a low cost hearing aid

Thread Starter

luc

Joined Mar 4, 2011
153
If your circuit is powered from +-5V then 0V is at "half the supply voltage". The output can swing equally up and down to make the undistorted AC signal.

You show a 'scope waveform that is rectified (extremely distorted) because your circuit is biased at 0V with a single polarity supply then it cannot produce the missing negative part of the waveform. It should be biased at half the supply voltage for the output to swing equally up and down.
ya, i am using a single polarity supply, because i think that if using +-5v, a 9V battery may not enough power to supply it. since the hearing aid should be portable and small,then i cant use a power supply for source instead of battery.Is that mean that it must use AC power instead of single polarity?



You don't have a highpass filter circuit. Your lowpass filter was biased wrong at 0V.
I do have a highpass filter before , but due to the waveform of output is rectified,so i decide to cancel it. the oscilloscope waveform that i show you just now is take from the highpass filter that attach below.

For my lowpass filter, it wont biased wrong at 0V when i set an offset for my input. This problem just happen when i using high pass filter.



Ps: do you know that a EM noise that genereated by mobile phone can or can not be filter off by filter?i think that EM noise is one type of radio frequency with high frequency( probably few MHz?)

Thanks!
 

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Audioguru

Joined Dec 20, 2007
11,248
ya, i am using a single polarity supply, because i think that if using +-5v, a 9V battery may not enough power to supply it.
9V is good. With my circuit then the output of your opamp will swing equally up 4.5V and down 4.5V to produce 9V peak to peak.

since the hearing aid should be portable and small,then i cant use a power supply for source instead of battery. Is that mean that it must use AC power instead of single polarity?
You can use one battery and make a single supply circuit that is biased at half the supply voltage or you can use two batteries to make a +-9V dual polarity supply circuit that is biased at 0V. 0V is where the two batteries join together. It is the ground in the circuit.
I have seen hearing aids that use a tiny low voltage battery cell, not a huge 9V battery.

I do have a highpass filter before , but due to the waveform of output is rectified,so i decide to cancel it. the oscilloscope waveform that i show you just now is take from the highpass filter that attach below.
Your highpass filter opamp input is connected to ground but is not connected to the battery. R2 must connect to a circuit that gives +2.5VDC which is half the supply voltage. It will be complicated to make +2.5V at a low impedance but the highpass filter circuit is not needed anyway.

For my lowpass filter, it wont biased wrong at 0V when i set an offset for my input. This problem just happen when i using high pass filter.
You call it an offset voltage and I call it a bias voltage which is the same thing. The lowpass filter will have an input at half the supply voltage when it is connected to the output of the preamp which is at half the supply voltage.

Ps: do you know that a EM noise that genereated by mobile phone can or can not be filter off by filter?i think that EM noise is one type of radio frequency with high frequency( probably few MHz?)
Look in Google! A cell phone operates at 824MHz to 1.99GHz. These are microwaves where a tiny piece of wire is an antenna (the power of a cell phone is high) and many things are inductors.
Your cable from the mic must be shielded and the entire circuit must be shielded inside a metal box connected to 0V. Also, RF capacitors are probably needed to reduce the interference.
 

Thread Starter

luc

Joined Mar 4, 2011
153
9V is good. With my circuit then the output of your opamp will swing equally up 4.5V and down 4.5V to produce 9V peak to peak.

You can use one battery and make a single supply circuit that is biased at half the supply voltage or you can use two batteries to make a +-9V dual polarity supply circuit that is biased at 0V. 0V is where the two batteries join together. It is the ground in the circuit.
ya,but it is quite large and heavy too.


I have seen hearing aids that use a tiny low voltage battery cell, not a huge 9V battery.
ya,but i think some of them does not use op-amp which need atleast 5v to power it up,They are using transistor,so i think we should have atleast 9v battery.



You call it an offset voltage and I call it a bias voltage which is the same thing. The lowpass filter will have an input at half the supply voltage when it is connected to the output of the preamp which is at half the supply voltage.
So the bias voltage is same as offset, mean the output should be above the 0v ?

I use ur sketch in a stimulator and the result 1 show that in 50Hz input, the output signal was not match with the input signal in -ve part, but in theory it should be match right?

I major 2 output which are : 1 from pre-amp ,another from low pass; the stimulator show me that they have same output waveform no matter how much frequency in the input but it should not be same. because the pre-amp does have a low pass filter ,then it should not filter the frequency that greather than 20kHz which mean that it should have almost same amplitude with input right? But i get totally same ouput as i major from lowpass. i wondering is that any wrong connection for my circuit?


Look in Google! A cell phone operates at 824MHz to 1.99GHz. These are microwaves where a tiny piece of wire is an antenna (the power of a cell phone is high) and many things are inductors.
Your cable from the mic must be shielded and the entire circuit must be shielded inside a metal box connected to 0V. Also, RF capacitors are probably needed to reduce the interference.
i do search fin google, but what i want to know is that the phone will generate how much frequency of EM noise to interfere the mic, not the operates frequency.. By the way, if cell phone have such a high frequency , why filter cant filter the noise that generated by it?( which mean that the "pu pu" sound that can hear from earphone) Is that not consider as a type of frequency that can be filter off?

About the shielded, i try it before. i just leave the mic outside the box to receive the voice signal, but when cell phone was near by ( make a call or received call) it will still affect by the the noise. Is that mean my mic receive frequency that generate by phone(EM noise) and my filter does't work to filter it( or unalbe to filter it)?
 

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Audioguru

Joined Dec 20, 2007
11,248
Am I wasting my time talking to you about biasing the opamp at HALF THE SUPPLY VOLTAGE? In the new simulation your schematic has the bias voltage floating. It does not have anything to do with the supply voltage. So your opamps will not work properly but your sim program doesn't know your opamps are not biased preperly.

Since you have a single supply voltage then everything grounded should connect to the negative terminal of the battery. C2 should not connect to the pin 3 input of the opamp, instead it should connect to ground.
The opamps are not biased properly because your circuit is missing a 100k resistor from the filtered supply at R1, R2 and C2 to the pin 3 input of the opamp.

Your filter is not a Butterworth second-order lowpass filter because the value of C6 is too high at 2.2nF. It is supposed to be 2.0nF. So the filter cuts 8kHz and 20kHz too much. I think it has a peak in its response around 17kHz.

Your questions about the sim display I think are because your sim program has not read the datasheet for the modern Cmos opamp. Instead it thinks the opamps are lousy old 741 opamps that work poorly above 9kHz.

The "pu pu" interference from a cell phone is difficult to stop unless you know microwave RF circuit design. A cell phone has a lot of power and is close by.
You don't have a microwaves RF filter at the input so the interference into the first opamp is very strong. Your 20kHz audio lowpass filter does not work above about 100kHz.
 

Thread Starter

luc

Joined Mar 4, 2011
153
Thanks for reply!

Am I wasting my time talking to you about biasing the opamp at HALF THE SUPPLY VOLTAGE?
Sorry for not get your meaning.. is that mean if my input is an 9v battery, then it should be bias at 4.5v? mean i set my offset to 4.5v?


In the new simulation your schematic has the bias voltage floating. It does not have anything to do with the supply voltage. So your opamps will not work properly but your sim program doesn't know your opamps are not biased preperly.
bias voltage floating due to my wrong connection? i just found out that i miss connected the C2 to R5 while there should have another resistor missing. i re-sketch my schematic already.

Since you have a single supply voltage then everything grounded should connect to the negative terminal of the battery. C2 should not connect to the pin 3 input of the opamp, instead it should connect to ground.The opamps are not biased properly because your circuit is missing a 100k resistor from the filtered supply at R1, R2 and C2 to the pin 3 input of the opamp.
“everything grounded should connect to negative terminal of battery ”do you mean that instead of connect to ground, i have to connect it with negative terminla of battery? For this schematic, i'm using an AC source, but i think i would change to dc source. For C2, i miss connected it to pin 3. i already resketch it.


Your filter is not a Butterworth second-order lowpass filter because the value of C6 is too high at 2.2nF. It is supposed to be 2.0nF. So the filter cuts 8kHz and 20kHz too much. I think it has a peak in its response around 17kHz.
so i need to change it by parallel two 1nF capacitor?


You don't have a microwaves RF filter at the input so the interference into the first opamp is very strong. Your 20kHz audio lowpass filter does not work above about 100kHz.
but i have test this lowpass circuit with the frequency input above 100kHz, it cant pass through my lowpass filter.

I redraw my circuit already, but i not sure whether correct or not. sorry for my poor understanding. and once again , thanks for your help!
 

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Audioguru

Joined Dec 20, 2007
11,248
Sorry for not get your meaning.. is that mean if my power supplyis a 9v battery, then it should be bias at 4.5v? mean i set my offset to 4.5v?
Yes, but from the same supply that powers the opamp. Two resistors in a voltage divider makes the "half the supply voltage" bias voltage. It is fed to the non-inverting input of the opamp.

bias voltage floating due to my wrong connection? i just found out that i miss connected the C2 to R5 while there should have another resistor missing. i re-sketch my schematic already.
Your new schematic is correct except the value of R1 is too low at 1k and it reduces the output level from the mic. Use 4.7k when the supply is 5V.

“everything grounded should connect to negative terminal of battery ”do you mean that instead of connect to ground, i have to connect it with negative terminla of battery? For this schematic, i'm using an AC source, but i think i would change to dc source.
With a single supply voltage, the negative terminal of the battery is connected to ground together with everything else that is connected to ground.

For C2, i miss connected it to pin 3. i already resketch it.
It is correct now.

so i need to change it by parallel two 1nF capacitor?
You need 2nF. You can look for one and buy it or use two 1nF in parallel.

but i have test this lowpass circuit with the frequency input above 100kHz, it cant pass through my lowpass filter.
The level for a 20kHz second-order Butterworth lowpass filter at 100kHz is reduced to about 1/15th the level of lower frequencies if the opamp has a high frequency response.
 

Thread Starter

luc

Joined Mar 4, 2011
153
Yes, but from the same supply that powers the opamp. Two resistors in a voltage divider makes the "half the supply voltage" bias voltage. It is fed to the non-inverting input of the opamp.
i was try to run the circuit again, but it convert my output from sin wave to square wave, when i increase the frequency , my lowpass does't work to reduce the amplitude of output. My output also have been amplifier by pre-amp which the peak-to-peak voltage is greather than input.


Your new schematic is correct except the value of R1 is too low at 1k and it reduces the output level from the mic. Use 4.7k when the supply is 5V.
I think i should using 9V supply because op-amp need atleast 5v to power up, if using 5v with +2.5v -2.5v is not enough right?
 

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Audioguru

Joined Dec 20, 2007
11,248
Maybe you do not understand that a microphone has an output level of only 0.01V, not 1V.
Maybe you do not understand that your preamp has a voltage gain of (R4/R3) + 1= 101 times.
Then its output is trying to be 101V! Impossible so it is a severely clipped square-wave.
Its max output is only 4.37V pk.

So with such a high amount of voltage gain (101 times) and such a high input level (1V) then of course the output is severely clipped into a square-wave.

You cannot measure frequency response when an amplifier is severely over-driven into a square-wave.

The opamp you selected has very low voltage gain for an opamp and its gain at high frequencies is even lower. Its datasheet shows that it works very well at low frequencies when its supply is only 5V.
 

Thread Starter

luc

Joined Mar 4, 2011
153
Maybe you do not understand that a microphone has an output level of only 0.01V, not 1V.
Maybe you do not understand that your preamp has a voltage gain of (R4/R3) + 1= 101 times.
Then its output is trying to be 101V! Impossible so it is a severely clipped square-wave.
Its max output is only 4.37V pk.

So with such a high amount of voltage gain (101 times) and such a high input level (1V) then of course the output is severely clipped into a square-wave.

You cannot measure frequency response when an amplifier is severely over-driven into a square-wave.

The opamp you selected has very low voltage gain for an opamp and its gain at high frequencies is even lower. Its datasheet shows that it works very well at low frequencies when its supply is only 5V.
I thought that due to my op-amp can support up to 40dB, so it is about 100 gain,so i am using my resistor value as R3=495k and R4 as 5k to obtain a gain of 100.
for the microphone part, may i know which part that i connect wrong so it become 1v? cause i really dnno how you can calculate out that 1v?
can you tell me what is the function for C1? to stop dc pass through?


Sorry, i really not familiar with the microphone part.Thanks for reply.
 
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Audioguru

Joined Dec 20, 2007
11,248
for the microphone part, may i know which part that i connect wrong so it become 1v? cause i really dnno how you can calculate out that 1v?
Why can't you see that your simulation circuit does not have a microphone, it has a function generator instead?
Why can't you see that you have the amplitude from the function generator set to 1V pk instead of 0.01V?
 

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Thread Starter

luc

Joined Mar 4, 2011
153
Why can't you see that your simulation circuit does not have a microphone, it has a function generator instead?
Why can't you see that you have the amplitude from the function generator set to 1V pk instead of 0.01V?
sorry, i did't see carefully, that is my fault..

by the way, do i need to change my gain? previously i set my gain to 100 in previous,now i change to 10. is that i need to reduce my gain?
i capture the output for my pre-amp.

but when i connect it to my low-pass filter, i found out that my low pass output have same value as the pre-amp output, and it cant filter the frequency higher than 20kHz, is that mean i have some connection problem?
 

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Audioguru

Joined Dec 20, 2007
11,248
Why did you change the gain to 10? You need a gain of 101.
If you adjust the settings of the oscilloscope correctly then you will see the input sine-wave and see the output sine-wave.

I don't know how you are measuring the frequency response of the lowpass filter.
With two 5.6k resistors, one 1nF capacitor, one 2nF capacitor and an opamp with a gain of 1 it should have its output response down -3dB at 20kHz, -12dB at 40kHz and -24dB at 80kHz.
 

Thread Starter

luc

Joined Mar 4, 2011
153
Why did you change the gain to 10? You need a gain of 101.
If you adjust the settings of the oscilloscope correctly then you will see the input sine-wave and see the output sine-wave.

I don't know how you are measuring the frequency response of the lowpass filter.
With two 5.6k resistors, one 1nF capacitor, one 2nF capacitor and an opamp with a gain of 1 it should have its output response down -3dB at 20kHz, -12dB at 40kHz and -24dB at 80kHz.
i change my gain because in previous post i saw ur reply

Maybe you do not understand that your preamp has a voltage gain of (R4/R3) + 1= 101 times.
Then its output is trying to be 101V! Impossible so it is a severely clipped square-wave.
so i think that maybe my gain too high already...
 

Audioguru

Joined Dec 20, 2007
11,248
I told you to reduce the input signal amplitude from the function generator, not to reduce the gain.
With a gain of 101 and an input of 10mV pk then the output will be 1.01V pk. Then a power amplifier with a gain of 1 or 2 can drive the earphone.
 

Thread Starter

luc

Joined Mar 4, 2011
153
I told you to reduce the input signal amplitude from the function generator, not to reduce the gain.
With a gain of 101 and an input of 10mV pk then the output will be 1.01V pk. Then a power amplifier with a gain of 1 or 2 can drive the earphone.

Thanks!! i will try it on my circuit.:)
 

Thread Starter

luc

Joined Mar 4, 2011
153
I told you to reduce the input signal amplitude from the function generator, not to reduce the gain.
With a gain of 101 and an input of 10mV pk then the output will be 1.01V pk. Then a power amplifier with a gain of 1 or 2 can drive the earphone.

Hi, i have some question would like to ask you.
1) about the ouput from attachement, my input voltage is 5v, and my bias voltage should be 2.5v right? But when i run the stimulator, the input is at 2v instead of 2.5v,and i use the multimeter to test it, it also show me the voltage around 2.05V. Is that because the R2 resistor there? R2 and C2 build as smoothing filter right? may i know how to design the value for this filter?

2) when mic have an input of 0.01v, and my gain set to 101 and have an output of 1.01v, is that mean the pre-amp amplify the signal about 0.99times grether than input?

Thanks!
 
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Audioguru

Joined Dec 20, 2007
11,248
Hi, i have some question would like to ask you.
1) about the ouput from attachement, my input voltage is 5v, and my bias voltage should be 2.5v right? But when i run the stimulator, the input is at 2v instead of 2.5v,and i use the multimeter to test it, it also show me the voltage around 2.05V. Is that because the R2 resistor there?
Your input voltage is not 5VDC, it is a 0.01V sine-wave. Your supply voltage is +5VDC.
You wrongly have the DC offset voltage of your signal generator set at 0V so it is shorting the mic DC input to ground. Then the 1k resistor feeds the 4.7k resistor that is connected to 0V and the bias voltage becomes only +2.0614VDC.

R2 and C2 build as smoothing filter right? may i know how to design the value for this filter?
An RC filter has an output that is at -3dB (0.707 times) at its cutoff frequency. The formula for the cutoff frequency is 1/(2 pi RC). The frequency for the 1k resistor and 47uF capacitor is 3.4Hz. Then any 3.4Hz ripple on the +5V is reduced -3dB. The lowest frequency of a man's voice is 80Hz which will be reduced -27dB and higher frequencies will be reduced more.

2) when mic have an input of 0.01v, and my gain set to 101 and have an output of 1.01v, is that mean the pre-amp amplify the signal about 0.99times grether than input?
The gain is the amount of amplification of the signal. It is 101 times not 0.99 times.
Simple arithmatic shows that 0.01V times 101= 1.01V.
 

Thread Starter

luc

Joined Mar 4, 2011
153
Your input voltage is not 5VDC, it is a 0.01V sine-wave. Your supply voltage is +5VDC.
You wrongly have the DC offset voltage of your signal generator set at 0V so it is shorting the mic DC input to ground. Then the 1k resistor feeds the 4.7k resistor that is connected to 0V and the bias voltage becomes only +2.0614VDC.
so i should not set the offset of signal generator to 0? then what value i have to set ? or just leave it blank? the circuit already have a bias voltage of 2.0614VDC, i still need to set offset value in signal generator? that 4.7k resistor is connected to 5v supply voltage instead of 0V right?
 

Audioguru

Joined Dec 20, 2007
11,248
The 4.7k resistor that powers the mic is not connected to +5V. It is connected to the 1k resistor in the smoothing filter.

I don't know your simulator software but maybe if you do not set any offset voltage then it will be correct. Or set the offset voltage to +2.5V.
 

Thread Starter

luc

Joined Mar 4, 2011
153
The 4.7k resistor that powers the mic is not connected to +5V. It is connected to the 1k resistor in the smoothing filter.
But from the circuit diagram, the R2(1k ohm resistor) is connected with +5v, then the R1(4.7k ohm resistor) is connected with R2. i have use multimeter to measure the the lead of R1 and C1which is connected to signal generator, the voltage is 5v. Is that mean there have some part with wrong connection?

I don't know your simulator software but maybe if you do not set any offset voltage then it will be correct. Or set the offset voltage to +2.5V.
If i set an offset to +2.5V, my signal will be 2.5V+2.0614V which would floating on +4.5614VDC. I think set a 2.5v offset in signal generator is no use in future because i need to replace it with mic as input,the mic wont generate a 2.5v offset, so i have to depend on the bias voltage.. the 1k ohm resistor in smoothing filter will reduce my 2.5V bias voltage to 2.0614V, is there any way to maintain the 2.5V bias voltage?
Thanks!
 

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