Suggested Input Protection AD8253

I am currently designing a breakout board for this amp and would like some suggestions on input protection.

My input is a piezo sensor(reverse speaker, home made, non powered) that puts out about 70mV full swing and no clue on the amps.

The data sheet says to use series resistors and low leakage diode clamps. I see the resistor examples but don't see any diode hook up examples.

My input voltage is -15,+15, any suggestion on where to start with resistor values and diode values, and placements.

Below is my current schematic, any tips or suggestions would be appreciated.

Thanks

  • 0
    •  Analog Employees 
    on Aug 21, 2012 7:08 PM

    Hello josiah47,

    For the resistor values, there is a trade-off between protection and performance. It really depends how much protection you think you need. Input resistors add additional noise and offset to the circuit. For example, 500 ohms at each input generates 4nV/rtHz of johnson noise, 3.5nV/rtHz from the AD8253 current noise, and 20uV input offset voltage from the AD8253 input offset current. This is going to give you 6 mA * 500 ohms = 3V beyond either supply of input protection. It also increases your G=1000 noise from 10nV/rtHz to 11.3nV/rtHz.

    If you also use diode clamps, we sometimes recommend something like the following configuration:

    If BAV199 are used, for example, they can handle 160mA continuous forward current. With a 500 ohm resistor, this is more than 80V. But the power dissipation in the resistor will limit you before that. But if you only need 10V beyond the supply, you could use a 0.5W 200 ohm resistor and less than 50mA would flow at 10V beyond the rail. Also a 33 ohm resistor is shown here to ensure that the current flows through the external diode and not the internal one. It may need to be slightly larger for a low leakage diode, which typically have higher forward voltages. To continue the previous example, Vf at 50mA for BAV199 is 1100mV. To be safe, you might want about 500mV drop across that resistor at 6mA to ensure that there is not excessive current in the AD8253 inputs. This would require 83 ohms.

    The schematic looks good as far as I can tell. Make sure that the input bias current has a ground return path when you hook it up. This probably involves grounding one input and possibly using a fairly high value shunt resistor between the inputs if your sensor is high impedance at dc.

    Best regards,

    Scott

  • I have updated my circuit to include the diode clamps, See any thing wrong with it?

    I fried my -15/+15 volt supply NMH0515SC

    http://www.murata-ps.com/data/power/ncl/kdc_nmh.pdf

    wondering maybe if its drawing to much current some how? no sensors where hooked up.

  • You need a 1meg-10meg resistor on one side to ground.  See fig 56 on AD8253 d.s.

    Harry

  • Wouldn't I also need some caps there as well, as shown in fig 56?

    Would not having this causing me power supply to blow?

  • 0
    •  Analog Employees 
    on Jan 17, 2013 7:44 PM

    Hello josiah47,

    The capacitors are only required if you want to AC couple the sensor. While the circuit at the bottom of figure 56 is meant to show a dc bias return path for this case, in general does not provide very good CMRR. If you can avoid the use of capacitors, you'll get much better performance.

    I don't know what is the mechanism that blew your DC to DC converter, but I do know that those can't sink current. When you overload the inputs and dump current into your voltage rail, you will be raising that voltage. I don't know which one is going to die first as the supply voltage goes beyond an absolute max rating (the amplifier or the supply) but you will need to clamp this voltage with a zener to be safe, otherwise your protection circuit isn't complete.

    Regards,

    Gustavo