hello ,
my interest is to design a ADC without DAC . So i continued my design with integration type ADC.
I will explain this design now. its quite simple . it has a comparator with one input being the analog value to be converted(given to the inverting pin). the other input being the voltage value from the integrator(given to the non inverting pin). And there is a counter present (up down counter) which keeps counting irrespective of the integrator and the comparator operation. The counter is connected to ide a shift register.
when the comparator output crosses , the zero line that is to +Vsat, it triggers a MOSFET and thus the integrator finds a path to discharge to zero. And not only it triggers the MOSFET but also it resets the counter and at the same time clocks the shift register so that the counter value is available for the user as the digital value(output)
The idea behind the design is that the clock for the counter is dependent on the resistor and the capacitor values used in the integrator. When the integrator gets the output as 1volt, the counter reads 000000000001(a 12 bit counter used). The clock for the counter is set in such a way for this design.
This is the idea.
I worked with proteus for designing this circuit.
but i could get the correct output. I couldnt fix the problem.
Can anyone help in this design please ???
I believe he is using R1 as an approximate current source to charge C1. When the voltage on C1 equals V3 the counter will give a digital value proportional to the voltage V3 at which point C1 is discharged and the process repeated. It would be better to use an actual current source to charge C1.
charliesixpack:
I believe he is using R1 as an approximate current source to charge C1. When the voltage on C1 equals V3 the counter will give a digital value proportional to the voltage V3 at which point C1 is discharged and the process repeated. It would be better to use an actual current source to charge C1.
okay , i will try using a current source.
but as of now , whenever this circuit is simulated , i could get the count value as 111111111111 . i could fix the problem . please help.
shriganesh:
And please suggest any way to clock the data into the shift register. why is this not a good idea ?
I don't know what type of shift register you are using but it looks as if you may need to connect to the 'Load' pin. Do any of the other pins of the shift register need to be connected, such as OE? Even so, when the output of your comparator goes high the counter will reset at exactly the same time (more-or-less) that you want its outputs to load into the shift register.
shriganesh:
but as of now , whenever this circuit is simulated , i could get the count value as 111111111111 . i could fix the problem . please help.
This is a simulation you are doing? You need to set the counter value to 0 when the voltage on C1 is 0 volts. Doing a quick calculation, it should take 200 microseconds to charge C1 to 1 volt. I am not familiar with the Proteus simulator but I assume that it must be good at doing mixed mode analog/digital simulation to do that length simulation.
why do you say that its not a integrator ? i really dont understand.please explain
If you integrate a constant, that is a DC voltage, you will get a function in X ( in this case time against voltage ). The output of an integrator will be a linear slope, or constantly increasing voltage..
What you have is a voltage that rises not linear but exponentially. If you want an integrator you have to charge a capacitor with a constant current not a constant voltage. The simplest way to do this is to have an op-amp with a capacitor connected between the output and the -ve input. The constant of integration should also be applied to the -ve input.
This is basically a single ramp A/D, have you seen the dual ramp A/D design? That does not need any precision on the clock, just a stable clock.
You do not want an opamp integrator. The opamp integrator will integrate the voltage on the input. You want a linear integrator that integrates at a constant rate regardless of the input voltage. You need a constant current and a linear capacitance.
Best to use a pnp current mirror instead of R1 to charge C1.
charliesixpack:
You do not want an opamp integrator. The opamp integrator will integrate the voltage on the input. You want a linear integrator that integrates at a constant rate regardless of the input voltage. You need a constant current and a linear capacitance.
Best to use a pnp current mirror instead of R1 to charge C1.
Either a classic operational amplifier integrator or a constant current source charging of a capacitor could be used to create the required voltage ramp. The constant current source approach is probably better in this instance, especially considering the need for a MOSFET to discharge the capacitor.
hello everyone , thanks for all your views .
I made minute changes to my design and it gives some output , a digital value. But the digital value which i get i not correct
And how to fix the resolution of the ADC because here i use 10 bit counter . how should i calculate the resolution for this design ??
And i have attached the image of the new design.
please help me with this circuit.
The voltage between the two inputs of your integrator 741 operational amplifier will be extremely small. That means the non-inverting input of your comparator will be very close to ground . . . . . . so your circuit will not work.
I suggest you connect R2 to your -12V supply and, to compensate for the extra voltage, increase its value to 12KΩ. That way you will not need a 2V supply (V5) and the output of your 741 operational amplifier will ramp linearly with increasing positive voltage. Also reverse the polarity of C2 or, preferably, make it non-electrolytic. Connect the 741 output to the non-inverting input of your comparator so the comparator gets the voltage ramp required. I believe you can put your MOSFET across C2 with its source connected to the inverting input of the operational amplifier, but I am slightly uncertain whether this would work satisfactorily.
If you build this circuit, I would like to see the addition of a small diode across C2 to prevent the possibility of the output of your operational amplifier going negative by more than 0.7V while you are messing about with the circuit.
Note your present circuit diagram shows no connection to the gate of the MOSFET.
What device are you using for your shift register?
EDIT: I suggest you use your Proteus software to check intermediate points in your circuit, such as the voltage ramp, function as expected (assuming Proteus can do that!).
Archibald:
The voltage between the two inputs of your integrator 741 operational amplifier will be extremely small. That means the non-inverting input of your comparator will be very close to ground . . . . . . so your circuit will not work.
I suggest you connect R2 to your -12V supply and, to compensate for the extra voltage, increase its value to 12KΩ. That way you will not need a 2V supply (V5) and the output of your 741 operational amplifier will ramp linearly with increasing positive voltage. Also reverse the polarity of C2 or, preferably, make it non-electrolytic. Connect the 741 output to the non-inverting input of your comparator so the comparator gets the voltage ramp required. I believe you can put your MOSFET across C2 with its source connected to the inverting input of the operational amplifier, but I am slightly uncertain whether this would work satisfactorily.
Actually , if connect the capacitor to the non inverting input , wont it be positive feedback ?? we need negative feedback to be present right ??
shriganesh:
Actually , if connect the capacitor to the non inverting input , wont it be positive feedback ?? we need negative feedback to be present right ??
I was not suggesting you connect the capacitor to the non-inverting input. Because you need a voltage ramp from the output of the operation amplifier that starts at zero volts and has increasing positive voltage, the positive side of your electrolytic capacitor (if it's electrolytic) will need to be connected to the output of the operational amplifier and its negative side connected to the inverting input of the operational amplifier.
Note the whole integrator circuit is inverting, so you need to connect a negative voltage to its input to get a ramp that has increasing positive voltage from its output. That's why I suggested you connect R2 to a negative voltage such as -12V.
