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I read the LT3573's datasheet, I have some questions would like to ask.

Based on this block diagram, enter image description here

1.) What's the main function of the below circuit (red circle).

2.) If I need to detect the output voltage, why I cannot just use R3 and R4 to do that.

3.) What's the main function of I2.

And I also read the MPS MP6004's datasheet, but the method MPS use is different way, they add the extra winding to detect the output voltage.

4.) I am curious about what's the difference between MPS and LT.

5.) Why MPS do not need to use the circuit like LT?

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1 Answer 1

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The voltage across L1A is referenced to Vin not ground. It can't be easily used for feedback in that form, Q2 and associated circuitry resolve that.

Just using a divider from L1A to ground would not give a voltage proportional to the output.

The purpose of R3, Q2, I2 and R4 is to create a voltage that is referenced to ground to feed the Flyback Error Amplifier.

Although Q2, Q3 and I2 look like a current mirror the purpose of Q3 and I2 is to create temperature compensated voltage source matching the Vbe of Q2 so that the emitter of Q2 is very close in voltage to that of Q3. The current injected into Q2 by R3 will therefore be proportional to the voltage across L1A.

The current through Q2 will be proportional to the voltage across L1A that in turn is proportional to the voltage on L1B.

The great majority of the injected emitted current of Q2 will flow through R4 creating a voltage across it that is proportional to the voltage on L1A and in turn proportional to the voltage at he output. A small proportion of the current fro R3 is lost (<1%) to the base current of Q2.

All of this allows the controller to provide feedback depending upon the output voltage without requiring additional components such as an opt-isolator or even additional windings on the transformer.

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  • \$\begingroup\$ Hi Kevin, Why Q2 I2 can create a voltage that is referenced to the ground? I still do not understand what's the function of this circuit. \$\endgroup\$
    – EEC
    Jan 23, 2022 at 8:25
  • \$\begingroup\$ @EEC I suggest you grab your favourite simulator, create a bare-bones equivalent schematic, run it, then follow the results of the simulation in parallel with what Kevin said, step by step. \$\endgroup\$ Jan 23, 2022 at 11:14
  • \$\begingroup\$ @EEC - I have added some text to the answer that hopefully helps. \$\endgroup\$ Jan 23, 2022 at 17:34
  • \$\begingroup\$ Hi Kevin, I think this is more clear to me, thanks. \$\endgroup\$
    – EEC
    Jan 24, 2022 at 16:33

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