Hello All,

Sorry if I put this question onto incorrect sub-forum. Could you please help me with understanding attached schematic. Is it a digitally controlled current source or something completely different?? I have doubts what is a function of this circuit.

Many thanks

 

Hello All,

Sorry if I put this question onto incorrect sub-forum. Could you please help me with understanding attached schematic. Is it a digitally controlled current source or something completely different?? I have doubts what is a function of this circuit.

Many thanksView attachment 152746

I'm sorry I can't understand it well enough to help, but I have to say:

[rant]
It drives me crazy when schematics don't clearly indicate when crossing wires connect. I see some crosses that must connect, and others that I'm pretty sure shouldn't connect, but I don't see dots or jumps anywhere.
[/rant]

 

That circuit would be a constant-current output controlled by the DAC if there was feedback from the transistor emitter to the op amp pin 2.

As it is shown, the op amp is DC open-loop, so will saturate at the positive rail.

 

Take a look at DAC08 datasheet, correct use of OpAmp conversion
to V is shown in application circuits.

http://www.analog.com/media/en/technical-documentation/data-sheets/DAC08.pdf


Regards, Dana.

 

Where did you get that circuit?
The normal circuit will be similar like this or this:
8 bits digital inputs → 8 bits DAC(current out Iout) → I to V converter
So the DAC is a constant current output and the last output should be a current to voltage converter, but in your circuit doesn't looks like a right connection.

 

Thank you for all answers. This circuit comes from the board which creates 2 very stable reference voltage (not depicted in picture which I attached) and this voltage FDRV which doesn't have to be very stable.
According to measurements this circuit works as I to U converter (here voltage on X10 FDRV is R8*Iout_DAC), but why there is V2 and capacitor C3 for AC feedback - this I don't understand that's why I asked. Could be a kind of protection for DAC (overcurrent protection) or is it something else? - this is not clear for me.

 

As the links that I posted in #5, the circuit only used one op amp and the Iout of DAC was connected to the Vin(-) of op amp, but in your circuit that the Iout was connected to the Vin(+) of op amp, because the output is the same phase of Iout, so added the V2 to inverts to as the phase inverter, and I think to added the R9 and C3 for the voltage changing slowly(smoothly) reason when the Iout current decreasing or increasing during the current converting to voltage.

 

As the links that I posted in #5 that the circuit only used one op amp and the Iout of DAC was connected to the Vin(-) of op amp, but in your circuit that the Iout was connected to the Vin(+) of op amp, because the output is the same phase of Iout, so added the V2 to inverted to as the phase inverting, and I think to added the R9 and C3 for the voltage changing slowly(smoothly) reason when the Iout current decreasing or increasing during the current converting to voltage.

In your opinion, for what reason we need invert a phase using V2?? Can we say that R9 and C3 forming low pass filter?

 

In your opinion, for what reason we need invert a phase using V2?

For the circuits that I linked were only used one op amp that the output is the inverting of input, in your circuit that the output of op amp is the same phase(positive output) of the input , so it needs the V2 as inverter to inverting it, is that clear?

Can we say that R9 and C3 forming low pass filter?

I don't know the input info of the DAC and its operation frequency, so I only guessed that the R9 and C3 is working on a very low frequency, probably lower than 10 Hz, and since the working frequency of the R9 and C3 is so low, so the R9 and C3 doesn't like as filter, and they just like the charging pump to slow down the voltage changing speed during the value of DAC is changing, if what I thought was right then they were used to make the voltage changing looks more linear, and it won't looks like ladder(or sawtooth).

 

For the circuits that I linked were only used one op amp that the output is the inverting of input, in your circuit that the output of op amp is the same phase(positive output) of the input , so it needs the V2 as inverter to inverting it, is that clear?

Yes, this is clear that circuit from your links inverts signal, but less assume we removed R3, R6, R2 and V2. In my opinion still DC signal FDRV would be the same sign (positive, equals R8*I_DAC), that's I don't exactly understand the role of V2.

I don't know the input info of the DAC and its operation frequency, so I only guessed that the R9 and C3 is working on a very low frequency, probably lower than 10 Hz, and since the working frequency of the R9 and C3 is so low, so the R9 and C3 doesn't like as filter, and they just like the charging pump to slow down the voltage changing speed during the value of DAC is changing, if what I thought was right then they were used to make the voltage changing looks more linear, and it won't looks like ladder(or sawtooth).

Thanks for this clarification.

 

Yes, this is clear that circuit from your links inverts signal, but less assume we removed R3, R6, R2 and V2. In my opinion still DC signal FDRV would be the same sign (positive, equals R8*I_DAC), that's I don't exactly understand the role of V2.

Sorry for duplicating message, but I'm afraid that info could be omitted in my previous message.