4
\$\begingroup\$

Hello fellow wanderers of the electromagnetic plains,

I'm looking to replicate the function of the below pictured reference design, just a simple 230VAC SMPS using an iW1700 chip (72kHz operation), in order to troubleshoot some failures I've had in some dLAN adapters (Devolo 550+ for the curious). I was thinking of etching a quick and dirty board, so was hoping to nix any unnecessary components that would still allow the circuit to function without them. Specifically I am looking at the red circled element, it is obviously some kind of snubber circuit of the transformer leakage inductance right? enter image description here Could someone explain how it works, and why there are diodes pointing in both directions? Specifically what is the function of the LL4148 in there?

Also, what is the need/function of C5, the capacitor in parallel to the aux coil rectifying diode? I assume it it to dampen some voltage spikes but would appreciate some clarification.

Furthermore, if anyone happens to have an inkling, the iW1700 (or almost identical iW1707) datasheet doesn't specify how it sets the final output voltage it is regulating to come out the secondary side. The Vsense pin is on a voltage divider, and there is a nominal Vsense specified at 1.533V. Am I to take it that the PWM chip regulates in order to maintain this 1.533V at the Vsense pin, giving more or less duty cycle as the sensed voltage goes down and up respectively?

The reference design gives instructions for the transformer to be 160 turns primary, 10 SEC and 12 AUX for 5V out. How strict are these winding ratio suggestions, as it's a PWM chip, it will still regulate to the adequate voltage right, even if the winding ratio is changed slightly (keeping the AUX to SEC ratio the same but varying both with respect to the primary, like 160 PRI, 20 SEC, 24 AUX)?

Do you think I'd get away with this thing working if I omitted the red circled element, C5, L1 and FB1?

I know this is a long winded question, but I'd appreciate those of you who take the time to nicely break things down for me. Cheers to you and many thanks!

Ref. design: https://datasheet.octopart.com/EBC10010-iWatt-datasheet-12503214.pdf

iW1707 datasheet: https://www.dialog-semiconductor.com/sites/default/files/iw1707_datasheet.pdf

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Not sure about the rest, but I assume D7 is to allow C2 to slowly discharge \$\endgroup\$
    – BeB00
    Dec 21, 2021 at 19:27
  • 1
    \$\begingroup\$ D7 is most likely labeled with the wrong part number. It should be a zener (most likely in the range of 100V to 200V). Omitting the snubber will destroy the switching transistor. \$\endgroup\$ Dec 21, 2021 at 20:11
  • 1
    \$\begingroup\$ This reference design is optimized for a no-load standby power of 5 mW. The best to achieve this result is to use a TVS instead of a RCD clamping network - not a snubber - but the cost prevents its usage in these low-cost applications. They have probably found a way to design the clamp a bit differently to provide the best standby power. \$R_{10}\$ typically damps oscillations for a smoother EMI signature. You can resort to a classical RCD structure if standby is not a concern. \$\endgroup\$ Dec 21, 2021 at 21:34

1 Answer 1

3
\$\begingroup\$

Yes, you are correct. The bit you have circled is indeed a type of snubber. There are many different types and topologies of snubbers, some entirely passive like this one, and others that are actively controlled.

The point of a snubber is to mitigate the effects of voltage transients and ringing that can result from switching current through inductive elements.

It is also worth keeping in mind that a flyback converter is really just a buck-boost converter whose inductor has multiple windings. The transformer is not really being used as a transformer per se, but rather simply inductors that share the same energy storage reservoir.

Put another way, transformers transfer energy and minimizing any energy that gets stored in the magnetic field instead is highly desired. The primary will directly transfer energy to the secondary with as little energy stored in stray inductance as possible.

A flyback transformer, on the other hand, stores energy magnetically and that is then transferred to the load. When the primary side switch is turned on, current ramps up through the primary winding, but the secondary's phase is opposite that of the primary, making the secondary winding reversed biased. As a result, the secondary-side diode prevents any current from flowing, and thus no energy is able to transfer directly from the primary to secondary.

Instead, energy is stored in the magnetic field, in the air gap and magnetic core of the flyback transformer.

Only when the primary-side switch is turned off, will the voltage across the primary winding reverse (and likewise, the secondary). No current can flow on the primary side, but now the diode is no longer reversed biased on the secondary side, and the energy stored in the magnetic field has somewhere to go, which is into your load on the secondary side.

This is no different form a buck-boost converter, the only difference being that the 'primary' and 'secondary' windings are instead just a single winding. They are otherwise identical, the only difference being galvanic isolation (or lack thereof) between them.

At least, that's how it would work in a perfect world.

There is, unfortunately, one additional difference. Any time you have more than one winding around a shared core, you will introduce something else: leakage inductance. Leakage inductance is stray magnetic flux that, when created by one winding, doesn't perfectly intersect the other winding or windings. Instead, it 'leaks' out of the core.

This is a problem because this leakage inductance is still inductance, it still represents energy stored in a magnetic field. But when the primary side switch is turned off, it can't conduct through the secondary side winding and it's diode because it doesn't intersect the secondary side winding. The only winding it will be able to return that stored energy to is the one that stored it originally: the primary.

But there is that big power transistor and it is not going to let any current flow on the primary side either. Or at least, it will try.

The thing about collapsing magnetic fields is the energy must go somewhere. Everything is closed circuit if the voltage is high enough (cloud-to-ground lightning strikes are a humbling reminder of this).

When the situation isn't just some open circuit theoretical inductor, what usually happens is that the voltage increases until whatever is preventing current from flowing is overcome by said voltage.

In this case, it is usually the power transistor that is overcome.

So generally, unlike standard buck-boost converters, when used in a split-winding flyback topology, you need a snubber across the primary to deal with these spikes. Otherwise, they can destroy your switching transistor (either immediately or gradually over time) as without the snubber, you'll see much higher voltages across the primary on each turn-off of the switch than the actual line voltage it is switching.

There will also be ringing which snubbers help to dampen out for EMI/EMC purposes. Let me show you:

Flyback that is properly snubbed:

w

And the same one, without a snubber. Note, the vertical scale is not the same. The grey line in the middle is the peak of the waveform in the snubbed waveform image above:

enter image description here

Granted, you may not necessarily need a snubber on every flyback, but those will be very specifically designed with that in mind. If the reference circuit includes a snubber, then you should assume you need a snubber.

I can, however, offer a compromise: the snubber in that reference circuit is a strange one designed to optimize low-load/idle power consumption. The LL4148 is being used as a zener diode (you don't actually have to use zener diodes for typical zener diode purposes, you can use any diode as long as you are careful to limit the current that can flow through it when the blocking voltage is exceeded, which is the purpose of the 10K resistor in series with it).

It is really just a fairly standard RC snubber combined with a zener snubber, but in a way that is somewhat poor for snubbing (but 'good enough') but minimizes the power lost at low loads.

To minimize components like you want, I suggest simply removing both diodes and connecting the RC leg directly across the primary, and possibly increasing the capacitance some too. This will make it burn a bit more power but you'll have a cleaner switching waveform with less ringing and can not have to worry that voltage transients are causing problems if you have any.

In this case, it is there to limit voltage spike transients that result from the inductance of the flyback transformer.

You are probably already familiar and have used snubbers, though you might not realize it. The simplest and arguably most widely used snubber is often referred to as a flyback diode or freewheeling diode.

These are the diodes that are placed in parallel with things like relay colors, solenoid coils, and other inductive loads that are going to be switched by a circuit. The diode gives the back-EFM/inductive kickback a path to conduct through, thus limiting the voltage spike that would otherwise occur when suddenly stopping current flow through the inductive load.

Current flowing through a given inductance (be it an intentional one like of an inductor, or something parasitic like a long cable) will create a magnetic field, and energy is stored in this field. The equation is identical to energy storage in an electric field (which is what capacitance is), as well as the kinetic energy equation, with only the units of each being different: \$E = \frac{1}{2}LI^{2} \$.

\$\endgroup\$
2
  • 2
    \$\begingroup\$ "The LL4148 is being used as a zener diode (you don't actually have to use zener diodes for typical zener diode purposes, you can use any diode as long as you are careful to limit the current that can flow through it when the blocking voltage is exceeded, which is the purpose of the 10K resistor in series with it)." - Fairchild LL4148 datasheet says absolute max. reverse voltage is 100V, but then contradicts it with a graph showing 100uA at ~145V - which is well below maximum power dissipation so it must be OK!. \$\endgroup\$ Dec 28, 2021 at 23:20
  • \$\begingroup\$ Thanks a bunch for the detailed answer! \$\endgroup\$
    – parkside
    Jan 2, 2022 at 20:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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