High Frequency opamp

Discussion in 'The Projects Forum' started by Nano001, Jan 30, 2010.

  1. Nano001

    Thread Starter Active Member

    Jan 12, 2010
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    Hi everyone. I ma looking for a single source high frequency opamp that can handle 10-13MHz clock input. Can anyone recommend a part that can satisfy this frequency input? Thanks for your help.
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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    Can you state the rest of the parameters? Gain, nature of the pulse, voltage limits?
     
  3. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Why do you need an op amp for a clock signal?
     
  4. Nano001

    Thread Starter Active Member

    Jan 12, 2010
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    The gain I have not thought about yet, maybe 10^4? The pulses come from a photodetector, in the 10-13MHz frequency range. Low voltage single source is the constraint for the supply.
     
  5. Ron H

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    Apr 14, 2005
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    You'll have to cascade several low gain stages, even with wideband op amps, to get that much gain.
    Do you have access to some good vendors?
     
  6. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Perhaps a ultra fast comparator like the LT1016 will do the trick.

    Greetings,
    Bertus
     
  7. Nano001

    Thread Starter Active Member

    Jan 12, 2010
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    Yeah I can order from anywhere. With a single opamp whats the max gain I can obtain? I would like to breadboard one opamp first to see what gain I am getting.
     
  8. Ron H

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    How much voltage (or current) are you expecting from the photodetector, or do you need to experiment to find out?
    If you are getting more than 10-20mV p-p, the comparator that bertus suggested is a good bet. Otherwise, you can use gain-bandwidth product (GBW) to estimate the maximum gain you can get at 13MHz. For example, a 130MHz GBW op amp would -3dB (0.707) down from the midband gain if you set the gain for 10. So, if you had 10mV in, you would get about 71mV out. If your GBW were 1300MHz (not a common part), you could get a gain of 100, giving you an output of 710mV (-3dB down from the gain at lower frequencies).
     
  9. Nano001

    Thread Starter Active Member

    Jan 12, 2010
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    I will have to experiment, but the current output will be somewhere in the low microamp range. Will this comparator Bertus suggested still do the trick? Once I know the product to use I will experimentally find out the gain. Is this correct? If so then this is something I can do. Someone also suggested the LF157.
     
  10. SgtWookie

    Expert

    Jul 17, 2007
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    The LT1016 would be a good choice.

    A comparator is designed to be operated in open-loop.
    If you try to use an opamp, you'll run into the gain bandwidth problem.

    The LF157 has a GBW of 20; not nearly fast enough, and besides that it's obsolete.
     
  11. Nano001

    Thread Starter Active Member

    Jan 12, 2010
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    Thanks SgtWookie. Even though this is marketed as a comparator, I can just add a feedback capacitor or resistor between the output and inverting input and it will operate in its linear region. Is this what you are saying?
     
  12. hgmjr

    Moderator

    Jan 28, 2005
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    There are not many photodetectors that can operate at frequencies of 10 to 13 MHz. Have you selected one already?

    hgmjr
     
  13. Nano001

    Thread Starter Active Member

    Jan 12, 2010
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    I have a PIN 10d silicon photodiode. Reversed biased at -10V should be able to generate the desired response.
     
  14. Nano001

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    Jan 12, 2010
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    Just bought the lt1016, thanks for the help.
     
  15. Ron H

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    Apr 14, 2005
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    I predict you will need a preamp. The 10D capacitance @-10V is a whopping 350pF.
    I designed a 3 transistor preamp that should give you enough signal amplitude to drive the comparator. Try it without a preamp, and if it works, let us know. If it doesn't, I can post my preamp.
     
  16. Nano001

    Thread Starter Active Member

    Jan 12, 2010
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    Thank Ron. While I wait to get the comparator in, I built a simple charge amplifier using a single supply LM324 and the PIN photodiode, however I am not sure if I am using this amp correctly. I attached my schematic. It consists of a simple feedback resistor (1K) and a DC blocking capacitor (guessed at the value, what do you think?) to extract only the LED pulse. I used an optical power meter (Newport) to determine the power generated by the PIN photodiode when a red LED is incident on the surface which was 61uW (94uW minus 33uW background power from the white light). Looking at the spectrum of the PIN diode at 650nm the diode should be supplying 26.5uA.

    With the LED blinking at a frequency of about 10Hz, output of the circuit with R1= 1K gives me a square wave with peaks of about +/- 20mV. When I change R1 to 100k the peaks drop to about +/- 2mV. Since the opamp supply is wired from 5V to ground, shouldn't the output not swing below ground? Also, why do I have no gain?

    I feel that I am not understanding this circuit correctly. Is it operating in the right way? Also, I will need to have an integrating capacitor in the feedback loop. How do I calculate this value?

    Also, this circuit has to be single supply, so I will have to think about another way of generating the desired photodiode frequency.
     
  17. Audioguru

    New Member

    Dec 20, 2007
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    The LM324 or any other opamp needs a dual-polarity supply including a negative supply in order for its output to go to a negative voltage like in your LM324 circuit.
     
  18. Ron H

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    Apr 14, 2005
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    1. GBW of the LM324 is way too low for 13MHz,
    2. Capacitance will be huge with zero bias (which you have).

    You need a common-base transistor preamp.
     
  19. Nano001

    Thread Starter Active Member

    Jan 12, 2010
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    Thanks for the help. I understand that the LM324 cannot handle the 10MHz frequency, I want to breadboard a test circuit with the 324 at low frequencies while I wait so I can understand how the circuit works. I thought the 324 is a single supply opamp which only requires a single positive supply? Isn't this how single supply opamps work?
     
  20. Ron H

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    Audioguru was, I believe, pointing out that, given the direction of current you indicated, the output voltage would have nowhere to go, since it is already at ground when the diode current is zero. I think you drew the current pointing the wrong way, but your diode is connected correctly. My understanding is that the photocurrent will flow from cathode to anode.See fig. 2-4(b), here.
     
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