Active current sinks? help!!!

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Hey gang,

I need to design a current sink that will follow the output of a MOSFET power follower to minimize DC offset... I will be using a IRFP140N as both the power follower and the CCS.

power supply is +12v and -12v split supply. Bias is around 3A.

Problem is that the 12v supplies are 17Ah SLA batteries. So as the amp is run the 12v will drop. They will probably drop non-symmetrical. the -12v will have a consistant 3A draw, while the +12v will draw depending on the music signal.

So, offset may become a problem.

yah I know the amp has a gain of like 0.95.

Any help? I have searched for CCS but couldn't find anything about active ones. Also is it possible to build one totally discreetly without any op-amps?

tia
Richard
 
The one and only
Joined 2001
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Offset drift won't be a problem, but if you drive
the Gate of the follower at ground DC potential, the
output will be at -Vgs plus the drop across any
Source resistance you might be using.

As a practical matter, you will have to bias the
Gate of the follower positively to correct this, and
this means a capacitively coupled input.
 
How about a 1:1 IT? Then on the secondary just connect the lower terminal to a positive bias, the upper to the gate of the mosfet?

I may even get a 1:2 or something to provide a little voltage gain to the gate.

Sowter has some nice 1:1's...

I think the upper MOSFET bias should be run from a seperate voltage source. Would there be any current draw for the bias??

Thanks!
Richard
 
Hi Richard,
I don't think your two batteries will discharge at different rates unless you are running dc into the speaker load (I assume this is for driving a speaker?). The average current drawn from both batteries should be about the same.

Have you considered using a BJT for the CCS? It will operate down below 0.5V and has lower capacitance (looks more like an ideal current source).
BAM
 
Well no actually I have not. Do you have a good circuit to use? I kind of like the idea of having:


+12v -> IRFP140N -> 0.2 ohm -> IRFP140N -> 0.2 ohm -> -12v

Its nice and symmetrical. Surely that is a good thing??

The ave. current draw should be the same, but over time one battery may drop faster than the other...

I guess its something I have to try.

p.s. the ouput is taken between the first 0.2 ohm resistor and the second IRFP140N.... but I guess you know that.

tia

Richard
 
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