1
\$\begingroup\$

i was looking for the Energy conversion efficiency of my circuit in Ltspice, but i find out that my Transformers is giving me strange power values i mean with K = 1 the P1 should be equel to P2 but in my case P2 is is way too small , i get equality 90% only if get rid of the Flyback diode, can you please help me understand the reason behind this

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

\$\endgroup\$
  • \$\begingroup\$ How do you calculate the powers? \$\endgroup\$ – Andy aka Nov 5 '15 at 9:12
  • \$\begingroup\$ Are you using a 1N4148 as a flyback diode with 2 kV source? I'm afraid it will be in avalanche effect! \$\endgroup\$ – CasaMich Nov 5 '15 at 9:59
  • \$\begingroup\$ no i made my own diode it can hold up to 6kv :) \$\endgroup\$ – ernd59 Nov 5 '15 at 13:51
  • \$\begingroup\$ @CasaMich no i made my own diode it can hold up to 6kv :) \$\endgroup\$ – ernd59 Nov 5 '15 at 13:53
  • \$\begingroup\$ Il1 * U = P1 , IL2 *U = P2 \$\endgroup\$ – ernd59 Nov 5 '15 at 13:54
1
\$\begingroup\$

Since your simulation schematic appears to have the transformer connected as a Flyback coupled inductor, I'm going to assume that you're trying to implement a Flyback converter. But, there are lots of problems with the circuit.

First, \$D_5\$ is directly across \$L_1\$. That's a problem because of the Volt-seconds it forces on \$L_1\$. When \$M_1\$ is conducting, \$L_1\$ has 2000V across it for time \$T_{\text{on}}\$. But when \$M_1\$ is turned off, \$L_1\$ can only have the forward voltage of \$D_5\$ (\$V_{\text{D5-forward}}\$) across it for time \$T_{\text{off}}\$. Let's say that \$V_{\text{D5-forward}}\$ is 4V (that's probably about right for a 6000V diode). That means that \$T_{\text{off}}\$ has to be 500 times \$T_{\text{on}}\$. Otherwise, in a real circuit \$L_1\$ would saturate. But, it's hard to say what a simulator will do here, since often this kind of thing isn't well modeled. This alone could cause your problem getting any kind of sensible results. If you want \$T_{\text{on}}\$ to be somewhere close to the same as \$T_{\text{off}}\$, then \$D_5\$ doesn't belong across \$L_1\$ at all. Of course, \$M_1\$ will have to withstand some multiple of 2000V.

Second, for a Flyback, \$D_2\$, \$D_3\$, and \$D_4\$ don't belong in the circuit either. Only \$D_1\$ would be used as rectifier in the output. When \$M_1\$ conducts, energy is stored in \$L_1\$. When \$M_1\$ is turned off that stored energy is dumped into the output load through \$D_1\$. So, \$L_2\$ dot would connect to \$D_{\text{1-anode}}\$, and \$L_2\$ non-dot connects to ground.

| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

My guess, (I'm still having troubles understanding everything with coupled inductors):

If you switch off the current through the transformer in L1, the stored magnetic energy will try to force it's way out.

If you place just a simple diode as a snubber, the current has a much easier way flowing on your primary side then on your secondary side, and will probably just do that.

To prevent that, your snubber circuit should be of much higher impedance than your secondary side, so the magnetic energy will find it more easily to discharge through the secondary side.

The snubber on the primary side must make sure that the voltage across the switch doesn't go above the breakdown voltage, but nothing more.

(and why do you use a full bridge rectifier at the output?)

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.