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Ear headphone with low sound

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burai

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hi every one
i want to amplifi audio signal from phone
i have ear headphone with low sound
to do that i want to use smd dual op amp
1/2 for right the rest for left
vin =0.7v , power in= 1mw
gain=3
power source : 2 coin cell 3v
speaker = 600 ohm (from some set)
could it work without lead from op amp to another
and is there any error
picture's attached
any advise are thankful
 

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It might work but I would advise capacitors to couple the input signal and possibly a load resistor across the input to keep the phone happy. It also depends what op-amp you plan to use. Some are only for very low level signals or have very restricted bandwidth.

Brian.
 
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    burai

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Hi,

I agree with the capacitors.
I recommend to use a dedicated earphone amplifier. Usual Opamps may have problems to drive that low impedance with output close to the supply rails.

Maybe use class D amplifier with low pass filter.

Klaus
 
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    burai

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It might work but I would advise capacitors to couple the input signal and possibly a load resistor across the input to keep the phone happy. It also depends what op-amp you plan to use. Some are only for very low level signals or have very restricted bandwidth.

Brian.

thanks mr Brian
ok i will use capacitors and load resistor
i want to use LM158 but from it data sheet
Input Voltage ( -0.3V to +32V )
but my Input Voltage go to (-0.7)
sorry im confuse about -0.3 and -0.7
is -0.3v mean the lowest input voltage ?
 

Hi,

I agree with the capacitors.
I recommend to use a dedicated earphone amplifier. Usual Opamps may have problems to drive that low impedance with output close to the supply rails.

Maybe use class D amplifier with low pass filter.

Klaus
thanks mr Klaus
ok i will do google search for dedicated earphone ic and after read it data sheet i will post my effort
about class D amplifier is it will be 2 unit ?
one for right and second for left ?
 

Hi,

Because humans have two ears, I assume most earphone amplifiers come with two channels.

Klaus
 

Hi,

"Wide Power Supply Range:
Application areas include transducer amplifiers, dc
– Single Supply: 3V to 32V gain blocks and all the conventional op-amp circuits
– Or Dual Supplies: ±1.5V to ±16V"

If the input only goes to -0.7V on +-3V, you're fine. The -0.3V means not exceeding an additional +-0.3V beyond the +-supply rails on the inputs.

I'm not sure but I'd check that the output can actually go to +-2.8V, if not then extend the supply or lower the output if possible, or just switch to a rail-to-rail (input and) output op amp. One of the experts here will know better if the op amp fits the input and output range.

LM158
 

sorry you say: Maybe use class D amplifier with low pass filter.
my project have 2 input so i guess you mean classic old class D amplifier one input one output
not ic based on class D amplifier technology
block-diagrame.jpg
Hi,

Because humans have two ears, I assume most earphone amplifiers come with two channels.

Klaus
someone set behind keyboard wait for reply from he's admire teacher
reply come that way!!!!
sadly,,,
sorry for being not know humans have two ears.
 

Hi,

I assume there are class D earphone 2 channel ICs.
Did you do a search? At manufacturers and/or distributors...

Klaus
 

Hi,

"Wide Power Supply Range:
Application areas include transducer amplifiers, dc
– Single Supply: 3V to 32V gain blocks and all the conventional op-amp circuits
– Or Dual Supplies: ±1.5V to ±16V"

If the input only goes to -0.7V on +-3V, you're fine. The -0.3V means not exceeding an additional +-0.3V beyond the +-supply rails on the inputs.

I'm not sure but I'd check that the output can actually go to +-2.8V, if not then extend the supply or lower the output if possible, or just switch to a rail-to-rail (input and) output op amp. One of the experts here will know better if the op amp fits the input and output range.

LM158
thanks mr d123
seriously you helped me
i want very few component and lite circuit
so i plan to use coin cell
first
a rail-to-rail
if not
i have to lower the output

- - - Updated - - -

Hi,

I assume there are class D earphone 2 channel ICs.
Did you do a search? At manufacturers and/or distributors...

Klaus
from post #5
thanks mr Klaus
ok i will do google search for dedicated earphone ic and after read it data sheet i will post my effort
about class D amplifier is it will be 2 unit ?
one for right and second for left ?
 

You want the 0.7V RMS amplified 3 times so the output must be 5.94V peak-to-peak.
Don't use a lousy old LM158 because it has crossover distortion and its output swing will be much less than the 6V p-p required so it will produce horrible clipping distortion.

I think any amplifier will not produce a rail-to-rail output swing when loaded with 600 ohms. Then increase the supply voltage or reduce the input and output levels.
I thought about a PAM8403 stereo class-D little amplifier but it is designed for a 5V supply and your 6V is too high for it.
 

You want the 0.7V RMS amplified 3 times so the output must be 5.94V peak-to-peak.
Don't use a lousy old LM158 because it has crossover distortion and its output swing will be much less than the 6V p-p required so it will produce horrible clipping distortion.

I think any amplifier will not produce a rail-to-rail output swing when loaded with 600 ohms. Then increase the supply voltage or reduce the input and output levels.
I thought about a PAM8403 stereo class-D little amplifier but it is designed for a 5V supply and your 6V is too high for it.
thanks mr Audioguru
i can do Switching Power supply unit that use XL6009
and search for another op amp
but do Switching noise here ignorable ?
also dose coin cell suit for Switching ?
so if Switching is not recommended i will reduce gain from 3 to 2
 

i want very few component and lite circuit

This class D amplifier is made from a 555 timer IC. It's simple, inexpensive and useful to experiment with.
The 555 is configured as a pulse generator. An audio signal is applied to the 'control' pin, causing the duty cycle to grow or shrink.
The output is PWM. By filtering it becomes a close match to the original audio.

555 is class D amplifier LC 2nd-order fil 60 ohm load gets 3V sines.png
 

This class D amplifier is made from a 555 timer IC. It's simple, inexpensive and useful to experiment with.
The 555 is configured as a pulse generator. An audio signal is applied to the 'control' pin, causing the duty cycle to grow or shrink.
The output is PWM. By filtering it becomes a close match to the original audio.

View attachment 148945

thanks mr BradtheRad
it will be my next option
if Switching Power supply not work

- - - Updated - - -

thanks every one

- - - Updated - - -

thanks every one
 

The minimum supply for a 555 is 4.5V so even a 6V battery will fail soon.
The output of a 555 does not swing anywhere near rail-to-rail.
 

picture's attached

Interested to know the mapping between these two pictures. Is differential Input of 0.7V is mapped to Left/Right of the universal 3.5mm Jack?
 

I thought about a PAM8403 stereo class-D little amplifier but it is designed for a 5V supply and your 6V is too high for it.
i read PAM8403 data sheet
Absolute Maximum Ratings : Supply Voltage =6v
how about make voltage divider and stay at 5v?
file Attached
- - - Updated - - -

Interested to know the mapping between these two pictures. Is differential Input of 0.7V is mapped to Left/Right of the universal 3.5mm Jack?
right to ground ~= 0.7v
left to ground ~= 0.7v
 

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The datasheet for the PAM8403 recommends a max supply of 5.5V. Your coin batteries will produce 6.4V when new.
You can use a voltage divider to reduce the battery voltage only if the current in the divider is very high that will kill your batteries in a few minutes because a voltage divider that powers something that has a variable current is a poor voltage regulator.
Its inputs are not differential, each amplifier is bridged so the outputs are differential. The bridged outputs cannot drive headphones that have a ground, coupling capacitors can feed half the bridge to each headphone.
 
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    burai

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The datasheet for the PAM8403 recommends a max supply of 5.5V. Your coin batteries will produce 6.4V when new.
You can use a voltage divider to reduce the battery voltage only if the current in the divider is very high that will kill your batteries in a few minutes because a voltage divider that powers something that has a variable current is a poor voltage regulator.
Its inputs are not differential, each amplifier is bridged so the outputs are differential. The bridged outputs cannot drive headphones that have a ground, coupling capacitors can feed half the bridge to each headphone.
thanks you
i decide to use one of pam8xxx
and forget the circuit from post#1
the main problem is the power source
it toke me in new way by use
mini Rechargeable battery »» Switching Power unit
and stabilize the volt
 

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