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I recently thought of a way to force the LM2596-ADJ buck converter IC to behave like a boost converter, just as a fun experiment. Let me explain the idea (I would like to get your opinions before I actually build it.)

This is the schematic block diagram of the LM2596:

enter image description here

Typically, Vin is connected to a fixed input voltage and an LC filter is connected to the switch emitter. If you look at the block diagram, you will see that the switch collector is connected to Vin. By connecting an inductor to Vin, the output to ground and a diode between Vin and an output capacitor, the buck converter should behave like a boost converter, right?

I went ahead with this idea and simulated it on LTspice. I fed 9V to the circuit and configured the feedback resistors to get 15V:

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And sure enough, it worked:

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(Blue waveform: voltage across C2, green waveform: voltage across C4)

enter image description here

Then, I realized something about a boost converter and took a look at the voltage at Vin:

enter image description here

As you can see, the voltage at Vin periodically changes between 0V and close to 15V. This this because the switching action of the internal BJT connects one end of the inductor to ground when the switch is on. This also means that Vin, which is not only connected to the switch collector, but is also the input to power the chip itself, is periodically connected to ground. Logically, this should cause the IC to shutdown. In fact, the chip should shut down even before the BJT manages to pull the Vin pin all the way to ground, but that doesn't seem to happen in the simulation. It is as though there is a diode and capacitor at the input that keeps the control circuit powered during the on time of the switch and recharges during the off time.

Please let me know what's going on and what you think would happen if I built it on the breadboard. Will the chip keep cycling? Will it function as a boost converter? Will the chip fail?

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  • \$\begingroup\$ I suspect issues with the internal 2.5 V Vref generation due to no hold up capacitance. \$\endgroup\$
    – winny
    Commented Mar 27, 2021 at 9:00
  • \$\begingroup\$ It depends on how the model is built. If the detection coincides with the internal clock, then it's possible that the voltage does turn off in-between cycles, but its value is then OR'ed with the clock ON time, and if that is ON, then you get a valid output. I don't know if that should be the behaviour of the chip, but I suspect it shouldn't (internal capacitances plus inherent delays will, most probably, cause hiccups and, eventually, shut down). \$\endgroup\$ Commented Mar 27, 2021 at 9:21

2 Answers 2

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Please let me know what's going on and what you think would happen if I built it on the breadboard. Will the chip keep cycling? Will it function as a boost converter? Will the chip fail?

enter image description here

Always use simulation models as they are intended to be used because in the real world, odd-ball uses of circuits will result in definitely different behaviour compared to when tried out in a simulator.

Logic tells you that as soon as the output transistor shorts its collector to ground, the whole internal supply rail for the chip is also shorted out. So, why does the simulation work? Answer: because it assumes that vital internal nodes remain connected to a valid supply rail.

  • If you built in on breadboard it will fail to work (not even a slight chance of it working).
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    \$\begingroup\$ Thanks for answering the post. It seems like a weird simulation implementation to fix all the vital nodes to a valid power rail instead of powering the IC from its actual input. Also, I'd like to point out that the blocks that you have marked aren't the only blocks that will fail. All the other blocks, like the oscillator, comparator, etc. will fail as well. The whole IC will shut down. \$\endgroup\$ Commented Mar 27, 2021 at 11:06
  • \$\begingroup\$ I only showed the blocks that were unambiguously and clearly connected to Vin but yes, it will be a total system fail at many levels @PrathikPrashanth. Sim models will cut corners if there are valid reasons to cut corners. On this occasion it's totally valid to do so. \$\endgroup\$
    – Andy aka
    Commented Mar 27, 2021 at 11:17
  • \$\begingroup\$ @PrathikPrashanth if you are done with this now please close down the Q and A session by accepting an answer. If you have any remaining questions, please leave a comment. \$\endgroup\$
    – Andy aka
    Commented Mar 29, 2021 at 8:34
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The above method has a flaw: when the internal switching transistor turns on, the IC looses power due to the fact that the collector is connected to the same power rail as the control circuit of the IC. This results in a very unpredictable behavior of the chip.

However, there is a solution to the above problem. Simply connect Vin to the input voltage source, as one would normally do with a buck converter IC, but use the output pin of the IC to drive an external transistor. Here is what I'm talking about:

enter image description here

The solution seems like a simple one, but is it? The answer is no. There is a very critical component in the above circuit which determines whether the circuit will work or not. Can you guess what it is?

The very critical component is the 1uF soft-start capacitor C3. Its absence will cause the whole circuit to fail. Also, choosing a very small value for C3 will also cause the circuit to fail. Why? Let's see.

The thing is that buck converter ICs can have a max duty cycle of 100%, which means that the switch can be permanently on. This is because such a condition causes no problem whatsoever to the circuit. However, in a boost converter, a duty cycle of 100% means that the inductor is permanently connected in parallel to the input voltage source, which shorts the voltage source, as you can see in the picture below:

enter image description here

Before I explain how C3 helps avoid this problem, let me just explain how the LM2596 attains and regulates the output voltage (when used as intended) as set by the feedback resistors R1 and R2.

A fraction of the output voltage is fed back to the error amp in the IC through R1 and R2. This is compared with an internal voltage reference to check if the output voltage is equal to, lower than or higher than the reference voltage. Initially, this feedback voltage is lower than the reference voltage. This results in the error amp telling the IC to increase the duty cycle all the way up to 100%. This cause the output voltage to rapidly rise up (happens during the on time of the switching transistor). As the output voltage approaches the desired value, the error amp tell the IC to slowly decrease the duty cycle and eventually, it reaches a steady-state value (0 < D < 100). This is how a buck converter controller works.

In a boost converter, on the other hand, the output voltage increases during the OFF time of the switch, and not the ON time. So what happens during the ON time? The inductor is charged up. It is during the OFF time of the switch when the inductor discharges into the output capacitors, which causes the output voltage to rise up. Due to this, it becomes critical to implement switch current limiting to ensure that the switch opens after the boost inductor is charged to a certain extent, or limit the duty cycle to 90-95% at most by some other method. In essence, the switching transistor in a boost converter MUST turn off within one clock cycle in order to allow the output voltage to rise and to prevent shoring the input through the inductor.

Now its time to explain how C3 helps achieve that. As you all know, initially, for a short period of time during the startup, C3 is almost like a short circuit. This causes the feedback pin to get effectively connected to the output momentarily. Since, initially, the output voltage of a boost converter is almost equal to the input voltage, the feedback pin sees a very large voltage and thinks that the output is much higher than necessary, and keeps the switch off. As time progresses, the capacitor C3 slowly charges up, with allows the voltage on the feedback pin to slowly fall. This results in the duty cycle increasing slowly. In case of this circuit, the soft-start must be very 'strong' in order to prevent the dutycycle from reaching the full 100%. This is why C3 must be very large. It must be large enough to ensure that the voltage on the feedback pin doesn't fall so rapidly that the feedback voltage falls below the reference voltage much before the output attains its desired voltage.

(Green: Voltage at f/b pin, Blue: Output voltage) enter image description here

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Inductor current (note the difference in the inductor current between this case and the previous one)

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Duty cycle of the switch (blue) and output voltage (green)

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Inductor current (red) and switch PWM (green)

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Output Voltage:

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But remember, this is a very dodgy solution with instability written all over it.

Edit: Adding a diode between GND and Feedback (as shown) will help 'reset' the soft-start mechanism by rapidly discharging C3 if the output voltage suddenly drops due to load transience or a short circuit

enter image description here

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  • \$\begingroup\$ With this configuration, you better hope that the power source powering the circuit is current limited, otherwise there will be a huge failure if the output is short-circuited, due to the fact that the current flowing through M1 is NOT sensed by the chip. Therefore, the switch current limit protection is lost. \$\endgroup\$ Commented Mar 27, 2021 at 12:19

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