The input and VCM of LTC6600-20

Hi imageradar,

Regarding your questions about LTC6600-20 operation, here are my thoughts:

1. Input Coupling: The inputs can be DC or AC coupled. Only thing is that the applied differential input CM voltage must be such that the device input pins (IN-, and IN+) do not dip below ground when operating from a single 5V supply, when using DC coupling. With AC coupling, you don't need to worry about that.

2. Output CM Voltage Range: The main issue with operating the output CM voltage at less than recommended values is the increase in distortion that you see in the attached plot from the datasheet:

One way you can improve your case of output CM (pin 2) = 0.9V is run VMID (pin 7) at 0.9V as well (instead of leaving it floating to mid-supply). This way, you'd be operating at x-axis (pin 2 relative to pin 7) = 0V in the plot above.

I've put this in the LTspice example circuit below. You can experiment with this circuit for your exact set of conditions:

Here is the simulation file for reference:

LTC6600-20 Example Single Supply EZ 2_28_22.asc

Regards,

Hooman

Thanks 

And i have read the datasheet, i find that the VMID is the V+/2( my V+ = 5V，V- = 0V)

So if i connect 0.9V to VMID, it means the sink current is nearly (5-0.9)/11k =  0.37mA

So how can check whether this will affect the ADC-AD9628. i do not find any specification of sink current in the datasheet of AD9628

Thanks.

Hi imageradar,

Regarding your concern about AD9628 ADC VCM pin being able to supply the current needed to run LTC6600-20 VMID at 0.9V:

That can be easily remedied if you add a shunt resistor to VMID pin in order to reduce the AD9628 VCM current to near 0uA. I've shown that in the image below where Rshunt (3.1k) reduces the VCM (V3) pin current (through Rsense) to uA:

Hope this helps.

Regards,

Hooman