Bipolar vs unipolar amplifiers and large slew rate change?

Discussion in 'General Electronics Chat' started by flu, Aug 12, 2009.

  1. flu

    Thread Starter New Member

    Aug 12, 2009
    1
    0
    Hi, I'm using the APEX PA84S power opamp in a simple inverting configuration with a gain of 30 (R2=300kOhm, R1=10kOhm). I used it successfully for driving a MEMS device (*very* small capacitive load) from 0-150V in 1 microsecond and used this opamp specifically for its high spec'd slew rate (~200 V/us). The rails were biased at + and - 150V using an Acopian dual regulated supply. I wanted to use it again with the same power supply, but this time at 0 V at the lower rail and +300 V at the upper rail. This is pretty easy to hook up with the power supply by grounding the minus terminal of second supply and tying the plus terminal to the minus terminal of the first supply; with the plus terminal of the first supply being +Vs. Overall, things work, however, the slew rate with the new rails is now around 15 V/us, about 10 times slower. Playing with the compensation resistor and capacitor values did not seem to change anything (which was somewhat a surprise), nor did changing out the R1 and R2 to have different values (maintaining and varying the gain). Does anyone know why the slew rate suddenly dropped with this configuration and how to bring it back up if possible? I was hoping this would be an easy problem. The power supply can source 100 mA according to its data sheet which should be more than enough for the opamp (according to its data sheet). The voltage terminals at the power supply have been checked using a voltmeter and offset trimpot specified by the datasheet (50 kOhm) is being used. Bringing back the power supply to a bipolar configuration (+/- 150V) showed that the opamp was still good so the opamp did not die. Any suggestions would be appreciated!
     
  2. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    Please post a schematic. Most likely there is a difference with the way the load is grounded or with the input voltages. For example, suppose you had the negative input terminal still tied to ground. That would put the input common mode voltage outside the recommended range.
     
  3. kkazem

    Active Member

    Jul 23, 2009
    160
    26
    HI,

    It could also be that you're power supply has very little output capacitance and you should have at least a 1uF cap in parallel with a 0.1uF cap in parallel with a 0.001uF cap directly across the op-amps supply terminals. Don't forget to make sure the caps can handle the voltage range of 300VDC and you're cap leads to the IC should be very short, especially for the two smaller value caps-less than 1/4" for each lead. Also, if you have long wires in the unipolar circuit whereas you had short hookup wires on the bipolar, the extra lead inductance could be your problem.

    1) Make sure the IC Psupply leads are well bypassed right at the IC leads with the shortest possible cap leads using multiple caps of a uF to several uF for low freq, 0.1uF for medium freq and 0.01uF or 0.001uF for high freq. Your very high slew rate is equivalent to a high frequency and the IC must be extremely well bypassed to obtain full slew rate and full bandwidth.

    2) Keep all sensitive IC leads (especially the inputs) as short as possible with regard to the R's and C's that connect to them. The R & C leads that go to the output of the op-amp can be longer. The load you're driving with the hi-slew-rate IC should be connected with either very short leads or with a transmission-line like coax or twisted pair.

    3) When troubleshooting, use a scope and not a voltmeter as you cannot tell if the supply voltage is dropping out using a V.M.

    4) Don't use the scope probe ground strap. Instead, remove the spring loaded tip clip and wrap some #18 AWG bus wire around the ground section near the probe tip positive lead. THen bend the bus wire to form a ground lead that's parallel to the short probe tip and clip the ground lead to roughly the same length as the positive pin of the probe. The two leads should be about 1/4" apart. You can bend this ground lead out a bit if needed so that the probe tip pin can touch the VCC pin at the same time as the newly formed probe short ground lead touches the IC ground (VEE or VSS) pin. WHen looking at very fast signals, the scope ground strap leads of 3" to 6" or more will completley distort the true signal, which is why it's critical to use the absolute shortest possible probe positive and ground leads, or you can use coax with a BNC connector instead of a scope probe, but now it will be a 1:1 and not a 10:1, and you need to keep the other end leads as short as possible, not more than 1/2" apart if you can help it.

    Good luck,
    Kamran Kazem
    kkazem
     
  4. rjenkins

    AAC Fanatic!

    Nov 6, 2005
    1,015
    69
    Is your output wiring and load arrangement, including load ground, still the same?

    The only things I can think of with the info you provide are, it's seeing a higher capacitance in the load circuit so the amp's current limit is limiting the slew rate,
    or,
    resistance / inductance in some wiring common to the input and load circuit is causing a voltage drop that affects the input circuit while the output is supplying current.

    As kkazem says, decoupling is critical but grounding is equally so - are you using a 'star point' ground? That's possibly the best & simplest approach for something like this.

    His mention of using a BNC connector for the 'scope is also a good idea, however you *must* still use the x10 scope probe to avoid loading the signal with the capacitance of an extra connecting cable.

    All decent scope probes come with a short push-on metal adapter sleeve which allows the probe to fit directly into a BNC socket and still maintain the low capacitance x10 input attenuation. The socket should be connected directly against the IC output, not on a trailing lead.

    For reference, a x1 scope probe (or using a bnc to bnc patch lead) is a waste of time for anything much above audio frequencies. The cable is quite high capacitance and the normal 'scope input impedance is 1M Ohm, so a complete mismatch to the coax.

    Fast edges cause all sort of ringing and resonance effects on the waveform and the circuit you are testing.

    (Unless all cables & connections are impedance matched and the 'scope has a 50 Ohm termination switch, eg. for RF work).
     
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    I suspect Steve had the answer. Flu, did you leave the noninverting input connected to GND? If so, the inputs are outside the common-mode range.
     
  6. rjenkins

    AAC Fanatic!

    Nov 6, 2005
    1,015
    69
    According to the data sheet, the common-mode range is +/-VS -8.5V

    It's designed to work with both single & bipolar supplies and the common mode range includes negative supply / single supply 0V.
     
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    I interpret that spec to mean that the CM range goes from 8.5 above the negative rail to 8.5 V below the positive rail. Look at the output voltage swing spec and you'll see why I came to that concluusion.
    Also, the schematic supports that conclusion, as I analyze it.
     
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