# TVS diode as snubber for ac-dc flyback converter

Discussion in 'Electronic Design' started by aravind, Jul 18, 2007.

1. ### aravindGuest

hi guys
I plan to use TVS diode as snubber for a ac-dc flyback converter, by
connecting TVS diode in parallel with the primary winding of the
flyback transformer so that TVS diode absorbs the transient energy
stored in the leakage inductance of the flyback transformer during
turn off of the switch. The rated output voltage of the converter is
50 V dc. The turns ratio of the transformer is 4:1. My rated power of
operation is 210 W. Switching frequency is 50KHz in CCM mode. During
transient my output voltage can go up to 80 V(worst case), and the
current in the switch might go up to 15 A(max during full load in
transient). Therefore I would require a TVS diode with a peak pulse
power dissipation (Ppp) of about 1.6*80*4*15 =7680 W with rated
working peak voltage (Vwm) of 80*4= 360. Since I could only get TVS
diodes with Ppp of 1500 W. I intend to put six TVS diodes of 1500 W
Ppp and Vwm 0f 60 V in series.
Can anyone suggest weather this idea will work, and will any special
measures have to be undertaken to for heat dissipation if the whole
ckt (other than flyback t/f) has to be implemented on a pcb.

2. ### Tim WilliamsGuest

So, um, er, why are you *dissipating* something which can be *saved*?

The usual snubber involves shunting the current into a holding capacitor,
from which it leaks back into the supply rail via resistor.

Is this a flyback or forward converter?

Tim

3. ### Terry GivenGuest

psst - guess what happens to the charge that "leaks back" thru the resistor?

it doesnt actually matter, the problem is still the same
so the flyback voltage is 200V nominal

ouch. thats crappy regulation, and bumps the flyback up to 320V

look at the TVS datasheets closely. Ppp is often specified for a
1.2/50us pulse (sometimes even for an 8.3ms haversine). convert it to
Joules, and do your calculation based on that.

there are also 5kW Ppp TVS available.

yes it will work. If you use a bidirectional TVS then you dont need a
series diode, *BUT* TVS have a LOT of capacitance, which dumps
C*Vpk^2*Fsmps into your FET. a series diode effectively isolates the TVS
capacitance.

for a 210W SMPS, you will have to throw away 20-30W in your switches,
transformer and rectifiers, which will require some heatsinking.

the clamp network basically dissipates the energy stored in the leakage
inductance, which should be < 2% of the primary inductance if the
transformer is well designed; if not, it can be a lot higher.

figure it out like this:

- the current in the leakage inductance is the peak primary current.
- the clamp network clamps the Drain to (Vflyback + Vextra) = Vtvs
- the reflected load voltage Vflyback appears across the magnetising
inductance, so
- the voltage across Lp_leak = Vextra = Vtvs - Vflyback
- the current thru the leakage inductance linearly (ish, assuming the
xfmr is wound for low losses) decays to zero, as:

Vextra = Lp_leak*Ip_peak/Tclamp

- re-arrange to solve for Tclamp
- now you know how long the current takes to ramp down to zero.
- the average power dumped into the TVS during Tclamp = 0.5*Vtvs*Ip_peak
(current waveform is a triangle, hence the half)
- the total energy dumped into the TVS is 0.5*Vtvs*Ip_peak*Tclamp
- the average power dumped into the TVS is 0.5*Vtvs*Ip_peak*Tclamp*Fsmps

So although the peak pulse power is perhaps high, the average power
should be pretty low (but 2% of 210W is still 4.2W). Years ago I made a
flyback with 3 x SOD-87 SMT zeners, on a vertical PCB. I didnt do this
calc, and during operation the zeners got hot enough to desolder
themselves - I found them on the bench after the SMPS blew up. oops.

If you have to implement it in a small space, I'd use a pair of FETs and
a pair of catch diodes as a diagonal half-bridge converter - a FET at
either end of the primary winding, connecting to each supply rail, forms
one diagonal half of a full bridge; the other diagonal half is done with
diodes, and both FETs switch simultaneously. That way almost all the
leakage energy is returned to the supply rails. And it also keeps the
FET voltage clamped to the DC bus, so slightly lower voltage FETs can be
used (which makes up for having 2 in series)

Cheers
Terry

4. ### Tim WilliamsGuest

resistor?

What about it? The charge is saved, looping it into the +V rail rather than
wasting it to GND.

Running the charge through a resistor does tend to dissipate some voltage
though, yes ;-)

Tim

5. ### Terry GivenGuest

Are you referring to an RCD clamp? cos if you are, all of the energy
gets dissipated in the R. e.g.:

+Vdc-------+------+-------+
| | )
[R] [C] ) Xfmr
| | )
+------+--K|---+
|
|
X FET etc
|
0V

all of the variant s of an RCD clamp achieve the same objective, viz.:

- at switch turn-off, current commutates thru diode into cap, thereby
- once thats done, the R bleeds off some or all of the charge stored on
the cap

the last one is worth thinking about. If you dont get rid of all the
charge dumped into the cap, the voltage across it rises. eventually,
something breaks.

in the case of a turn-off snubber, whose purpose is to shape the load
line (very common with bipolars, not so much with FETs), the cap is
mostly discharged, is the current commutating into it sets the switch
dV/dt = I/C. In this case, the cap is connected to 0V.

for a voltage clamp, the cap is not fully discharged, and sits at the
+Vdc rail, at some voltage Vc_min. At turn off, the switch voltage rises
to Vdc + Vc_min, then the diode conducts, and the leakage energy dE_lk
is dumped into the cap, causing its voltage to rise, calculated as:
Ec_final = Ec_start + dE_lk = 0.5*C*Vc_final^2

then the resistor is sized to discharge the cap back down to Vc_min.

If you draw the cap charge and discharge current paths, its easy to see
that the resistor dissipates (some or all of) the energy stored in the cap.

Que?! interesting (read as: wrong) use of terminology, Tim.
Cheers
Terry

6. ### Tim WilliamsGuest

Charge as in coulombs. The cap fills up with some discrete amount of
charge, then the charge bleeds through a resistor (at some rate and
therefore some current). Charge is a thing that tends to be conserved.
However, a voltage associated with that charge flow corresponds to power
dissipation. When I originally said "something that can be saved", I meant
the charge, which is misleading since I also used "dissipated" (power by
definition) in the same sentence...

Tim

7. ### aravindGuest

its a flyback converter