A question about the current detection

Dear friends,

I have got an AC current detecting circuit. But I failed to understand. Can anyone help me? Thanks a lot in advance! [url=

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It seems that the circuit between the input node SI and node 8 is useless. Anyone agree with me?

And what is the capacitors C19 and C24 for?

Reply to
zhenyuanwu
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Reply to
zhenyuanwu

:Dear friends, : :I have got an AC current detecting circuit. But I failed to :understand. Can anyone help me? Thanks a lot in advance! : [url=

formatting link
: :It seems that the circuit between the input node SI and node 8 is :useless. Anyone agree with me? : :And what is the capacitors C19 and C24 for?

You need to learn to copy and paste the correct url link.

Reply to
Ross Herbert

On Fri, 30 Jan 2009 00:07:55 -0800 (PST), " snipped-for-privacy@gmail.com" put finger to keyboard and composed:

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The voltage at U4A-1 = SI/2 since R1 and R2 form a potential divider.

The voltage at U4C-9 = voltage at U4C-10 (if op-amp is operating in linear region).

Therefore voltage at U4C-8 = SI.

So it would seem that U4A and U4C are functioning as a unity gain buffer with a 10K input impedance, in which case I don't understand why they are needed, either.

They reduce the gain of the high frequency components in the signal.

- Franc Zabkar

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Please remove one \'i\' from my address when replying by email.
Reply to
Franc Zabkar

Thank you very much!

The Capacitors C19 and C24 are for the stability of the > On Fri, 30 Jan 2009 00:07:55 -0800 (PST), " snipped-for-privacy@gmail.com"

=C2=B7=C3=96, " snipped-for-privacy@gmail.com"

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Reply to
zhenyuanwu

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