A circuit for controlling a resistor current

Thread Starter

arman74

Joined Jul 1, 2017
41
Hi everybody.
I need a circuit to control a high watt resistor for some purpose.I have a circuit which includes an opamp and a mosfet but in that the Vgs would be 2.5 v and thus high Rds on that would burn the mosfet.The power supply is 5 volts max and the current varies between 0 to 4 Amps.
upload_2019-6-24_9-19-45.png
This is my own circuit.
 

dl324

Joined Mar 30, 2015
11,230
The circuit won't work. The output of the LM324 is only guaranteed to get to about Vcc-2V.

If you can't increase the supply voltage, you need some combination of an opamp with a rail to rail output, a MOSFET with a lower threshold voltage, or a smaller current set resistor.

This is my own circuit.
It's a classic design.
 
You also need a gate series resistor and a gate resistor to ground (10K would work).

Rds(on) may be of less significance here because your operating the MOSFET in the linear region. It's going to get warm.
 

dl324

Joined Mar 30, 2015
11,230
What is the necessity of the pull down(for nmos)and pull up(for pmos) and the series resistors?
Opamps will sometimes oscillate when driving capacitive loads. A small resistor in series with the base will prevent that. A series resistor could also be used to limit gate charging current.

Pull up/down resistors might be used in a conservative design. They're more important if there are cases where the gate might not be driven (floating). That isn't the case in your circuit.
 

BobaMosfet

Joined Jul 1, 2009
1,171
Of practical concern with your circuit- You have placed R1 between the circuit and ground, raising the ground-voltage level. Essentially your circuit becomes one half of a voltage divider with R1. Thus you risk creating a groundloop if you bypass R1 when grounding anything else in your circuit (if you bypass R1). Don't do this.
 

Vinnie90

Joined Jul 7, 2016
86
Of practical concern with your circuit- You have placed R1 between the circuit and ground, raising the ground-voltage level. Essentially your circuit becomes one half of a voltage divider with R1. Thus you risk creating a groundloop if you bypass R1 when grounding anything else in your circuit (if you bypass R1). Don't do this.
I think his purpose is to generate a voltage to current converter to adjust the current through the load. R1 sets the current he wants
 

crutschow

Joined Mar 14, 2008
25,247
I need a circuit to control a high watt resistor for some purpose.
If you tell us exactly what this mysterious "some purpose" is, along with the resistor value, we can give you better answers.

We don't need a 20-question game before we determine what you actually need. :rolleyes:
 

TeeKay6

Joined Apr 20, 2019
572
Why to swap the oppamp inputs?Is the positive feedback stable then?

You mean this?

View attachment 180309
If you tell us exactly what this mysterious "some purpose" is, along with the resistor value, we can give you better answers.

We don't need a 20-question game before we determine what you actually need. :rolleyes:
The resistor value is given as 0.5 (ohms) in the TS's first post.

@arman74:
As I understand your post, you wish to cause a current in the range 0-4A to flow through R1. Thus, the voltage across R1 could be as high as 2.0V. In order to turn Q1 on sufficiently to conduct that current, the gate of Q1 must be held at least several volts higher than it's source. That is, the gate drive must be able to reach 2 + 3 = 5V or higher. You can make this circuit work only if you can provide that gate drive. You could lessen the drive requirement a little bit by using a low threshold MOSFET, but at least 5V would likely still be required. Thus, a safe solution is to raise the power supply voltage.

@crutschow
How about using a bipolar Darlington in place of a MOSFET?
 
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TeeKay6

Joined Apr 20, 2019
572
The resistor value is given as 0.5 (ohms) in the TS's first post.

@arman74:
As I understand your post, you wish to cause a current in the range 0-4A to flow through R1. Thus, the voltage across R1 could be as high as 2.0V. In order to turn Q1 on sufficiently to conduct that current, the gate of Q1 must be held at least several volts higher than it's source. That is, the gate drive must be able to reach 2 + 3 = 5V or higher. You can make this circuit work only if you can provide that gate drive. You could lessen the drive requirement a little bit by using a low threshold MOSFET, but at least 5V would likely still be required. Thus, a safe solution is to raise the power supply voltage.

@crutschow
How about using a bipolar Darlington in place of a MOSFET?
@arman74
Take a look at this thread https://forum.allaboutcircuits.com/threads/mosfet-having-the-lowest-rds-on.161398/page-2#post-1413261, post #34, for a suggested low threshold MOSFET. Adding a resistor (e.g. 2K) between +5V and the MOSFET gate will enable the LM358/324 output to swing higher, perhaps enough to make this work.
 

danadak

Joined Mar 10, 2018
4,057
I did not sweep gate drive far enough, revised -

upload_2019-7-17_19-4-9.png

I looked at transient analysis a little, you will need to compensate this
as it looks like phase margin is not adequate. I played with R between
OpAmp out and MOSFET gate, more R helps but thats diminishing
returns. You will have to examine this more thoroughly.


Regards, Dana.
 
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MrAl

Joined Jun 17, 2014
7,748
Hi,

One thing for sure you dont go by the Rds spec to figure out the power the mosfet will have to dissipate.
You go by the voltage across the mosfet and the current through it.
Since the R is 0.5 and the max current is 4 amps, the voltage across the mosfet is:
Vm=5-4*0.5=5-2=3 volts.
So the max power is 3*4=12 watts.
That's quite a bit of power which requires a decent size heat sink.
Interestingly, if you go up to 10 amps the max power is just a little over 12 watts still (12.5 watts). because this occurs at 5 amps.
 

TeeKay6

Joined Apr 20, 2019
572
I did not sweep gate drive far enough, revised -

View attachment 181878

I looked at transient analysis a little, you will need to compensate this
as it looks like phase margin is not adequate. I played with R between
OpAmp out and MOSFET gate, more R helps but thats diminishing
returns. You will have to examine this more thoroughly.


Regards, Dana.
I question whether this circuit and the IRF9310 are a good choice for these reasons. (1) This circuit is no longer the well-behaved source-follower of the original circuit; this circuit is a two-stage amplifier: the LM324 and the P-ch MOSFET and will likely be difficult to stabilize. (2) The IRF9310 is in a tiny SO-8 package with a <2.5W max dissipation limit. (3) It is problematic whether the LM324 can drive the gate positive enough to guarantee MOSFET control, considering variations in MOSFET threshold, etc.
 

TeeKay6

Joined Apr 20, 2019
572
Hi,

One thing for sure you dont go by the Rds spec to figure out the power the mosfet will have to dissipate.
You go by the voltage across the mosfet and the current through it.
Since the R is 0.5 and the max current is 4 amps, the voltage across the mosfet is:
Vm=5-4*0.5=5-2=3 volts.
So the max power is 3*4=12 watts.
That's quite a bit of power which requires a decent size heat sink.
Interestingly, if you go up to 10 amps the max power is just a little over 12 watts still (12.5 watts). because this occurs at 5 amps.
@MrAl
You say "Interestingly, if you go up to 10 amps the max power is just a little over 12 watts still (12.5 watts). because this occurs at 5 amps." I do not understand. If you get 10A by lowering the resistor to 0.2ohms (assuming the MOSFET can pass 10A with the same 3V Vgs), the MOSFET dissipation is 3V*10A=30W. What am I misunderstanding?
 

crutschow

Joined Mar 14, 2008
25,247
Here's an LTspice simulation of the constant-current circuit using a Sziklai pair (complementary Darlington) and a low value sense resistor to maximize voltage headroom for the load.

upload_2019-7-17_21-7-46.png
 
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