There is an application note from ON-semiconductors which provides a few diagrams to use a well known MC34063 switching regulator as LED driver. it's around the idea of setting the 1.25v feedback value using the voltage drop of a series, current sense resistor :

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further, it suggests using a transistor to "shift levels"(as it says) so the current sense resistor doesn't have to be that big to provide 1.25v drop; now it needs a voltage drop of about 0.65v, and less power is wasted in the resistor:

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  1. how this transistor provides 0.6v (1.25-0.65) so the resistor is less burdened? does it have anything to do with base-emitter voltage drop ?
  2. there are many adjustable regulators with the same feedback setup (e.g. LM2596 or the linear LM317). does this idea (either the one with the transistor or the one without) work on them too?
  3. does it work on the same MC34063 configured as step up? to drive LEDs from a single cell lithium ion battery for example?
  4. can we use another component to provide more drop so the power dissipation of current sense resistor is decreased even further?



1) Indeed the Base-Emitter voltage is added to the voltage across the current sense resistor. The sum of \$V_{BE}\$ and \$V(R_{sense})\$ still needs to be 1.25 V but if we consider that \$V_{BE}\$ is more or less constant then this works.

In practice the \$V_{BE}\$ isn't constant and will vary over temperature for example. Still, the solution might be good enough for driving the LED. The current will probably still be constant enough for the LED.

2) Yes, the same solution should work with those kind of regulators as well. The voltage (or current) regulation loop works in the same way.

3) Yes, the voltage (or current) regulation loop doesn't care if you made a buck or a boost converter. If implemented correctly the idea from the application note should also work when the MC34063 is used as a boost converter.

4) If you have such a component then yes (I challenge you to find one though that drops more than 0.7 V but less than 1.1 V and is very stable). Beware though that the more voltage you add, the more stable it needs to be. Extreme example: If you'd use a component that drop 1.20 V, that only leaves 0.05 V or 50 mV across the sense resistor. Suppose that the 1.20 V increases with 25 mV over temperature, what will your current be? I think it will be halved compared to what it should be.

A more usable implementation would be to make a separate reference voltage, divide that down to for example 200 mV using a voltage divider and then use an opamp to compare that voltage to the voltage across the reference resistor. Care must then be taken to keep the regulation loop stable though, added an opamp in the loop isn't always trivial. Get it wrong and you made an oscillator.


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