Electronics > Beginners
op-amp, TL081 - 400ST Ultrasonics, Headbanging...
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rdbanks:
Hi guys, I'll keep this short, its a bit of a noob question however i'm trying to get my head around this to explain it the most accurately...  :palm:

I have a simple TL081 op-amp acting as an Astable multivibrator +/- 12V supplies - this 40kHz output is then fed into another TL081 set up as a non-inverting unity buffer which feeds an ultrasonic transducer (looking for 40kHz (receiver tuned) transmitting).

Now, my question....

   I understand, without the buffer, the transducer would 'load' the first op-amp however I cant explain this the way I understand it... Simply put, as i recall, the transducer acts as a capacitance in the op-amp circuit, if that right? I tried measuring DC current through the transducer and it came up at around 9.1uA - am I measuring this correctly - AC current is around 3mA. AC or DC? I was happy with my understanding until now, perhaps its exam prep and i'm thinking too much...  |O |O |O

I think my issue is i'm confusing my basic understanding due to the circuit involving op-amps, I've confused a lot with BJTs lately and was wondering if someone could elaberate on my question which was:

Is it more important to use a buffer to 'remove the load' or 'remove the capacitance'.

This isn't a coursework/exam question, just more of a understanding exercise.

I've found this post hard to explain, I hope you smarter bunch can shed some light... Thanks. Ryan!

// the clean sq wave is Vout of the first TL081, with a buffer after connecting to the transducer.
// the deformed wave, is Vout of the first TL081 with transducer connected directly - this is the same wave form as seen at Vout2 (second TL081 (buffer)) with the transducer connected to that.
//Frequency reading is different, before and after tuning, it doesnt really matter for the question).
mikerj:

--- Quote from: rdbanks on May 07, 2018, 11:40:15 am ---Is it more important to use a buffer to 'remove the load' or 'remove the capacitance'.

--- End quote ---

When you are dealing with AC (with or without a DC offset), capacitance IS a load.
rdbanks:
I see! I've just taken 5mins away from analog and its all clicking!
Thanks Mikerj!
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