# Need help wih snubber

Discussion in 'Electronic Projects' started by bonedoc, Feb 26, 2012.

1. ### bonedocVIP Member

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There is some ringing on the primary side of my DC to DC converter that happens when the switch closes. I thought I understood how to make the snubber, but no luck yet. All I am doing is burning resistors. Here are my calculations for the RC value. For L, I used the leakage inductance of the transformer, and for C I used the interwinding capacitance of the transformer. I dont know if this is what I need or not. Also, I think I need a RCD snubber so the snubber is not conducting white the mosfet/switch is conducting. Here are my values:

.144uH and 76pF transformer:

R = (L/C)^(1/2) = (.000000144/.000000000076)^(1/2) = 1894.73^(1/2) = 12.37 Ohm

C = 2Pi(LC)^(1/2)/R = 2 x 3.14 x (.000000144 x .000000000076)^(1/2) / 12.37 = 1.13uF

I attached a pic of the snubber I soldered together to test. The anode of the uf4007 is placed where the drain and "low" side of the transformer primary meet. The resistor and cap go to the "high" side where the 12V comes into the primary. So, I am pulsing 12v through the transformer primary and need a snubber for this. So confusing...

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bonedoc, Feb 26, 2012

2. ### GonzoEngineerVIP Member

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How much current are you switching?

Too often, the ringing is blamed on the Inductive load, when the truth is that the ringing
is actually caused by the Inductance of the DC Link at turn off.

(DC Link refers to the input side DC voltage to the switch.)

Think of it this way, when you turn off the current, the commutating Diode of the device sends the power back to the input capacitors.

If you have high Inductance, both in the wiring and capacitors themselves, you will see
large spikes at turn off.

No snubber in the world is going to help that!

I switch 600VDC through IGBT H-Bridges at over 2000 Amps. It took me a year to discover that the spikes weren't from the transformer. I tried every snubber combination you could imagine, but still had spikes reaching 1200V or more.

When I got rid of the Litz Wires, and went to a laminated Buss, the spikes practically dissapeared! And finding capacitors with much lower inductance removed the problem altogether.

Can you provide photos of the input capacitors, and the wiring to the device?

I would bet I can provide some solutions to solve the problem.

GonzoEngineer, Feb 26, 2012

3. ### bonedocVIP Member

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Hmmm.....I dont believe I have an input capacitor. Never heard of one ;-)

Here is a schematic of the primary side of my transformer. Driver and mosfet on the left. You can see where the 12V goes into the transformer and then through the switch. The snubber is on the right. I missing something.

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bonedoc, Feb 26, 2012
4. ### GonzoEngineerVIP Member

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OK.....now that I see what you are trying to do........

First of all, remove the capacitor and resistor that are in series with the diode.

Then connect the Cathode of the diode to the 12V input.

Where are you measuring the spike, referenced to what?

Measure the spike across the FET at turn off. how much is it!

The diode is designed to route any voltage more than 12V back to the 12V source.

You need a low inductance/low esr capacitor connected to the juncion of the diode and the 12V. The wires also neet to be short.

Let me know how that works.

GonzoEngineer, Feb 26, 2012
5. ### bonedocVIP Member

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Ok. What value and rating of cap should I use? I am not at my desk right now but can post a pic later. The noise only occurs during switching, and I can see it anywhere in the circuit :-( I figured it was from spikes when switching closes. It is less than a volt at the drain with reference ti ground, but enough to cause some erratic behavior. Once I get this primary side dampened and I see the effects, I will probably have questions about the high voltage secondary side.

bonedoc, Feb 26, 2012
6. ### bonedocVIP Member

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Is a small ceramic disc cap fine? Are you saying the diode goes anode the drain and cathode to 12v, and then connect the cap from cathode to 12v but not in series?

bonedoc, Feb 26, 2012
7. ### bonedocVIP Member

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Ok, here are a couple of pics. They are at 1us/cm and .1v/cm. The probe is on the drain and the ground clip is at the source. I tried the uf4007 alone, with no cap and things got hot. See the second picture. It is crappy pic, but that is what happens with the diode anode at drain and cathode at the 12v going into the transformer.

I need to add the cap, but I am a little confused. Do you mean circuit A or B below? Seems like B would be sending electricity to the cap when the switch was on or off.

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bonedoc, Feb 26, 2012
8. ### davennModerator

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no you need to remove the cap from between the diode and supply as gonzo said
I suspect he meant the cap to go between + and - rails of the 12V supply and probably should be at least a 100uF electrolytic of ~ 25V.

Your circuit A stops the diode from its back EMF protection
your circuit B just puts a short cct across the capacitor and so it doesnt do anything

The inductor... what is it ? it isnt the coil of a relay is it ?

Dave

davenn, Feb 27, 2012
9. ### bonedocVIP Member

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Thanks! The inductor is the primary side of a transformer in a dc to dc converter.

bonedoc, Feb 27, 2012
10. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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B has no capacitor in it at all (it is shorted), It is also the circuit I would recommend trying first.

edit: but perhaps not for a DC-DC converter.

Last edited: Feb 27, 2012
(*steve*), Feb 27, 2012
11. ### bonedocVIP Member

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Ok, here is what I stumbled on, thanks to everyone here. Placing a 100uF/35V electrolytic cap on the 12v rail to ground eliminates all my noise problems on the entire circuit! However, I believe the secondary side winding is seeing slightly higher voltages, and I think it may be due to this capacitor. But, there is no distortion on the circuit. I placed it between the transformer and the drain...dont know if this is right.

I believe it may be smart to add the diode, but yes, I noticed it wasnt performing, and as you pointed out, it must b short ;-)

What is the final solution? Put the diode in series with the electrolytic cap? In other words, have the diode anode on the sink and put it in series with the cap from its cathode to the ground?

Last edited: Feb 27, 2012
bonedoc, Feb 27, 2012
12. ### bonedocVIP Member

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So, do i:

-place the cap is series with the diode like A

-place the cap positive between the 12v and transformer like B

-place the cap between the transformer and the drain like C

If the answer is B or C, where do I place the diode?

Thanks for the help!

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bonedoc, Feb 27, 2012
13. ### GonzoEngineerVIP Member

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Sorry I haven't been back to you. I am laid up with 3 broken ribs and a ractured Vertabra.

a,b and c are wrong.

The positive of the cap connects to 12V.
The negative of the cap connects to the Ground.
The diode is placed in parallel with the inductor, with the cathode
side to 12V.

What you are seeing are turn off spikes. When you turn off the transformer primary, the magnetic field collapses, causing the voltage at ther bottom of the inductor to rise above 12V. The diode is used to conduct that voltage back to the 12V rail. That is why you need a capacitor on the 12v rail, to absorb the energy.
If the cap is a crappy electrolytic cap, or if the inductance of the wires is to high, it won't do a good job of absorbing that spike.
(Google snubber capacitors. WIMA has a good site for information.)
Are you in the U.S.? If you are, I can send you some appropriate caps. Free, you just pay for shipping.
I would really like to see some pics of the actual setup, because the wiring inductance is very critical here. Moving a few connections can make a world of difference.
Hope this helps, Cheers!

GonzoEngineer, Feb 27, 2012
14. ### bonedocVIP Member

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Ouch! Was it a compression fracture? Lumbar or thoracic? Have a hard fall?

Here is am embarrassing picture of the circuit. It may not help. I have it in perf board since the output side is HV. I appreciate the offer for the capacitors! Let me look to see what I have here before you have to send anything. The capacitor helps a lot. I took the 100uF electrolytic off and replaced it with a 1.0uF metalized film. It does not seem to do as good. Probably because it is too small.

The only issue I have is with the diode. I always have a meter on the output side of the converter. Well, when I have the diode on, the converter doesnt work at all. It is almost like there is a short. I have the anode on the drain and the cathode on the 12v supply. So, it is basically reverse biased on the transformers primary legs.

The capacitor connects 12V to source.

So, my final questions are:

-Is metalized film a good cap choice?

-why is the diode causing symptoms of a short?

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bonedoc, Feb 27, 2012
15. ### bonedocVIP Member

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Ok, I found some equations!

For the snubber voltage rating:

Vsn = Vout(Np/Ns) = 300(1/10) = 30V

Now for the resistor value:

Rsn = 2 x (Vsn^2 - (Vsn x Vout x (Np/Ns)) / Ipk^2 x Lleak x f)

Rsn = 2 x (30^2 - (30 x 300 x (1/10) / 5^2 x .000000144 x 125,000)

Rsn = 2 x (900 - (900) / 25 - .018) = 2x (0 / 24.982) = ??

So....according to the equation, my resistor value is ZERO. Does this mean I dont need one? It is confusing is to then calculate the capacitor capacitance, because is needs to know the snubber resistance:

Csn = Vin / Vripple x Rsn x f = 12 / 3 x 1 x 125,000 = 32uF

Or, do I just pretend the 0 is a 1 in the Rsn calculation to give 2/24.982 = .08 Ohm

and therefore

Csn = Vin / Vripple x Rsn x f = 12 / 3 x .08 x 125,000 = 400uF

It also says that diode selection needs to be a fast diode with a repetitive voltage of:

Vrr = (Ns/Np) x Vin + Vout = 10 x 12 + 300 = 420V which would = uf4004-4007

So, as far as capacitance calculations, the subber resistor must be known. However, I dont know how to handle the equation if it comes to zero!

bonedoc, Feb 27, 2012
16. ### duke37VIP Member

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I understand that this is a DC-DC convertor. The energy in the inductance should go to the load, thus a snubber which absorbs the energy is a bad thing.
The inductor or transformer is part of the circuit and is critical. If it is a transformer, you will need close coupling between primary and secondary. Also the load on the PSU will be important. If there is nowhere for the energy to go, something will go pop.

duke37, Feb 27, 2012

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