# Why is the gain of the op amp in a ultrasound transmitter important?

Discussion in 'Homework Help' started by Denny1234, Apr 19, 2009.

1. ### Denny1234 Thread Starter Member

Feb 17, 2008
27
0
Hi

I know that you have 2 stage op amps and thats so you can get a high gain (30 x 30 - or whatever the gain of each op amp is) but why does the gain have to be so high? If you have a comparator comparing these resultant voltage cant it just be compared with another small voltage (from a pic or whatever) and the result be obtained to give a view of distance.

I know its a such a noob question but i dont see why? I dont know what kind of decisions have to be made and why when choosing the gain of the op amp in the ultrasound circuit?

Thanks for any help.

2. ### Audioguru New Member

Dec 20, 2007
9,411
896
You didn't post your ultrasonic transmitter circuit so I don't know why it needs a gain of 900. An ultrasonic receiver needs to have a high gain.

An ultrasonic transmitter does not have a comparator.

An opamp has a gain of 200,000 at DC and low frequencies. It rolls-off its gain at high frequencies so that it doesn't oscillate when it has negative feedback added and its phase-shift causes the feedback to be positive at a high frequency.

A TL07x opamp has a max gain of about 50 at 40khz.

3. ### Denny1234 Thread Starter Member

Feb 17, 2008
27
0
Oh hi, sorry yeh I meant to say receiver, thnx for your help.

Im refering to this circuit (below) trying to understand how it works because it seems pretty straightfoward. Its just im still unsure where it says

Each gain stage is set to 24 for a total gain of 576-ish. This is close the 25 maximum gain available using the LM1458

I dont understand why you would want a gain of 576ish? How you can come to that kind of decision?

From what you say does this mean that its chosen because at this gain it won't oscillate and amplify noise as much? But why not chose an even lower gain? You wont amplify noise and as its just going into a comparator the output you get is just a comparison of that and the MAX232(whatever it is) signal - if both voltages are small than you get the same result - a comparative signal that shows how far away the object is.

http://www.robot-electronics.co.uk/htm/srf1.shtml

4. ### Audioguru New Member

Dec 20, 2007
9,411
896
I couldn't find a datasheet for the N1081 ultrasonic receiver transducer.
If it is piezo like most of them then its impedance is about 10k ohms and is loaded down by the 2.2k ohms input impedance of the first opamp circuit. The gain of the first opamp is 36k/(2.2k + 10k)= 3.0. The gain of the second opamp is 36k/2.2k= 16.4. The total gain is only 49.2.

5. ### Denny1234 Thread Starter Member

Feb 17, 2008
27
0
Thanks a lot mate, I couldnt fathom how they got those values seems they are wrong.

I like the look of that design, its easy to make and straighfoward and so i have chosen it for my project.

Just one thing. I have inputted my ultrasonic wave as shown in the attachment as 40khz (obviously) and 2mvp Amplitude - im sure I read that ultrasonic waves in these type of rangefinders vary in level from 1mv to 1 volt on entering the transducer (cant remember where though). So with my design on entering 2mvp into my2 stage op amp i obtain an output value of 850mV ptp as shown and 1 volt gets me about 5 or 6 volts output does this seem the right kind of voltage for my 2 stage op amp output?

Also just to ask, i missed the impedance of the transducer till you pointed it out for me, is there anything else I have omitted from my design that I should have taken in to account?

(I have calculated the gan badnwidth for my op amps and it gives me a max gain of 35 - im below that so im ok)

Appreciate the help, Im on my own with this - with very little knowledge.

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6. ### Denny1234 Thread Starter Member

Feb 17, 2008
27
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Also sorry for asking so much.

I was wondering what design decisions are made with C6, C9 and R7 (in terms of values and why they are there!).

Im pretty sure i know C9, its just to remove residual DC (which i think they mention)

The way to calculate C9 i assume would be just to say....

1 / (2π f RIN)
with f being 40000 and Rin being 10000ohms as you told me?

Not sure how C6 and R7 help the comparator and how those values are chosen?

Again sorry to ask so much.

7. ### Audioguru New Member

Dec 20, 2007
9,411
896
Now you are going around in circles and I will not follow you.
One schematic says C6 and the other schematic says C9.
One schematic has 36k and 2.2k and the other schematic has 62k and 1.8k.