A question about the current detection

Do you have a question? Post it now! No Registration Necessary

Translate This Thread From English to

Threaded View
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=
http://images.elektroda.net/17_1233301094.jpg][img]http://
images.elektroda.net/17_1233301094_thumb.jpg[/img][/url]

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?

Re: A question about the current detection
By the way the current has been converted to be voltage at the input
node SI.

On 1D4%C230C8%D5, CF%C2CE%E74CA%B100B7%D6, " snipped-for-privacy@gmail.com" <zh=
snipped-for-privacy@gmail.com>
wrote:
Quoted text here. Click to load it


Re: A question about the current detection
On Fri, 30 Jan 2009 00:07:55 -0800 (PST), " snipped-for-privacy@gmail.com"

Quoted text here. Click to load it

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

Quoted text here. Click to load it

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.

Quoted text here. Click to load it

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.

Re: A question about the current detection
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 1E6%9C88%31E6%97A5%, E4%B88A%E58D%885E6%97B6%04E5%8886%, Franc Za=
Quoted text here. Click to load it


Re: A question about the current detection
On Fri, 30 Jan 2009 00:00:06 -0800 (PST), " snipped-for-privacy@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!
: [url=
http://images.elektroda.net/17_1233301094.jpg][img]http://
:images.elektroda.net/17_1233301094_thumb.jpg[/img][/url]
:
: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.

Site Timeline