Re: boost converter with current mirror feedback and type II compensator?

Aug 29, 2008 4 Replies


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>
>I wonder why to use a current mirror here? the internal ref. V = 1.26.
>the current mirror calculation is correct I verified.
>But why use a current mirror here? keep from transient voltage?
>temperature compensation?
>
>if add this current mirror, how to calculate the compensator? here is
>a R-C in series, is't a simplified type II ?
>
>calculation for the compensator is welcomed.

This is not so much a current mirror, as a level-shifting inverter. In this circuit the U2 boost converter produces +5V from a pre-existing


-12V source.



This 5v is expressed across R8, sensed by Q7A which has a base biased one EB voltage drop below ground by Q7B and R11. Due to simple transistor action, R8 current (minus Q7A base current) flows in R10, providing a feedback voltage for U2 that is referenced to the -12V rail, on which the internal U2 voltage reference sits.



RL


Doing so would introduce gross regulation effects from the -12v raail onto the +5V output. Normally, regulation requirements on a 5V rail are pretty tight.

RL

Because the ground of the LM5002 is actually at -12V not at ground. As the other poster mentions the circuit is to change the reference point. If you just used a resistive divider any variation of the -12V rail would affect the 5V rail as well.

kevin

.26.

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