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help with Class AB amplifier

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Jay_Diddy_B:
Hi,

This little sketch should help show you get from a standard current mirror to the one used in this amplifier:




The circuit on the left is the standard implementation. This works best if the two transistors are at the same junction temperature. The transistor on the left is actually wired as a diode.

In the centre circuit the transistor has been replaced by a diode.
This still has most of the properties of a current mirror, but the mirror ratio will not be 1:1 especially over temperature.

In the third circuit the positions of the diode and resistor have swapped. This doesn't change the function in any way.
We are now at the circuit in the AB amplifier.

Regards,
Jay_Diddy_B

David Hess:

--- Quote from: rex1232 on November 24, 2019, 09:48:08 pm ---Hey, the current mirrors I have seen always have 2 transistors where they are connected by the base and also to their collector.
How does it work when there only is when transitor? (Q6)

Does it have something to do with D10's temperature compensation and Q16, Q17 pulling the Current mirror up and down?
--- End quote ---

In a current mirror, one of the transistors is wired as a diode with its base and collector shorted.  So if close matching is not required, one of the transistors can be replaced with a diode.

rex1232:
Hey guys thank you for all the help  :)

I made this block diagram is the signal ways/connection of block to block correct?

Thank you in advance

rex1232:
So basically its a cheap version of the "normal" type of current mirror

Thank you for the response

David Hess:

--- Quote from: rex1232 on November 25, 2019, 12:31:17 pm ---So basically its a cheap version of the "normal" type of current mirror
--- End quote ---

I suppose.  If the transistors were 2N4401s then the savings from replacing one with a diode would be insignificant but with a more expensive BD140, it is a common optimization.  In this case the lowered precision is not important.

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