1
\$\begingroup\$

I have seen many times on the Internet how a boost converter can be used to step up the voltage of a low voltage battery (like 1.5 V) to a higher level to light up an LED which normally wouldn't operate (it would not emit any light we can see) at this lower voltage.

I understand that the current-voltage relationship of an LED is exponential since an LED is essentially a diode.

But what confuses me is this: to conserve energy, if we step up the voltage with a boost converter, the current must decrease. So if we connect an LED to the boost converter, why do we see it light up?

I mean, the LED says that if the voltage across it increases, the current through it increases exponentially (and it lights up). But on the other hand, the converter says that if the voltage on the output increases, the available current output decreases. To me this seems that the LED and the converter "contradict" each other, for the lack of a better word.

I would much appreciate if somebody could clarify this situation for me. The type of converter I'm talking about is pictured below. I'm imagining the LED in place of "load" in this picture.

enter image description here

\$\endgroup\$
4
  • \$\begingroup\$ Have a look at the IV diagram of a diode/LED and you will see a region where little to no current flows. You need to first reach enough voltage until current starts to flow, then things will kinda level out \$\endgroup\$
    – PlasmaHH
    Apr 24, 2017 at 19:47
  • 1
    \$\begingroup\$ The question is, the current decreases from what?. Look at the left hand side of the circuit ... you'll see a voltage source, an inductor, and a switch to ground. Close the switch and the current increases ... potentially to infinity. Open the switch, and the current decreases ... to whatever you need to run the LED. (In practice, you don't close the switch long enough to reach infinite current, but boost convertors can be tough on batteries) \$\endgroup\$
    – user16324
    Apr 24, 2017 at 19:49
  • 1
    \$\begingroup\$ You're not conserving energy - the circuit with the boost converter draws more energy from the battery than would have been drawn by the load without the boost converter. \$\endgroup\$
    – brhans
    Apr 24, 2017 at 19:59
  • \$\begingroup\$ You see the LED light up because the energy is delivered in pulses. The conservation of energy law that you are thinking of is expressed in terms of averages. If you want, you could simulate the Joule Thief in LTSpice, and you'll see what I mean. In this question I asked, there are two LTSpice files you can use to try it out. Just use a text editor and copy and paste the LTSpice "code" into a text file with an (*.asc) file extension (on Windows), then LTSpice can see the file to load it. It's fun to play around with. Enjoy. \$\endgroup\$ Jan 7, 2023 at 21:12

4 Answers 4

3
\$\begingroup\$

In a switching supply including a boost converter, the output power will equal the input power minus losses.

So suppose you have a 2V 1A input. A good boost converter might be 90% efficient, so 2W of input would result in 1.8W of output power.

If your LED forward voltage needed is say 3V, you could get 600mA out of the boost converter. That's plenty of current to power a typical LED, and in actual practice you would want to boost to a higher voltage and limit the current with a resistor. (Or build a current regulating converter rather than a voltage regulator.) Your diagram doesn't show any control scheme, so it's hard to tell what you have in mind.

So you do get less current out of the boost than you put in, but if there's enough output power to power your LED you should be fine.

\$\endgroup\$
3
  • \$\begingroup\$ You do not need the current-limiting resistor in this case. The boost is regulated on the input, and the inductor stops delivering current to the LED once its energy is depleted. The inductor is the current-limiting device in this case. The Joule Thief is a simple LTSpice simulatable example where this can be easily seen, and the regulation is on the input, based on the incoming voltage, the resistor, and the particular transistor and coupled-inductor used. Once the Joule Thief inductor is charged, that same energy is delivered to the LED. No need for any resistor right above the LED. \$\endgroup\$ Jan 7, 2023 at 20:55
  • \$\begingroup\$ Actually, I think I see what you're saying. I've considered doing that for LED's that are near their current limit. But efficiency is definitely lost that way, and I didn't want others to get the impression that the resistor is necessary. For flashlight circuits and the priority on efficiency, I definitely recommend against any current-limiting resistor, and rather just add a large (or larger) capacitor past the Schottky catch diode (and around the Load) if the boost pulses are putting the LED(s) over-spec. \$\endgroup\$ Jan 7, 2023 at 21:03
  • \$\begingroup\$ @MicroservicesOnDDD Sure, the most efficient topology is a switching current source or current-limited supply that delivers the current that the LED needs. However, if you have a voltage output DC-DC capable of delivering more current than your LED can handle you need the resistor. It's all down to the design choices that you make. \$\endgroup\$
    – John D
    Jan 8, 2023 at 21:39
2
\$\begingroup\$

Boost convertors work by using an inductor to take energy from the source side and pile it up at a different voltage on the load side, usually in a capacitor. The load can remove energy from that capacitor but not any faster than the source can supply it. Further, an additional amount of energy is lost in order to perform the translation.

However, what you have indicated in your schematic is actually a current pump more than a boost regulator. The switching sets up a current in the coil, storing energy in it, when the switch changes that energy is transferred to the LED.

When properly controlled the energy being drawn from the battery remains fairly constant and is higher than would be drawn without the pump.

\$\endgroup\$
2
\$\begingroup\$

A boost converter produces a bigger voltage at the output compared to the input and in doing so it must diminish the output current compared to input current. Regard it as a dc transformer but not quite as energy efficient.

If the output load requires (say) 100 mA at 3 volts, then a 1.5 volt battery has to be able to supply 200 mA plus another 15% or so that gets wasted in heat/losses.

\$\endgroup\$
1
\$\begingroup\$

You are talking about the circuit running around with the name, Joule Thief, and various related incarnations of same. Your behavioral schematic is pretty close. You don't need the diode there if the load is an LED. (In a typical Joule Thief circuit, a BJT provides the switch but also requires a driving circuit which is what slightly complicates the schematic.)

Sticking with the behavioral schematic: Without the switch closed, the battery source, the coil, and the LED are in series. But no appreciable current flows because the LED requires a substantial voltage in order for any useful current to flow. There may be tiny current too small to matter. But that's the circumstance without an active switch toggling on and off.

When the switch closes, the current builds up in the coil according to the usual formula for that:

$$\frac{\textrm{d} I}{\textrm{d} t}=\frac{V_t}{L}$$

With \$V_t\$ being the battery voltage (which if this is a real battery will probably vary a little as the load changes over time.) As the current increases, so does the energy stored in the magnetic field. When the switch opens, the direction of the current through the coil continues but because the current is now declining the sign of the voltage across it changes. (Just look at the above equation and imagine that \$\frac{\textrm{d} I}{\textrm{d} t}\$ changes from increasing (positive) to decreasing (negative.) You should be able to see why the sign of \$V_t\$ must therefore change. (It will change enough so that the LED will be forward biased enough to conduct the necessary current.)

The current will continue to flow through, now, the LED. But it will now be declining towards zero. If the switch is left open long enough, the current will decline to zero. This would be the right time to engage the switch again and start the cycle all over again.

The Joule Thief itself uses a particularly oriented winding in the BJT base circuit so that it aids the battery voltage and drives harder on the base while the other side of that transformer (a coil, too) in the collector leg allows the current in it to rise up. Eventually, one of two things stops this process: (1) the BJT's \$\beta\$ gives out and the BJT's collector voltage starts to rise, triggering a reversal; or, (2) the transformer core saturates causing the collector current to suddenly rise beyond what the base can support (again exhausting the BJT's \$\beta\$) and causing the BJT's collector voltage to start to rise and triggering a reversal. Either way, one that process starts the rising current cannot any longer be supported and is forced to start declining. And when it declines, the voltage on the coils reverse, which causes the coil in the base portion to also reverse and now oppose the battery, pretty much forcing the BJT to turn off (until the cycle can reverse when the magnetic energy has expired.)

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.