[SOLVED] Constant current source driver circuit rc filter

[jovin555]

In the attached image,What is the application of R3 and C1 in the feedback back.Is it used as an integrator circuit or to filter any noise?Will this have any impact on the rise time of the output driver circuit?

Also how can we calculate the output voltage of opamp if we vary the input voltage at the non-inverting pin of opamp?

 

[jovin555]

i tried a lot to calculate the output.but still i don,t know how to get the right answer with theoretical calculation.

Please give me the method to calculate the output.

 

[jovin555]

I am using ad8601

 

[jovin555]

I have read the document.But i still confused with the output.

Instead of giving a resistor divider at the non-inverting input,i want to give a dac input.

So if i give a dac input of 1V at the non-inverting input,what will be the output of op-amp?Can you please give me one example.

 

[FvM]

I feel that the questions in your post #1 are essentially unanswered.

1. Yes the RC circuit will affect the rise time if the circuit is operated with a respective fast control voltage. Obviously the circuit is designed for static constant current, so it doesn't matter here. The purpose of the RC circuit is loop compensation. Without it, the feedback loop could become unstable due to the large MOSFET input capacitance. The right dimensioning depends on the OP speed, a 100 pF - 1 nF capacitor should be sufficient for general purpose OP with MHz bandwidth.

2. The current source gain is only set by the shunt resistor, Iout = Vin/Rshunt

Having a small shunt resistor value is reasonable to reduce the shunt power rating, if you want to control the circuit with a DAC, you'll probably keep a voltage divider in front of the OP.

 

[jovin555]

So for 1V input , Current will be I = 1V/0.1R = 10A.At this point what will be the output voltage at the gate of the mosfet?Actually the output of op-amp?

is there any way to claculate the output voltage of opamp that goes to the gate of the mosfet with respect to each dac input voltage?
I actually want to know how to calculate the output voltage of opamp with each input?

I feel that the questions in your post #1 are essentially unanswered.

1. Yes the RC circuit will affect the rise time if the circuit is operated with a respective fast control voltage. Obviously the circuit is designed for static constant current, so it doesn't matter here. The purpose of the RC circuit is loop compensation. Without it, the feedback loop could become unstable due to the large MOSFET input capacitance. The right dimensioning depends on the OP speed, a 100 pF - 1 nF capacitor should be sufficient for general purpose OP with MHz bandwidth.

2. The current source gain is only set by the shunt resistor, Iout = Vin/Rshunt

Having a small shunt resistor value is reasonable to reduce the shunt power rating, if you want to control the circuit with a DAC, you'll probably keep a voltage divider in front of the OP.

 

[KlausST]

Hi,

Your calculation is correct.

The circuit is a current source. It's job is to generate a constant current.
Why do you bother about the opamp output voltage.

But I try to answer.
The shown circuit generates a voltage of about 73mV at the non inverting input.
The opamp regulates it's output voltage in a way that the voltage at the inverting input is 73mV also.
According the circuit the 73mV are across the shunt also. This generates a constant load current of 730mA.

The fet needs a gate-source voltage to become conductive. The higher the voltage, the lower the fet resistance the higher the load current.
If you want to know the gate-source voltage you need the 730mA and the expected drain source voltage. With both values you need to search the datasheet and find out gate source voltage. Maybe you find it in the chart, and sometimes you additionally need to interpolate.
Let's say you find out a gate source voltage of 2.5V (just as an example).
To know the Opamp output voltage you need to add the source-Gnd voltage (73mV) and the gate source voltage (2.5V).
It gives 2.573V.

But this value is not much of interest - at least for me. The circuit is a current regulator.
Replace the fet with another fet...and find out that the current is the same, but the opamp output voltage has changed.

Klaus

 

[jovin555]

Your explanation is clear.i understood the calculation.
Just one more doubt.
is the circuit reverse.Like.
First the non inverting input gets a voltage ->goes to inverting input->causes a current flow through the sense resistor->causes a voltage source at the gate for the required current.

 

[Audioguru]

You wanted to know how to calculate the output voltage of the opamp. It is impossible to calculate because Mosfets with the same part number have a range of gate to source voltages for them to produce the same current. The datasheet for the Mosfet states only the gate source voltage to barely turn it on (the threshold voltage which is also a range of voltages) and that 10V turns it on very well. The graphs in datasheets are for a "typical" device but yours might not be typical.

You understand it backwards. The opamp does not directly cause a current to flow in the sense resistor. The opamp creates a voltage at the gate of the Mosfet which turns on the Mosfet a little then the Mosfet causes a current to flow in the sense resistor.