Limiting current WITHOUT dropping voltage

If you did have 1A the 10 ohms + 2 ohm drop 12V leaving nothing for MOSFET.
Also LM358 has serious output swing limitations. It wants to put out ~ 4.5V to
enable MOSFET Id 1A. LM358 limited to ~ 1.5V from positive rail, use a CMOS
RRIO OpAmp .

Control loop signedness is wrong. If current goes up V- OpAmp input drops so its output
rises which is turning on to draw even more current. Swap V- and V+ OpAmp....?


Regards, Dana.

Hi,

I agree with the above.

2 Ohms and 10 ohm´s won´t work for 12V and 1A.

Besides this BSS84 is a P-Ch MOSFET, while you used a N-Ch MOSFET


Klaus

Your circuit has many mistakes.
You need a rail-rail op amp and a P_MOSFET as was in the reference schematic.

Below is the LTspice simulation with such an op amp and a P-MOSFET.
Also reduced the current sense resistor (R1 in my schematic) value to 1Ω to give some voltage headroom with a 10Ω load, and changed the reference resistors to give a 1A limit.

Note that you can regulate current or regulate voltage, but you can't do both at the same time.

Hi,

@OP:
In your circuit R4 is the load. What is it´s range in Ohms?

What´s the input voltage range.

Mind that the current is only as accurate as the input voltage. When the input voltage changes by x% then the ouput current also changes by x%

Klaus

Thanks, This schematic works in proteus simulation. I hope it will work in practice also,

Hi,

I see no need for a true RR OPAMP.

Some things to consider: (if it works or not)
* noise (reglation) ... maybe add a filter. Maybe a capacitor across the 1k.
* loop stability ... maybe add local HF feedback for the OPAMP
* and it depends on load resistance range. (output voltage range)
* and about load impedance (is it capacitive, purely ohmic or inductive?
.. if inductive: consider to add some overvoltage / reverse voltage protection.

I personally think that the 10R is useless. Either use a higher value or omit it.

Klaus

khatus said:

This schematic works in proteus simulation. I hope it will work in practice also,

Click to expand...

If R4 is you load, try varying it's value and see if the current stays constant.

KlausST said:

I see no need for a true RR OPAMP.

Click to expand...

It's dicey without one.
The op amp output voltage has to to within a couple volts of the supply to turn off the P-MOSFET (The MOSFET Vgs(th) plus the voltage across the sense resistor).
A standard op amp will be marginal to reach that voltage.

khatus said:

I'm trying to limit a current of a 12V power supply to 1A. I am using this schematic from an online source. While simulating it on proteus simulation software the current in the resistor R4 does not give 1A output current. What is the problem with my circuit?

Click to expand...

based on post #6:
the feedback to the op amp is coming from the drain of the FET, so the circuit will regulate that point.
V+ into the op amp is 10.9 V
if the circuit is working, that's what you should have at the drain of the FET,
which leaves 1.09 V shared across the FET and R4
since the voltage across R$ is 1.09V, the voltage at the source of the FET is 1.09 V,
thus the FET is ON, and the circuit is not regulating.

if you want to regulate the output current, you need to use the current, or a sample thereof, in the feedback loop

Hi

crutschow said:

It's dicey without one.
The op amp output voltage has to to within a couple volts of the supply to turn off the P-MOSFET (The MOSFET Vgs(th) plus the voltage across the sense resistor).
A standard op amp will be marginal to reach that voltage.

Click to expand...

From my understanding:
* to turn OFF: the OPAMP output needs to be VCC - V_GS_cutoff. But the Mosfet never needs to be turned off
* to turn ON: the OPAMP output needs to be VCC - V_shunt - V_GS.... in the simulation it shows 4.8V. I expect the Opamp output to stay close to this value as long as a suitable small load resistor is used = as long as not saturated.

****
In opposite to wwfeldman I think the circuit is working as it should.
The voltage divider generates about -1V wrt VCC.
The Opamp regulates to get the same voltage drop across the 1R shunt ... giving a load current of about 1A.

Klaus