Notices
Results 1 to 3 of 3

Thread: Hydrophone Amplifier Circuit

  1. #1 Hydrophone Amplifier Circuit 
    New Member
    Join Date
    May 2007
    Posts
    2
    Hi,

    This is a circuit I'm working on for a hydrophone. I have blueprints for a circuit using op-amps but was wondering if the circuit below would be any good?

    It is a bias applied to a transistor using a voltage divider. I was wondering how to apply the dc bias to the voltage divider on the left without affecting the incoming signal. Could I just couple the top junction of the voltage divider to a dc voltage? Would that give me a quiescent point dependant on the voltage divider that would simply show a rise or drop from half the applied dc voltage for the incoming signals, or am I barking up the wrong tree here?

    The bit on the right is just a low pass filter. I'm listening for low frequency sound.



    Thanks, kindly in advance,

    Chris


    Reply With Quote  
     

  2.  
     

  3. #2  
    Forum Freshman wonkothesane's Avatar
    Join Date
    Feb 2007
    Location
    Brisbane, Australia
    Posts
    47
    I think it is generally better to use an op-amp voltage-follower (buffer) between the instrument and amplifier/filter so as not to load the hydrophone perhaps with a series capacitor to decouple any DC. This means you need a power supply though, is that what you were wanting to avoid? You can still bias the op-amp input with a resistor divider.


    Fry me a kipper skipper, I'll be back for breakfast!
    Reply With Quote  
     

  4. #3  
    New Member
    Join Date
    May 2007
    Posts
    2
    I understand exactly what you mean. Thank you for the reply. Perhaps in a pm also you could tell me where your avatar comes from it looks rather like an amiga game to me

    I am to be honest trying to avoid a power supply which involves anything more than a few batteries.

    Perhaps I could use another transistor as an emitter/follower between hydrophone and voltage divider to lower the impedance of the signal on the voltage divider side and yet get the signal input impedance high so it didn't take much current to drive it?

    I wonder how that is possible though given how would it cope with signals close to or below ground. I can't remember if an emitter/follower easily allows negative signals?

    I'd prefer that if it is possible as I don't yet understand op amp that well

    Please if my explanation doesn't make sense tell me and I will post another picture thanks. Also please dont hesitate to tell me if I'm talking nonsense as I'm new to electronics.
    Reply With Quote  
     

Bookmarks
Bookmarks
Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •