Current Sense Opamp Circuit Problem

[jdhar]

I have implemented a slight variant on the circuit in this datasheet,
Page 21, Figure 69:

http://www.analog.com/static/imported-files/data_sheets/AD8551_8552_8554.pdf

The problem I am having is that as I lower the voltage at the node
Rsense/R1 are connected to (lets called that Vreg), and increase the
load current past a certain point, the output voltage Vout eventually
maxes out and stays constant regardless of the load current (IL). This
is because the voltage at the source is probably low enough that the
MOSFET can't operate in saturation anymore.

To give some numbers... at about Vreg of 0.8V, I start having problems
when Vsense (drop across Rsense) is 100mV. This would put the Voltage
at Opamp Pin 2 at 0.7V, and the Vgs of the MOSFET is now very close to
the threshold since the Opamp output is near zero. Increasing the
current even further will not work since the opamp has no more room to
move.

My question, is there any way I can get this to work for a Vreg of
about 0.6V? I have seen some variants on this circuit with a JFET
instead of a MOSFEt? Would that help? If I used a dual supply Opamp so
the output can go negative, would taht help?

Thanks

 

[Robert Monsen]

I have implemented a slight variant on the circuit in this datasheet,
Page 21, Figure 69:

http://www.analog.com/static/imported-files/data_sheets/AD8551_8552_8554.pdf

The problem I am having is that as I lower the voltage at the node
Rsense/R1 are connected to (lets called that Vreg), and increase the
load current past a certain point, the output voltage Vout eventually
maxes out and stays constant regardless of the load current (IL). This
is because the voltage at the source is probably low enough that the
MOSFET can't operate in saturation anymore.

To give some numbers... at about Vreg of 0.8V, I start having problems
when Vsense (drop across Rsense) is 100mV. This would put the Voltage
at Opamp Pin 2 at 0.7V, and the Vgs of the MOSFET is now very close to
the threshold since the Opamp output is near zero. Increasing the
current even further will not work since the opamp has no more room to
move.

My question, is there any way I can get this to work for a Vreg of
about 0.6V? I have seen some variants on this circuit with a JFET
instead of a MOSFEt? Would that help? If I used a dual supply Opamp so
the output can go negative, would taht help?

Thanks

Use an instrumentation amp (or instrumentation amp configuration) and
measure the difference across the sense resistor. Multiply by some
suitable factor.

If you aren't too picky, you could use this:

3.0V
|
|
|
10k | 100k
0.6V ___ | ___
o-------------o---o--|___|----------------)--------|___|----.
| | | |
| | |\| |
.-. .---------------------|-\ |
| | 0.1 ohm | >----------------'
| | .----------------|+/
'-' | |/|
| | .
| | |
o--------' |
| |
| |
|
LOAD GND



Vout = 0.6 - Iload

(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

Regards,
Bob Monsen

 

[jdhar]

I may be viewing that circuit wrong, but if you have both sides of the
sense resistor going to the opamp, how will it work since both nodes
at the opamps input can't be the same? (due to the drop across the
resistor). It won't be able to force the nodes to same value, so it
will act as a comparator.....

 

[jdhar]

I may be viewing that circuit wrong, but if you have both sides of the
sense resistor going to the opamp, how will it work since both nodes
at the opamps input can't be the same? (due to the drop across the
resistor). It won't be able to force the nodes to same value, so it
will act as a comparator.....

The other problem with using an instrument amp is that I can't seem to
find one with a low input bias current. I need this amp to be able to
have a gain of >= 5 @ 2MHz to be useful, and the only Instrument Amp
with a GBP that high has a input bias current larger than the current
I am trying to measure, which will no doubt throw the measurement off
quite a bit.

The nice thing with the configuration I linked to is that bias current
is sub microamp.

 

[Robert Monsen]

I may be viewing that circuit wrong, but if you have both sides of the
sense resistor going to the opamp, how will it work since both nodes
at the opamps input can't be the same? (due to the drop across the
resistor). It won't be able to force the nodes to same value, so it
will act as a comparator.....

Sorry, I drew the feedback is in the wrong place. Put it between the
10k and 100k resistor.

Here is an LTSpice simulation I did for this circuit. It uses a
subtractor, so the output voltage is equal to the current draw through
the sense resistor:

Version 4
SHEET 1 880 680
WIRE -112 -32 -256 -32
WIRE 80 -32 -112 -32
WIRE 448 -32 80 -32
WIRE -112 16 -112 -32
WIRE -112 144 -112 96
WIRE -64 144 -112 144
WIRE 32 144 16 144
WIRE 112 144 32 144
WIRE 240 144 192 144
WIRE -112 160 -112 144
WIRE 400 192 368 192
WIRE 528 192 480 192
WIRE -256 208 -256 -32
WIRE 80 224 80 -32
WIRE 32 240 32 144
WIRE 48 240 32 240
WIRE 448 240 448 -32
WIRE 240 256 240 144
WIRE 240 256 112 256
WIRE 272 256 240 256
WIRE 368 256 368 192
WIRE 368 256 352 256
WIRE 416 256 368 256
WIRE -112 272 -112 240
WIRE 48 272 -112 272
WIRE 528 272 528 192
WIRE 528 272 480 272
WIRE -112 336 -112 272
WIRE 272 336 -112 336
WIRE 368 336 352 336
WIRE 416 336 416 288
WIRE 416 336 368 336
WIRE -112 352 -112 336
WIRE 368 368 368 336
WIRE -256 464 -256 288
WIRE -240 464 -256 464
WIRE -112 464 -112 432
WIRE -112 464 -240 464
WIRE 80 464 80 288
WIRE 80 464 -112 464
WIRE 368 464 368 448
WIRE 368 464 80 464
WIRE 448 464 448 304
WIRE 448 464 368 464
FLAG -240 464 0
SYMBOL res -96 256 R180
WINDOW 0 36 76 Left 0
WINDOW 3 36 40 Left 0
SYMATTR InstName R1
SYMATTR Value 0.1
SYMBOL current -112 352 R0
WINDOW 0 24 88 Left 0
WINDOW 3 24 0 Left 0
WINDOW 123 0 0 Left 0
WINDOW 39 24 28 Left 0
SYMATTR InstName I1
SYMATTR Value 10mA
SYMATTR SpiceLine load
SYMBOL voltage -256 192 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value 3
SYMBOL Opamps\\LT1218 80 192 R0
SYMATTR InstName U1
SYMBOL voltage -112 0 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value 2.4V
SYMBOL res 32 128 R90
WINDOW 0 0 56 VBottom 0
WINDOW 3 32 56 VTop 0
SYMATTR InstName R2
SYMATTR Value 10k
SYMBOL res 208 128 R90
WINDOW 0 0 56 VBottom 0
WINDOW 3 32 56 VTop 0
SYMATTR InstName R3
SYMATTR Value 100k
SYMBOL Opamps\\LT1218 448 208 R0
SYMATTR InstName U2
SYMBOL res 496 176 R90
WINDOW 0 0 56 VBottom 0
WINDOW 3 32 56 VTop 0
SYMATTR InstName R4
SYMATTR Value 100k
SYMBOL res 368 240 R90
WINDOW 0 0 56 VBottom 0
WINDOW 3 32 56 VTop 0
SYMATTR InstName R5
SYMATTR Value 100k
SYMBOL res 368 320 R90
WINDOW 0 0 56 VBottom 0
WINDOW 3 32 56 VTop 0
SYMATTR InstName R6
SYMATTR Value 100k
SYMBOL res 384 464 R180
WINDOW 0 36 76 Left 0
WINDOW 3 36 40 Left 0
SYMATTR InstName R7
SYMATTR Value 100k
TEXT -256 488 Left 0 !.dc I1 0 1 1m

 

[Robert Monsen]

I may be viewing that circuit wrong, but if you have both sides of the
sense resistor going to the opamp, how will it work since both nodes
at the opamps input can't be the same? (due to the drop across the
resistor). It won't be able to force the nodes to same value, so it
will act as a comparator.....

the message I just posted has an LTSpice simulation, where there are
two opamps. I realized after posting that you can do the function with
a single opamp, since the first one just inverts and generates 10x
gain.

So, it looks like this:

3V -----------------------------------.
|
0.6V --o-------------. |
| | |
| | |
| | |
.-. .-. |
0.1R | | 100k | | 1MEG |
| | | | ___ |
'-' '-' .--|___|---)------.
| | | | |
| ___ | | |\| |
o-----|___|---)-----o--------|-\ |
| | | >-----o---output
| 100k o--------------|+/
| | |/|
| | |
.---o---. .-. |
| | | | 1MEG |
| LOAD | | | |
| | '-' |
'---o---' | |
| | |
| | |
GND ---o-------------o----------------'
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

The opamp should be a single supply variety.

Regards,
Bob Monsen