycle of the secondary voltage is nearly non existent. Is it because i have connected the -ve output to ground? I know i need to have a ground on the secondary loop of the circuit. Thanks again.

Hi again,


Since there was a gross error in the original connections (not that it is that hard to make that kind of mistake and it is after all quite common) i would suggest that you post the new circuit the way you posted the original circuit and we can take a direct look at that. it may be something simple again. Seeing the file will help solve this problem right away.

 

[BobW]

In order for a transformer to work properly, the coupling coefficient k needs to be very close to 1. A value of 0.95 is reasonable. The coupling coefficient is the ratio of the mutual inductance to the square root of product of primary and secondary self inductances:

k=M/sqrt(Lp*Ls)
which gives:
M=k*sqrt(Lp*Ls)

If you adjust your parameters so that k is in the range of 0.9 to 1.0 you'll be on the right track.

Another thing to keep in mind is a general rule of thumb that the reactance of the primary should be sufficient that it doesn't load down the power supply. A reactance of 1000 ohms is good; a resistance of 1000 ohms is bad. Reactance X is given by:

X=2*pi*F*L
which gives:
L=X/(2*pi*F)

So to get a primary reactance of 1000 ohms at 1 Hz you will have:
Lp=1000/(2*pi*1) = 1000/(2*3.14*1) = 159 Henries

In your first post you have your turns ratio and inductance ratios backwards. If the turns ratio is 2 then the inductance ratio will be 4, opposite to what you have shown. If you want a 4:1 turns ratio then the inductance ratio will be 16, not 2. Using a value of 16, you will get a secondary inductance of:
Ls=159/16 = 9.94 Henries

Knowing that you want a coupling coefficient of 0.95, you can determine the mutual inductance from the previous formula:
M=k*sqrt(Lp*Ls) = 0.95*sqrt(159*9.94) = 37.77 Henries

Primary resistance should be very small, say 50 ohms.

 

[Wond3rboy]

Hi,

Thanks for the replies MrAL and Bob,

@MrAl, yes thats something i realized when i had already posted the reply and was out on the road.

Thanks for the mathematical calculations bob, they are really helpful. The problem off the -ve cycle being cliipped is still there. I believe that it is because of the ground terminal on the secondary loop, because of that i cant get the bridge rectifier waveform.

 

[MrAl]

Hi,

Are you looking at the lower output of the bridge (not the grounded side)? When i look at the lower output of the bridge i see the waveform i would expect: close to 12.6v peak output full wave rectified sine. It appears to be working, so maybe you're looking at the wrong node. Note that if you point at a node and then later change the circuit you may then be looking at the wrong node. Get rid of all the waveforms and then click on the lower bridge output node and see if that helps. You should see a full wave rectified sine now.

The way it works is the sine wave is converted into a full wave rectified sine, which is a bunch of positive half cycles with no negative parts. They may be slightly spaced apart rather than exactly next to each other because the diodes have a finite voltage drop that has to be overcome before there is any conduction to the output. The point where the diodes begin to conduct is related to the peak of the sine, and for this circuit it would be roughly 4 degrees. That's 4 degrees on each side of the zero crossing too for a total of 8 degrees apparent spacing.

 

[Wond3rboy]

Hi again, thanks. It was the node problem. I was checking the voltage at the wrong node. Thanks for all the help

 

[MrAl]

Hi,

Hey great to hear. Now maybe you can get what you need from the simulations. If it turns out interesting, post some results if you get a chance.

 

[Wond3rboy]

Hi again,

I'll do that, it would be my pleasure. I am going for modeling of diode for a modeling and simulation course. Will ask/upload results as i move on. Thanks again.

 

[ericgibbs]

hi W3
Look at this txr zip.

E.

 

[Wond3rboy]

Hi eric, really helpful. Thanks a mil