Is there Any schematic to connect 38R headphone from 8R OPT of type 45 SE 2 watts amp?
Thank you in advance
Thank you in advance
I would suggest that you do NOT attempt putting 2 Watts (or even 1 watt, using TR100's idea) into a pair of headphones.
There are two possible outcomes:
1) Crank up the volume and you go deaf from driving a headphone rated for maybe 250 milliwatts with 1 watt of power
2) Crank up the volume and instantly burn out the voice coils of your headphones
Please, use one of the many schematics found online for an appropriate headphone-level attenuator.
I like to use the circuit that J.M.Fahey suggests.
And please, do not skimp out and use 1/4 watt resistors here, especially for the first two.
2 watt resistors for R1 and R2 would be fine.
I did this when I was around 17 and made lots of smoke come out of the resistor box!
There are two possible outcomes:
1) Crank up the volume and you go deaf from driving a headphone rated for maybe 250 milliwatts with 1 watt of power
2) Crank up the volume and instantly burn out the voice coils of your headphones
Please, use one of the many schematics found online for an appropriate headphone-level attenuator.
I like to use the circuit that J.M.Fahey suggests.
And please, do not skimp out and use 1/4 watt resistors here, especially for the first two.
2 watt resistors for R1 and R2 would be fine.
I did this when I was around 17 and made lots of smoke come out of the resistor box!
first resistor in serie is to drop voltage to protect headphone. How much watt and value of 1 st resistor?
The second resistor will be 10R in parallel with headphone to make 38R Headphone to be approx 8 ohms to match OPT.
Suppose normal level volume is around 1 watts out of 2 watts of 45. How much value of series First R to drop voltage . I listen to 45 around 60-67 db in small bedroom
The second resistor will be 10R in parallel with headphone to make 38R Headphone to be approx 8 ohms to match OPT.
Suppose normal level volume is around 1 watts out of 2 watts of 45. How much value of series First R to drop voltage . I listen to 45 around 60-67 db in small bedroom
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You don't PUT 2 watts into a load.
An amplifier is just a voltage source (we a relatively high RI with tube amplifiers)
2 Watts ar 8ohm means 4Vrms.
4Vrms at 38ohm is around 420mW.
But that's only when you crack the amplifier fully. We don't know what the max power is of the headphones, but there quite some out there that easily need (or can handle) 200-350mW.
In any of the cases, it's essential to match the output impedance of the amplifier.
Just an L-pad attenuator should be enough in this case to match the impedance and attenuate the signal.
Far to lazy to do all calculations by hand, so I very quickly put it in LTSpice.
See schematic below.
This will attenuate the power in half (210mW) while keeping the total resistance around 8ohm.
It's not a perfect match with these E-value resistors, but close enough.
Power dissipation for R1 is around 550mW and for R2 around 1.23W.
Just take double the value to be extra super safe, so two 2W resistors will be fine (plenty) probably.
See schematic below.
This will attenuate the power in half (210mW) while keeping the total resistance around 8ohm.
It's not a perfect match with these E-value resistors, but close enough.
Power dissipation for R1 is around 550mW and for R2 around 1.23W.
Just take double the value to be extra super safe, so two 2W resistors will be fine (plenty) probably.
Attachments
This L pad calculator might be handy to both present correct load to the opt; as well as give your volume control more range by adding some attenuation.
https://robrobinette.com/HeadphoneResistorNetworkCalculator.htm
https://robrobinette.com/HeadphoneResistorNetworkCalculator.htm