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I read how boost converters work and that got me thinking: Can I use a pass transistor to achieve currents that may be limited by the inductor used in the circuit? This is a bare version of a boost converter as I understood it (correct me if I'm wrong, I'm still learning!)

schematic

simulate this circuit – Schematic created using CircuitLab

So what if I did this? Would this give me a current higher than the inductor would have been able to put out?

schematic

simulate this circuit

Like the way the LM317/LM78xx series do?

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    \$\begingroup\$ No, that won't work. They work on a linear regulator because the voltage is being reduced. Your circuit is a booster and if it worked at all after adding Q1 you would find that the collector is at a higher voltage than the emitter and base so it couldn't turn on. \$\endgroup\$ – Transistor Aug 25 '17 at 20:28
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    \$\begingroup\$ Can I use a pass transistor to achieve currents that may be limited by the inductor used in the circuit? If the inductor cannot handle the current, it will saturate and behave like a short circuit ! The only way around that is to use an inductor which can handle the current you need. It is a simple as that, there is no solution to work around that as the inductor plays an essential role in a step-up converter. \$\endgroup\$ – Bimpelrekkie Aug 25 '17 at 21:15
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I read how boost converters work

But it doesn't appear you understood what you read, as I'll point out below.

and that got me thinking:

That's actually a good thing.

Can I use a pass transistor to achieve currents that may be limited by the inductor used in the circuit?

No. And the reason is because you clearly don't yet understand why the first circuit might work.

Your topology is designed for cases where you need an output voltage that is higher than the input voltage. So the BJT's collector would just become forward-biased relative to its base and would drain charge from the output capacitors. That's assuming the output actually did reach a higher voltage. The BJT actually makes things worse, not better.

Let's look at an even simpler behavioral version of it:

schematic

simulate this circuit – Schematic created using CircuitLab

Here, the first thing that takes place while the switch stays open is the diode is forward biased because the capacitors are discharged, initially. As the capacitors charge up to a voltage somewhat below the source voltage, the voltage across the inductor declines and eventually reaches zero volts. At this moment, there is already a current flowing through the inductor and onto the capacitor, through the diode. But now the inductor has stored energy in a magnetic field and it refuses to stop the current flow, instantly.

Instead, the voltage across it reverses polarity (now the lead towards the diode is still more positive than before) and the current in the inductor declines towards zero. In the meantime, current continues to charge up the output capacitors to a value well above the source voltage as the magnetic field loses energy and eventually is completely collapsed so that the current is zero. But the voltage across the inductor is now at a peak value nearly opposite in polarity to the source voltage and this voltage suddenly disappears (self-generated while there was a magnetic field to support it, but gone now.)

The moment this occurs some oscillation (ringing) starts to take place. That part, given the load you included, damps out pretty quickly with the result that the capacitors are now charged to a voltage higher than the source voltage and now the load just starts drawing current from the output capacitors.

Now you apply the switch for a moment. This starts the build-up of a new magnetic field in the inductor (while also reverse biasing the diode and preventing discharge of the output capacitors by way of the diode.) Since the voltage across the inductor is relatively fixed, the current ramps up along a very simple linear rate that is \$\frac{\textrm{d}I}{\textrm{d}t}=\frac{V_{IN}}{L_1}\$.

At some point later, the switch opens. But then the inductor (once again) has a peak current now that it demands to continue. Because now the current needs to decline, the voltage across the inductor reverses polarity (now adds to the input voltage source) and once again the declining current flows through the diode and adds charge to the output capacitors, raising the output voltage.

There are several different modes, at this point. You could wait until the magnetic field completely collapses, before starting another cycle. This would be "discontinuous mode." Or you could close the switch before that happens and start increasing the current once again. This would be "continuous mode." Either way, current rises and declines in cycles and adds energy to the output capacitors.

Adding a BJT from input source to output source would just ruin this entire process. It takes an inductor's polarity reversal (or else stacking up charged capacitors) to increase the output voltage. Just pasting a BJT there only damages the circuit. It doesn't help it.

I think you need to spend some time thinking more. I applaud the idea that the whole question made you think. But you need to spend more time thinking closely about the behaviors. And you may need to spend more time understanding inductors, too. (Just as a guess.)

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  • \$\begingroup\$ Great! A better explanation on how the boost topology works than I read on some (now I assume crappy) website. I will read up again and this time elsewhere reputable. Thanks for being blunt in your reply. The reason I asked this question is because I saw an ATX PSU using the tny279 which according to the datasheet doesn't go over 18W!! but the PSU is rated for 700W!! I guess more reading is required to understand all that \$\endgroup\$ – jm11011 Aug 27 '17 at 18:22
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No, your second circuit won't work as you intend. The transistor doesn't help pass more current from input to output because the output is at a higher voltage than the input. If the transistor were a dead short, current would flow thru it from output to input, having exactly the opposite of the desired effect.

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Active power supplies that convert a voltage from one value to another always have either a capacitive or inductive element that has to pass 100% of the energy going from the input to the output. If you hold the input frequency constant (60 Hz, 100 kHz, whatever), the only way to get more energy through it is to increase it so it can pass more energy. This is true for linear supply power line transformers, flyback transformers, boost or buck converter inductors, charge pump capacitors, SEPIC, CUK, etc.

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