# Insufficient gain power amp and Pre amp

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#### audioguru

##### Well-Known Member
If the output of the preamp is clipping then either reduce its input level or reduce its gain.
If the preamp is not clipping but the power amp is, then turn down the volume control because that is what it is for.

The output of the power amp when it has a 9V supply and a 16 ohm load is 250mW at clipping. That is enough to blow up headphones and destroy your hearing.
If both 16 ohm ear transducers are connected in parallel then the total power is 800mW at clipping.

#### elnino86

##### New Member
If the output of the preamp is clipping then either reduce its input level or reduce its gain.
If the preamp is not clipping but the power amp is, then turn down the volume control because that is what it is for.

The output of the power amp when it has a 9V supply and a 16 ohm load is 250mW at clipping. That is enough to blow up headphones and destroy your hearing.
If both 16 ohm ear transducers are connected in parallel then the total power is 800mW at clipping.
If i put the resistor in series like Nigel suggest,the total power will increase right?

#### Nigel Goodwin

##### Super Moderator
What is the different between speaker 16Ω and headphone 16Ω?Doesnt they have a same function but different in size?
A speaker is designed to accept a number of watts (perhaps MANY watts), the headphones are only designed to accept milliwatts - feeding them watts will destroy them.

Placing a suitable resistor in series with the headphones will lose most of the power in the resistor, leaving enough not to blow the headphones.

#### audioguru

##### Well-Known Member
If i put the resistor in series like Nigel suggest,the total power will increase right?
No.
Use Ohm's law:
1) 5.6V p-p at the output of the TDA2822M= 1.98V RMS.
2) 1.98V RMS into 16 ohms= 245mW.
3) 1.98V RMS into 22 ohms plus 16 ohms= 103mW.
4) 103mW into 22 plus 16= 43.4mW in the 16 ohms.

The headphone might have a max allowed power of only 50mW.

#### elnino86

##### New Member
No.
Use Ohm's law:
1) 5.6V p-p at the output of the TDA2822M= 1.98V RMS.
2) 1.98V RMS into 16 ohms= 245mW.
3) 1.98V RMS into 22 ohms plus 16 ohms= 103mW.
4) 103mW into 22 plus 16= 43.4mW in the 16 ohms.

The headphone might have a max allowed power of only 50mW.
Owh..i got it..I read my earphone datasheet it had a 3mW power rating..mean that i need to put a resistor to reduce it to 3mW!!!

thanks for the lesson!!!You too Nigel.Happy weekend you guys!!!Rest my brain for 2 days!!

#### audioguru

##### Well-Known Member
I hope that the 3mW earphone was not in your ear when you fed it 245mW or more.

I think I made a mistake before.
With a 9V supply, the output power per channel at clipping into an 8 ohm load is 800mW and is about 450mW into 16 ohms. So the output is about 7.6V p-p.
450mW into an earphone that has a max allowed input of only 3mW is an overload of 150 times!

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#### elnino86

##### New Member
I hope that the 3mW earphone was not in your ear when you fed it 245mW or more
I dont have my earphone right now.It blow up few days ago.
I think I made a mistake before.
Its ok Audioguru.Anyone made a mistake.

With a 9V supply, the output power per channel at clipping into an 8 ohm load is 800mW and is about 450mW into 16 ohms. So the output is about 7.6V p-p.
450mW into an earphone that has a max allowed input of only 3mW is an overload of 150 times!
I get it.As my calculation,

7.6 V p-p produce 450mW and the earphone is only 3mW power rating.I need about 2.2KΩ to reduce the power into 16Ω.

#### audioguru

##### Well-Known Member
As my calculation,
7.6 V p-p produce 450mW and the earphone is only 3mW power rating.I need about 2.2KΩ to reduce the power into 16Ω.
That is correct.

#### elnino86

##### New Member
That is correct.
Thanks Audioguru.Have a nice weekend!!

Im updated my circuit as recommended by Audioguru and Nigel.

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#### audioguru

##### Well-Known Member
El Nino,
You have forgotten many things we discussed:

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#### elnino86

##### New Member
El Nino,
You have forgotten many things we discussed:
Thanks for remind me~

Hai Audioguru, is it ok if im replace the 10k variable resistor with DS1802.DS1802 is a digi pot,can be configure with push button but only built in 45kΩ.

#### audioguru

##### Well-Known Member
Hai Audioguru, is it ok if im replace the 10k variable resistor with DS1802.DS1802 is a digi pot,can be configure with push button but only built in 45kΩ.
I don't think you understand how to bias the resistor ladder and to limit the input signal level.

#### elnino86

##### New Member
I don't think you understand how to bias the resistor ladder and to limit the input signal level.
Huh?sound really complicated..
I found 1 IC from intersil which can be configure using push button,X9511.At the 1st sight,it really similar to DS1666.I think it should work well as a replacement for variable resistor.Only autostore button look strange there.

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#### audioguru

##### Well-Known Member
The X9511 is a digital linear pot, not a logarithmic volume control.
It and the DS1666 use 0V to 5V for logic or pushbuttons and an analog input from -5V to +5V.

#### elnino86

##### New Member
The X9511 is a digital linear pot, not a logarithmic volume control.
It and the DS1666 use 0V to 5V for logic or pushbuttons and an analog input from -5V to +5V.
Finally,I found 2 circuit which able to control DS1666 without using any microcontroller.They use switch debouncer as a controller.

Im prefer the 1st shcematic but i dunno what is the switch symbol.Is that symbol similar the momentary push button symbol or else?
The logic output is same with DS1666 mode selection(datasheet).Only the symbol confuse me here.

The 2nd schematic is quite ok but i dun think switch debouncer IC is cheap as logic gate.

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#### elnino86

##### New Member
I need some opinion from the expert here.
Refer to the figure attached,

1st schematic:

-Very easy to configure.
-cheap
-the switch symbol very confusing(what kind of symbol is that?)

2nd schematic.

- probably work very well without any problem
- expensive
- use watchdog timer
- consume more power supply

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