INA826: How to calculate input impedance when adding a bidirectional TVS between differential Inputs

Hello Noam,

The INA826 differential input impedance (zid) is extremely high, typically around 20 GΩ in parallel with 1 pF. Shunting the differential inputs with the impedance of a high-power TVS will result in the parallel combination of the two; therefore, the input impedance to the circuit will be lower than that of the INA826 alone.

You would need know the off state impedance of the TVS to determine the overall input impedance. That may be difficult to find because it isn't specified in the SMBJ15A TVS datasheet. It is a bidirectional TVS that looks somewhat like two back-to-back zener diodes. The primary components of the TVS impedance will be some high resistance associated with the reverse leakage current of the diode junctions, and the capacitance of the junction. Note that the reverse leakage will be highly temperature dependent and will likely vary much over different production runs; both making it is difficult to nail down a specific value. I do expect the TVS will be the dominant player in determining the input impedance to the amplifier circuit.

You might consider contacting the TVS manufacturer regarding the TVS characteristics. They may be able to provide more information about the device when it is in the passive, off state.

Regards, Thomas

Precision Amplifiers Applications Engineering

Thomas Hi,

Thank you for the answer. That bring me to a second question I'll appreciate your help.

I am using an analog MUX to select the analog input to the INA826.

I am looking for a way to identify if the selected analog input of the MUX is disconnected from some reason. One way that was suggested is to add a 10 M OHM pull-up on the "+" input of the INA826 (See schematic). The INA826 "-" is connected to GND.

1. How it going to effect the input impedance?

2. How it will effect the measurement accuracy?

3. Is there a better way to identify a disconnected channel?

BR

Noam

4863.Analog In.pdf

Hey Noam,

I’ve been looking at this with Thomas and have some suggestions/comments.

1. Connecting a 10 MOhm pull-up resistor will shunt the source impedance of your mux. Thus, the effective impedance will be the parallel combination of the mux source resistance and the 10M resistor (R943).

2. The pull-up resistor will present some error in the Vout measurement due to the voltage divider of the 10M (R943) and the 249Ohm (R940) resistor. This will increase the offset voltage (Vos) between the inverting and non-inverting inputs of the INA826, which will be gained up and reflected on your output.

3. There may be better options for your application although it is up to you on how you want to interpret a disconnection.

With your current circuit, if a disconnection occurs on the analog mux pin and the non-inverting input of the amp is pulled up to 15V, then Vout will saturate close to the positive rail. Consider using our Vcm vs. Vout calculator here to see expected outputs: www.ti.com/.../ina-cmv-calc.

Instrumentation amplifiers usually need input bias current paths to ground, basically pull-down resistors. These prevent one of the inputs to float and saturate the output of the op amp. See section 8.3.6 of the INA826 datasheet. If you insert these resistors and remove the 10M pull-up resistor, your Vout will move to ground (or the reference) in the case of a mux disconnection.