# circuit senses high-side current

R

#### raf

Jan 1, 1970
0
http://rapidshare.com/files/15616508/current_sensor.doc.html
Dear all,
Here i have a current sensor (high-side) to measure the current and to
offset the common mode voltages. I would like to know how the circuit
is working. If i give 300mV as input (assumed the shunt resistor with
0.5m ohm and current 60A) then the output is measured at the
transistor (IRF150) which is connected to the op amp. So how the
circuit is removing the common mode voltage. for this the transistors
(dual type) are connected which are doing the common mode voltage
offset. The circuit i am giving is PSPICE model in which i do not have
exact components so i took available from PSPICE.
Any suggestions are appreciated
RAF
http://rapidshare.com/files/15616508/current_sensor.doc.html

W

#### Winfield Hill

Jan 1, 1970
0
http://rapidshare.com/files/15616508/current_sensor.doc.html
Dear all,
Here i have a current sensor (high-side) to measure the current and to
offset the common mode voltages. I would like to know how the circuit
is working. If i give 300mV as input (assumed the shunt resistor with
0.5m ohm and current 60A) then the output is measured at the
transistor (IRF150) which is connected to the op amp. So how the
circuit is removing the common mode voltage. for this the transistors
(dual type) are connected which are doing the common mode voltage
offset. The circuit i am giving is PSPICE model in which i do not have
exact components so i took available from PSPICE.
Any suggestions are appreciated
RAFhttp://rapidshare.com/files/15616508/current_sensor.doc.html

Was that circuit an ED Ideas for Design, or similar? Ref, please?

The high-side circuit is a two-stage current mirror, fed from a low-
side current source. Any voltage drop across the current-sensing
shunt resistor unbalances the mirror by a current Ix = Is Rs/R1.

The low-side circuit is a servo that senses the difference between
the current source and the mirror's output, and adds a current Ix
through the MOSFET to rebalance the mirror. This current is the
aforementioned Ix = Is Rs/R1, and is sensed by an output resistor
Rg, so the output voltage Vo = Is Rs Rg/R1 shows the current Is.

The MOSFET (which could have been a BJT instead) is operated at
very low currents, and should therefore be a small-die part (such
as the suggested 2n7000, not a large-die part like an IRF150), to
keep the capacitance down.

Contrary to the drawing, the opamp does not need to be a rail-rail
opamp. Its inputs should work to the Vee rail, e.g., like an LM324.

R

#### raf

Jan 1, 1970
0
Hi Mr. Hill,
thanks for u r reply. Yes, exactly it is from ED design ideas. In that
the operation is not explained in detail. If u could expalin more in
detail it would be appreciated.
RAF

W

#### Winfield Hill

Jan 1, 1970
0
raf said:
Hi Mr. Hill,
thanks for u r reply. Yes, exactly it is from ED design ideas.
In that the operation is not explained in detail. If u could
expalin more in detail it would be appreciated.

I'm sorry, RAF, I thought I had done so, e.g., Vo = Is Rs Rg/R1,
etc. Could you tell us exactly what issue of ED's Design Ideas
it came from? Thanks!

R

#### raf

Jan 1, 1970
0
Hi ,
I am searching for a current sensor for motor application (high side)
which can withstand high common mode voltages( 100V). The Circuit in
the ED's design ideas is meant for avoiding high common mode voltages
which the current mirrors do. So i need detailed operation of each
component so that i can proceed further.
Thanks and Regards
RAF

R

#### raf

Jan 1, 1970
0
In the ED'd design on March1,2001, EDN123 was published.

Replies
7
Views
175
Replies
17
Views
354
Replies
7
Views
365
Replies
4
Views
328
Replies
8
Views
370