AD8610 supply current

Hi!

I am trying to use an AD8610 opamp in an active loop filter for a PLL, but after applying a 25V supply voltage my power supply indicates 25mA of supply current instead of the 3-4mA described in the datasheet. How is it possible? The PLL was either off or with the charge pump configured as a three-state output, so there wasn't any input and I checked the board, there are no faults, short-circuits etc. Is it normal operation for this opamp?

Thanks,
Bence

  • Hi Bence,

    There may be instances that a certain component is pulling too much current from the supply or from the device. A schematic would greatly help us in analyzing the cause of the problem.

    Regards,

    Jino

  • Dear Jino,

    The enclosed schematic from ADISimPLL is my schematic. The negative supply is grounded, the positive supply is 25V.

    Thanks!

    Bence

  • Hi Bence,

    May I know what's is the source of your schematic, could you attach the ADIsimPLL please? There is only one thing I see that causes the issue, that is the CP pin of ADF410x is current out, 5 mADC max to be specific in the current configuration. Also, there is no DC path for that current on the AD8610 side being AC-coupled thru C1 and C2. In this case, the AD8610 output may have been saturated or worse damaged. Imagine a 5 mA multiplied by the input resistance of AD8610, let's say 1 Mohms, that would yield a 5kV at the input of the device.

    Regards,

    Jino

  • Dear Jino,

    I haven't even thought of that, but the ADISimPLL gave this schematic and also I have never seen an active loop filter schematic where a DC path was being made before the filter. I would think that the current loads the capacitors, effectively transforming the current into voltage which then will be amplified, as the opamp presents a bigger impedance than the first two capacitors and resistor of the loop filter.

    I would also like to add that this big supply current was present even when I first gave supply voltage to the amplifier and not to the PLL, so no current could have been present on the CP output - and this happened with two amplifiers on two PLL boards in the past weeks.

  • Dear Jino,

    I have an update. I made some debugging measurements on my PLL and the opamp is HOT, especially on the side of the supply. I really have no idea why, because the supply current was big even when the PLL was never ever turned on. And I also have never seen a loop filter schematic where there would have been a DC path to the ground at the input, however, if it is needed, how would you make it and wouldn't it interfere with the filter characteristics? Would a large (few hundred kiloohms) resistor from the CP output to the ground (parallel with the first integrating capacitor) solve this problem provided I replace the amplifier with a new one? Or would a series resistor be better for opamp current limiting? That woud seriously complicate the filter. Also it would add a lot of thermal noise but it is just a prototype and a peripheral circuit for a microwave VCO I've been working on so noise is not my primary concern now, just to make it work somehow. Or maybe using the opamp as an inverting amplifier, then the DC path would be through the feedback resistor. What would you suggest?

    Thank you very much,

    Bence