A question about the current detection

[zhenyuanwu@gmail.com]

Dear friends,

I have got an AC current detecting circuit. But I failed to
understand. Can anyone help me? Thanks a lot in advance!
http://
images.elektroda.net/17_1233301094_thumb.jpg

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?

 

[zhenyuanwu@gmail.com]

By the way the current has been converted to be voltage at the input
node SI.

On 1ÔÂ30ČŐ, ĎÂÎç4Ęą00ˇÖ, "zhenyua...@gmail.com" <zhenyua...@gmail.com>
wrote:

Dear friends,

I have got an AC current detecting circuit. But I failed to
understand. Can anyone help me? Thanks a lot in advance!
http://
images.elektroda.net/17_1233301094_thumb.jpg

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?

 

[Ross Herbert]

On Fri, 30 Jan 2009 00:00:06 -0800 (PST), "zhenyuanwu@gmail.com"
<zhenyuanwu@gmail.com> wrote:

ear friends,
:
:I have got an AC current detecting circuit. But I failed to
:understand. Can anyone help me? Thanks a lot in advance!
: http://
:images.elektroda.net/17_1233301094_thumb.jpg
:
: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.

 

[Franc Zabkar]

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

By the way the current has been converted to be voltage at the input
node SI.

On 1ÔÂ30ČŐ, ĎÂÎç4Ęą00ˇÖ, "zhenyua...@gmail.com" <zhenyua...@gmail.com
wrote:
Dear friends,

I have got an AC current detecting circuit. But I failed to
understand. Can anyone help me? Thanks a lot in advance!
http://
images.elektroda.net/17_1233301094_thumb.jpg

http://images.elektroda.net/17_1233301094.jpg

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

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.

And what is the capacitors C19 and C24 for?

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

- Franc Zabkar
--
Please remove one 'i' from my address when replying by email.

 

[zhenyuanwu@gmail.com]

Thank you very much!

The Capacitors C19 and C24 are for the stability of the opamp, and
filter for the higher frequency. They make the opamp a lower unity
gain frequency, and a phase margin of 90 degree.

On 1月31日, 上午5时04分, Franc Zabkar <fzab...@iinternode.on.net> wrote:

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

By the way the current has been converted to be voltage at the input
node SI.

On 1ÔÂ30ÈÕ, ÏÂÎç4ʱ00·Ö, "zhenyua...@gmail.com" <zhenyua...@gmail.com
wrote:
Dear friends,

I have got an AC current detecting circuit. But I failed to
understand. Can anyone help me? Thanks a lot in advance!
 http://
images.elektroda.net/17_1233301094_thumb.jpg

http://images.elektroda.net/17_1233301094.jpg

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

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.

And what is the capacitors C19 and C24 for?

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

- Franc Zabkar
--
Please remove one 'i' from my address when replying by email.