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This circuit kills the driver (an IR2104s) when under normal conditions.
Bat+ goes to a 12V lead acid battery.
In+ goes to a 20V laptop power supply, though it is intended to be a solar panel instead.

The circuit takes Vin, passes it through a SMPS with two MOSFETS (driven by the ir2104 that dies) and into Bat+. The LM321 is just a latch that allow the MCU to control and read the ir2104). The connector in the left comes from the MCU, the right one holds the power lines. PWM(PWM signal for the 2104) CS (current sense pin, for safety -it can stop the ir2104-) SHDN (a shutdown signal)

Using a purely resistive load instead of the battery works. Then the battery is connected and the driver gives up. The MOSFET doesn't die, though. This using a current-limited (to 30mA) power supply.

There is a power regulation loop (that feeds the PWM that goes into SMPS_PWM) which is not really good or responsive (it is a very simple ramp algo), but even then, why would the driver die?
My guess is that there's a voltage that goes too high in the wrong pin, but how or where?
One running theory is that, if the PWM were to be 100% you'd have an upper MOSFET that's permanently closed, and that'd be a problem, but I don't see why? (and sure a 100% duty cycle won't work in a charge pump driver, but that should affect the MOSFET, not the driver...)

Some misc data:

  • The driver dies very quickly (before any heat is noticeable) but builds heat until unplugged
  • The power supply is a laptop-type SMPS.
  • This same setup supposedly works with a solar panel, could it be some detail on the supplied power makes it break?
  • The MOSFETs are not broken in any visible way. Could a MOSFET bridge so it would kill something through its gate at a high voltage but look okay when testing?

clearer picture of schematic

NOTE: I found a document from Texas Instruments that, in page 9, deals exactly with the issue and provides a number of solutions and trade-offs.

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    \$\begingroup\$ Please take a screen-shot or print your file as a PDF and take a screenshot of it and post it here. It's very hard to see what's going on in the currently posted screenshot of your screen. \$\endgroup\$
    – Saad
    Dec 7, 2012 at 15:45
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    \$\begingroup\$ Have you taken a measurement of the voltage at pin 6? It's not supposed to exceed VB + 0.3V I think. Also when you say it dies what does that mean to you? \$\endgroup\$ Dec 7, 2012 at 15:49
  • \$\begingroup\$ Post your PWM section as well. We cannot see the complete picture without it. \$\endgroup\$ Dec 7, 2012 at 15:57
  • \$\begingroup\$ I did upload a fairly high-res version, I am surprised the shown image looks so bad. If you zoom the page rendering (ctrl/command +) or download the image you'll get the high res version. \$\endgroup\$ Dec 7, 2012 at 22:05
  • \$\begingroup\$ @Madmanguruman There's no PWM section as such. The MCU samples voltage and tries to keep it constant. It will become poles and zeroes and XF and have proper digital compensation. For now it is a simple staircase algo. \$\endgroup\$ Dec 7, 2012 at 22:13

5 Answers 5

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Madmanguruman and David Tweed are probably on the right track here.

Here is a scenario:

When the circuit starts the synchronous FET (T2) will be on. It has to be this way because there is no bias voltage for the high side drive for T1 except through T2. So, T2 sinks current from the battery through the inductor. Then finally T2 turns off. It doesn't matter if T1 turns on or not, the inductor dumps current back through T1 (or its body diode). The current has to go some place. SMPS are 1 quadrant devices (almost always), so the Laptop supply sinks nothing. The current can only go into C5 and Vcc of the IR2104, which has a max supply voltage rating of 25V. Boom.

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One obvious difference between a resistive load and a battery is that on startup, the resistive load is zero and the battery obviously isn't - it's a source after all. This is called a "precharged load" and can be messy without some form of active or passive OR-ing control.

You don't have any isolation between your synchronous buck and the battery, so if your PWM commands low duty cycle, your "T2" (it really should be Q2 or U2, since T is generally used for transformers) will be on with a wide duty cycle and sink current through the inductor. Your synchronous buck may start acting like a synchronous boost, or even saturate out and short the battery. As "Some Hardware Guy" stated, when the battery is connected you may see some interesting things on pin 6.

Also, there are some intelligent buck controllers out there that can sense the low-side switch drop and shut things off if heavy current is detected. The IR2104 is a high-side / low-side half bridge driver, which can be used to drive a buck, sure, but it's not the best choice.

Try a big diode between your buck and battery, and see if the behaviour changes.

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  • \$\begingroup\$ I'll try your ideas out, great overview! \$\endgroup\$ Dec 7, 2012 at 22:16
  • \$\begingroup\$ I really like your comment on the buck becoming a buck, I can see how that would happen and thus how it'd blow the ir2104 out. \$\endgroup\$ Dec 11, 2012 at 12:40
  • \$\begingroup\$ Seeing the buck as boost converter was very clever, I hadn't seen it until you mentioned it. You mention there might be better options than the 2104, any part suggestions? \$\endgroup\$ Dec 11, 2012 at 12:42
  • \$\begingroup\$ There are literally hundreds of parts out there. If you specifically look for synchronous buck controllers or synchronous buck drivers, you'll find a multitude of options. Many of them have nice features like low-side MOSFET sensing (automatic overcurrent shutoff) and keeping both FETs off in the presence of a precharged load until duty cycle is needed. \$\endgroup\$ Dec 11, 2012 at 13:38
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Application note AN-978 has some notes on sizing the bootstrap capacitor C3 which may be relevant. The examples use a considerably larger value than 0.1uf

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  • \$\begingroup\$ Awesome tip, I am checking that out may it be the solution to this particular issue or not! \$\endgroup\$ Dec 7, 2012 at 22:15
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You're not really supposed to connect the Vcc pin of the IR2104 directly to the high-voltage side of the output circuit. Among other things, this pin determines the drive voltage applied to the power FETs — directly, for the low side, and via the charge pump for the high side. (See the functional block diagram on p. 4 of the datasheet.) While your laptop supply does not technically violate the maximum voltage spec. for this pin, I'm wondering whether spikes that do are being fed back from the inductor via the high-side transistor (or its body diode).

I would recommend regulating Vcc to this chip to no more than +15V or so.

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Why don't you have any resistors on your MOSFET gates? Try 22 ohms. The gates will ring at fairly high frequencies and rather high Voltages making a fair amount of RF interference and MOSFET heat without the resistors. The ringing spikes can be up to double the supply Voltage, which likely exceeds some Maximum Datasheet values.

There is a condition that happens in CMOS logic gates where a spike on the output that goes over the Vcc Voltage can cause the upper MOSFET to get stuck on. I forget the name of it, but if that's happening it would explain exactly what is happening here.

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    \$\begingroup\$ Wow! Awesome catch! You are totally right, they are even there in the spec sheet! \$\endgroup\$ Sep 9, 2020 at 8:02

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