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I'm currently making a DC-DC SMPS power supply that converts 331VDC (rectified from 240VAC) to 60VDC with a maximum output current of 10 amps. I've asked once on this forum about a general problem on my circuit, specifically why my MOSFET keeps dying, and what I'm about to ask is a kind of a continuation of the problem. If you want to see my previous question, here it is. The circuit I'm currently using can be seen below

Gate Driver Circuit Fig 1. Gate driver circuit Buck Circuit Fig 2. Buck converter circuit

So with a more adequate testing equipment in my university lab I've figured out that what's actually been killing my MOSFETS are actually the current spikes across the D-S of my MOSFET due to L1's transient response. Without any snubber circuit, the spikes can reach as high as 10 times the average output current. There's also some voltage spikes visible on the scope, but said spikes are easily remedied by putting a simple RC snubber in parallel with the MOSFET (Node A-B in Fig 2).

Seeing that the RC snubber works, I then remove the RC snubber from Node A-B, and put a simple RLD snubber in series with the MOSFET (Node B-C in Fig 2) to try to minimize the current spikes in my MOSFET. With a 300uH inductor, the current spikes is also reduced by a lot. The current spikes are now just 1.2 times the average output current.

Now the problem is when I try to combine both snubbers (RC and RLD), it messes up with the buck converter operating cycle. In a normal buck converter circuit, when the switch is in the OFF state, the voltage at Node C (in fig 2) is 0V, and thus making the diode (D3 in Fig 2) operate in forward biased mode. But if I put both my RLD and RC circuits (in the corresponding nodes I explained above), when the switch is OFF, the inductor in the RLD circuit can still cause a current flow from node C to node A via my RC circuit. As a result, D3 will have to wait for the inductor in the RLD to fully finish discharging for it to go in forward biased mode (if my explanation is not making any sense, you can also see what I'm describing in Fig 3). In doing so, the maximum output voltage is reduced significantly, and it's not proportionally linear with the duty cycle.

ON and OFF State with RC and RLD Snubber Fig 3. Current flow when the switch is in the on and off state after adding RC and RLD snubbers to circuit in Fig 2

In order to still use the RC snubber circuit to clip excess voltage, where should I put the RC and the RLD? My RC and RLD configuration and component values is the same as shown in Fig 3.

Edit: I've also attached my gate voltage when there's a 7V DC in the drain, if it helps enter image description here Fig 4. Gate voltage when there's 7VDC in drain

enter image description here Fig 5. Gate-source voltage (yellow) and source-ground voltage (blue)

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  • \$\begingroup\$ Is your VB holding up (solid 12 volts)? I ask because I don't think the IR2110 can use the high side driver without also using the Low side driver. It is when the low side driver is enabled that the VB gets refreshed. \$\endgroup\$
    – Marla
    May 11, 2022 at 18:22
  • \$\begingroup\$ When there's no input voltage, the output is a perfect 12V square wave with the same duty cycle as my pwm controller output. However, when there's a voltage, the pwm signal is no longer a constant 12V. I think it's because the bootstrap capacitor ramping up the voltage when the switch is in the ON state. However, if there's a problem with my IR2110 like you're suggesting, shouldn't the output be always wrong regardless of my RC and RLD placements? \$\endgroup\$
    – Kevin
    May 12, 2022 at 1:40
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    \$\begingroup\$ I don't understand. "When there's no input voltage, the output is a perfect 12V square wave". Are you saying that VB is solid constant 12 volts ? \$\endgroup\$
    – Marla
    May 12, 2022 at 2:37
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    \$\begingroup\$ my gate voltage is a 12V square wave with a 50% duty cycle. to what reference? MOSFET's source or GND? \$\endgroup\$ May 12, 2022 at 6:48
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    \$\begingroup\$ @Kevin the high side drive output is 12V w.r.t. GND. This indicates that there's something wrong i.e. the bootstrap is not working. Now look at the schematic again. The source of the MOSFET is supposed to be connected to the inductor side, not the input supply side. Or in other words, the drain should be connected to the input voltage side. Your schematic looks wrong. Because, according to your schematic, the body diode of the MOSFET will be forward biased when you apply the input so you should see input voltage minus a diode forward drop at drain, not zero. Fix the circuit and try again. \$\endgroup\$ May 17, 2022 at 14:14

1 Answer 1

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Forget about the snubbers for a moment and focus on the initial design and basics.

First of all, I kindly advise you to get rid of IR2110 and use a gate transformer instead. Or, if you really want to use IR2110 then use it as a half bridge, not as a hi-side driver only. Remember that it's a bootstrap driver so the bootstrap capacitor should be charged prior to drive the high side MOSFET. This can only be done by driving the low side because the bootstrap capacitor, which has its bottom end connected to the bridge's mid point, cannot be charged while the low-side is left floating.

Plus, the output power and output current are relatively high, so you "should" go for synchronous buck:

enter image description here

IR2110 can be a good fit for that purpose but be careful with dead time and possible shoot through.

One more thing I noticed is the inductor: For 600 W power output, unless you keep the switching frequency too low (e.g. less than 20 kHz), 1 mH is way too high.

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  • \$\begingroup\$ Sorry for the late reply, but my reference for the IR2110 circuit can be found here (tahmidmc.blogspot.com/2013/01/…). In that website, there's a schematic that shows IR2110 being used for high side switching only. Is the website wrong? And also, my switching frequency is currently 20kHz. From what I've read, using a low switching frequency means that I need to use a higher inductance value. With that switching frequency in mind, how low should my inductance value be? \$\endgroup\$
    – Kevin
    May 17, 2022 at 13:19
  • \$\begingroup\$ @Kevin I will not say that the schematic given on the website is correct or not. What I say is that the IR2110 is a bootstrap driver so it needs the bootstrap capacitor to be charged to drive the high side MOSFET. Look at the schematics and think about how the bootstrap capacitor can be charged. Plus, I put a question/comment under your original question. Please have a look and answer it. \$\endgroup\$ May 17, 2022 at 13:34
  • \$\begingroup\$ @Kevin About the inductor, what you have read is true but you may want to consult one of the buck converter design tutorials out there, just to be sure and prevent the things get more complicated. I just made a rough calculation and found that the inductor can be at least 200 microhenries (uH). So you can use 330, 390 or 470 uH safely. \$\endgroup\$ May 17, 2022 at 13:35
  • \$\begingroup\$ Sorry, I didn't see your question above. The gate output is 12V in reference to ground when Drain voltage is 0 \$\endgroup\$
    – Kevin
    May 17, 2022 at 13:59
  • \$\begingroup\$ From what I've read, the bootstrap capacitor is charged when the low side switch is ON and the high switch is OFF, so the capacitor is charged to VB volts. Wouldn't it be the same if there's a load connected to the VS node and GND? The bootstrap can't charge only if there's no load connected. This is true in my case, when I forget to connect the load, the gate driver output is 0V, otherwise it's working correctly. Can you explain a bit more why my current configuration is bad? \$\endgroup\$
    – Kevin
    May 18, 2022 at 3:19

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