I-V circuit question

I have a question that I hope someone has time to answer. In the output stage of the AD1853 demo board there is what is almost a standard opamp I-V converter combined with some filtering. However, the noninverting input is connected to VREF through a 402R resistor rather than to ground.

For some more background on this DAC, the output current is 3mA p-p with 1mA into the DAC's IREF. 1mA is achieved by applying 2.7V (VREF) to FILTR on the DAC and connecting IREF to ground by 2k7 resistor.

Is this arrangement intended to account for variations in VREF?

I would guess that the differential outputs of this circuit are 6.18V p-p. Is this correct?

[To avoid confusion, note that I have specified 'I' in before the current-source inputs and the Vout designators are mine. Is the polarity correct?]

Thanks very much,
Tyler
 

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paulb

Member
2001-06-01 4:53 pm
Calgary
I don't have a lot of background in I-V converters, but it seems to me that even though they are current outputs they still have a limit on their voltage swing, and my guess is that the DAC has a single +5V supply, so the output is centred around 2.7V for Vref. The op-amps run from a higher-voltage, bipolar supply so that they won't be handling voltages anywhere near their supply rails, but their reference has to be the same as the DAC.
Does that make sense?
 
resistor

Paul is completely right. The caps provide AC ground, but the reference output must have a DC bias because the DAC is single supply.

I am a little puzzled about the 400 Ohm resistor. It could be meant to provide the same impedance as on the other op - amp input. This will lower offset voltage in high input current amps such as the AD797 but it does not really make sense for the OP275 (which is junk, but the way).

The other idea would be to decouple left an right channels. Then the resistors would be perfectly suited. However, then a another capacitor to ground at each op-amp non-inverting input would make even more sense...
 
I/V input resistor

My guess is it is there to protect the input stage of the op amps when the power supplies are coming up at turn on. Op amps don't like large differential voltages on the input pins. During normal operation this voltage is very small. The resistor will limit the current into the input stage during fault or power up conditions.

H.H.