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circuit of precision 1A current source, how thus this circuit work?
Posted by aidilismadi on 08 Mar, 2018 10:01 -
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circuit of precision 1A. can anyone explain me about the voltage and current flow. and what is the function of op amp and mosfet at the circuit.
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circuit of precision 1A. can anyone explain me about the voltage and current flow. and what is the function of op amp and mosfet at the circuit.
V = I x R, we use 1 ohm resistor, so for 1A to flow, there will be 1V across the resistor V = 1A x 1ohm = 1V. thats why Vref is set to 1V there. if less current flowing, there will be less than 1V across resistor, opamp output will go high to close (switch) (or turn on) the mosfet to allow more current, but when current exceeding 1A, bigger than 1V across resistor, opamp will sense this and output will go low to shutdown the mosfet. this will go on and on until power supply is removed. so opamp job is to control mosfet to make sure 1V across resistor, ie 1A current. if more current, switch off mosfet, if less current switch on mosfet. faham?
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circuit of precision 1A. can anyone explain me about the voltage and current flow. and what is the function of op amp and mosfet at the circuit.
V = I x R, we use 1 ohm resistor, so for 1A to flow, there will be 1V across the resistor V = 1A x 1ohm = 1V. thats why Vref is set to 1V there. if less current flowing, there will be less than 1V across resistor, opamp output will go high to close (switch) (or turn on) the mosfet to allow more current, but when current exceeding 1A, bigger than 1V across resistor, opamp will sense this and output will go low to shutdown the mosfet. this will go on and on until power supply is removed. so opamp job is to control mosfet to make sure 1V across resistor, ie 1A current. if more current, switch off mosfet, if less current switch on mosfet. faham?
Right. In addition to that you have to understand how opamps work.
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Right. In addition to that you have to understand how opamps work.
opamp will try to make both of its +ve and -ve pin equal. +ve pin is fixed to 1V, so it will do whatever necessary to make sure -ve pin is also 1V. so if -ve pin is lower than 1V, it will output higher voltage with hope the -ve pin will rise accordingly, when -ve pin rise higher than 1V, opamp output will go lower. this is called negative feedback system. -
thank you. will definitely learn op amp
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heyy, sorry to bother again. can you explain to me about the important of resistor at the output end of op-amp?. and if the output value of op amp 1A then what is the suitable value of resistor over there.
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That circuit can be very difficult to stabilise. It's very easy for it to oscilate. I'd recommend a Darlington transistor, such as the TIP121, rather than a MOSFET.
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heyy, sorry to bother again. can you explain to me about the important of resistor at the output end of op-amp?. and if the output value of op amp 1A then what is the suitable value of resistor over there.
very seldomly you'll find an opamp that can source a 1A current, to be safe let it be realistic like 100mA, or maybe 0.2-0.3A pulsed.. to decide which R to use, you have to work out V = IR formula or R=V / I, I is 0.3A, V is whatever voltage you apply to the opamp which you havent provided..ymmv. -
That circuit can be very difficult to stabilise. It's very easy for it to oscilate. I'd recommend a Darlington transistor, such as the TIP121, rather than a MOSFET.
No need to change transistor. You can add capacitor to opamp feedback loop. In current sink/source application you canovercompensateoverstabilize circuit heavily because most likely fast response of regulation is not required.
Note 100pf capacitor:
Circuit from Linear appnote 105
Further reading about opampcompensationstabilization -
thank you for the recommendation
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thank you for the recommendation. sorry to bother again, thus the v+ on mosfet connected with same voltage as the amp-omp dc source. in this circuit what is the preference dc source on v+ and vin.
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Is this just a theoretical question or do you have some specific application in mind? If so, it may help to give you more specific advice.
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sorry to bother again, i clearly had a few problem with the circuit. can u guys give an opinion on the basic connection. if there is a problem with the dc supply to the mosfet or the ideal dc supply volt to the mosfet. the second one is why the dc supply on the om-amp did not give any current supply. thank you.
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just doing an experiment on specific 1A current source for my college project. i did not get 1A current at the Resistor.
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One thing that immediately comes to mind is the breadboard layout and stability. The MOSFET will be very sensitive to stray coupling at its gate. I would advise putting a 100 ohm resistor right at the Gate pin - no jumpers, plug the resistor in directly at the package. Also check that you are only sensing voltage across the current sense resistor and not measuring the voltage across the jumper leads too.
It would be best (and easier to follow) to tidy up the layout and use short lengths of solid core wire rather a long jumper leads.
Also, I don't see any supply decoupling capacitors for the the opamp and MOSFET, this will significantly affect stability too. -
thank you for the help. sorry late reply
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hey everyone, thank you for the help so far. one more things to ask, how thus the mosfet working in this circuit, like specifically how the mosfet function and interface with op-amp. how can we know that specific current will flow through the mosfet when 1V is applied at the gate to source. in simply form, what is the main function of mosfet in this circuit.
the second question, if the resistor R1 which is 1ohms is change to 2ohms thus the value of current change at R2?. next can u guys explain the current flow through dc source to mosfet and to the node at the source of the mosfet and inverting input of op-amp. and from the node to the R1 and ground.
the third question, how to determine if we have to put value of Rs at the input of voltage reference? if not stated in the datasheet. secondly if we have to put Rs how to determine the value of Rs. noted that im using ltc1798.
ps thank you so much guys for giving a wonderful explanation so far. i hope to have an understanding of the circuit like u guys one day.
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hey everyone, thank you for the help so far. one more things to ask, how thus the mosfet working in this circuit, like specifically how the mosfet function and interface with op-amp. how can we know that specific current will flow through the mosfet when 1V is applied at the gate to source. in simply form, what is the main function of mosfet in this circuit.
The MOSFET simply passes current based on its gate voltage. The Op Amp tells the MOSFET how hard to turn on based on the difference between the feedback voltage (the MOSFET current times the 1 Ohm resistor) and the reference voltage 1V. The output of the op amp will keep rising, turning the MOSFET on harder and harder, until the voltage drop across the 1 Ohm resistor is exactly the same as the voltage on the + input.Quote
the second question, if the resistor R1 which is 1ohms is change to 2ohms thus the value of current change at R2?. next can u guys explain the current flow through dc source to mosfet and to the node at the source of the mosfet and inverting input of op-amp. and from the node to the R1 and ground.
As long as the reference voltage is fixed at 1V, the current flow through the sense resistor has to produce the same voltage. If the sense resistor is set to 2 Ohms, it takes 500 mA to create the 1V feedback.
The simple way to think about this is to separate the MOSFET gate issue from the drain-source issue. The drain-source current wants to be 1A. At some level of gate voltage the MOSFET will deliver that amount of current. We don't know what the gate voltage has to be and we don't care. That's the problem of the op amp. If will put out as much voltage as necessary to get the MOSFET to turn on hard enough to get 1V across the sense resistor.
Remember, the open loop gain of an op amp is HUGE! If there is a voltage difference at the input terminals, the output will change enough to reduce the difference to 0V. That's why these circuits oscillate. The output voltage goes all the way to the op amp rail, the MOSFET turns on harder, the sense voltage increases, the op amp swings all the way to the other rail and the MOSFET turns off. This is the very definition of oscillation. That's why that little extra capacitor slows things down. The math of what it is doing is a little hairy but basically, it is anticipating a change in the sense voltage before it happens based on how much the op amp output changes. It does this to provide a little extra sense voltage (or less voltage if the op amp is going low) in anticipation of the larger change across the sense resistor.Quote
the third question, how to determine if we have to put value of Rs at the input of voltage reference? if not stated in the datasheet. secondly if we have to put Rs how to determine the value of Rs. noted that im using ltc1798.
That resistor on the + input is just there to limit any fault current that might get involved with the input. It's value is pretty much irrelevant. No current flows into either of the op amp inputs (ideal op amp laws). As a result, there is no voltage drop across that resistor under any reasonable condition. Since there is no voltage drop and no current flowing, the value isn't important. There really is SOME current flow into the + input so keep the value reasonable - 1k seems ok, 100 Ohms seems ok, 1 gigaohm doesn't seem ok.
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thank you so much the mosfet part is a little complicated but i'll try to understand
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The simple way to think of the MOSFET is as a water valve. More water flows when the valve is open and in the case of the MOSFET, the water valve is the gate voltage. As the gate voltage goes higher, the valve is more open and more water/current flows. When the gate voltage goes down, the valve is more closed and the water/current flow decreases.
There are many details like Vgs and the various gate capacitances, in fact, there is an entire datasheet full of numbers. In the end, the MOSFET is nothing more than a voltage controlled water valve with some details around the edges.
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thank you so much for the explanation on mosfet
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That MOSFET explanation does not cover the fact that a MOSFET has polarity. An N type wants a positive gate voltage and a P type wants negative with respect to source.
Its very important in your circuit to have the correct MOSFET both polarity and current rating.
Do you know that a MOSFET (or transistor) will simply self-destruct at high current if not heat-sinked? -
The datasheet for the TL082 opamp shows that it has two opamps in it. The unused opamp might be interfering with the opamp in it that you are using. Use a single opamp instead.
The datasheet for the TL082 opamp also shows that its Common Mode Input Voltage Range is 4V more positive than its negative supply voltage so its inputs DO NOT WORK if their voltage is less than +4V in your circuit. The TL08x opamps have the Phase Inversion Problem then its output will go as high as it can if an input voltage is within a few volts from 0V in your circuit. It needs an additional negative supply of -3V or more.
Instead you could use an opamp that has inputs that work as low as +1V in your circuit.