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-   -   Will this work as Error Correction? (http://www.diyaudio.com/forums/solid-state/41931-will-work-error-correction.html)

lumanauw 3rd October 2004 01:49 AM

Will this work as Error Correction?
 
2 Attachment(s)
What do you think, will this work as EC? The signal is injected from one point from VAS (already got full swing) to the junction of emitors of T1 and T2. The feedback from output is injected to bases of T1 and T2.
I think it will balance/level input (signal to the emitors) and output (output signal feedback to bases). But is this an EC?

lumanauw 4th October 2004 03:18 AM

This idea is not working?

Eva 4th October 2004 09:59 AM

Both the error correcting current and the drive current applied to the bases of the predrivers are taken from the VAS output and typical VAS show huge output impedances [1Mohm?] so the circuit won't be able to correct anything in these circumstances

However, If you drive it from a low impedance source then it should work fine as a buffer, I think

lumanauw 5th October 2004 01:29 AM

So I have to make an emitor follower first after the VAS before this VBE multiplier? Or is it possible to make VAS with heavy bias (like 10-20mA) without emitor follower?

EVA, is there any difference if I put or omit 2 R's that is going to the bases of VBE multiplier from the output? I tought putting those 2 R there makes it an EC, or there is no difference at all?

bocka 8th October 2004 10:52 AM

Quote:

So I have to make an emitor follower first after the VAS before this VBE multiplier?
Yes

Quote:

Or is it possible to make VAS with heavy bias (like 10-20mA) without emitor follower?
No. You have to transform the high output resistance of the VAS to a low and well defined resistance for the error correction. When using miller compensation the output resistance of the VAS depends heavily on the frequency. So maybe some resistors from the output stage to grond as well as large emitter/source degeneration resistors of the LTP with reduced miller capacitance are necessary to maintain a constant output resistance of the VAS stage. But this is only a short idea.


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