Hello
Is there any way to reduce the gain on a project headbox s headphone amp. I have it connected to a denon tu260l vhf fm tuner and find its gain is too high and the vol knob is almost at zero. The phones are old ross re260l. Is there any facility internally if possible.
Thanks
thyristor44
Is there any way to reduce the gain on a project headbox s headphone amp. I have it connected to a denon tu260l vhf fm tuner and find its gain is too high and the vol knob is almost at zero. The phones are old ross re260l. Is there any facility internally if possible.
Thanks
thyristor44
It's easy to lose gain. The simplest way is to inset a series resistor between the source (tuner)
and the headphone amp. Find out what the input impedance of the amp is. Adding a series resistor
equal to that, loses 6dB. Adding a series resistor that is ten times that, loses more than 20dB.
I suspect you need the latter. The series resistor can be inside either the tuner or the amp.
Or it can be external, in a small box, or inside one of the cable plugs for each channel.
It is usually much preferred to use a voltage divider ("L-pad") setup for an attenuator - rather than a plain series resistor - to keep funny things from happening. TU-260L is a pretty basic tuna so I'll assume an output impedance of 3.3 kOhms and desired load impedance of at least 10 kOhms, if not even higher. For starters, I'd try 22 kOhms in series followed by 3.3 kOhms in parallel to the input, to be mounted close to the input side.
It occurred to me that any reduction may be better in the output as it would be much lower impedance and also less susceptible to interference. A series resistor on each headphone channel.
Yes I will take my chance with the output of the headbox rather than the quad, much lower power. I could do it with a plug, resistors and jack socket and be all external.
Hi. Ohm's law tells us that power is inversely proportional to resistance, ie P=V^2/R. The resistance you put in the divider network reduces the power demand, all of the excess power available doesn't have to be dissipated, it is not demanded by the load. The only power that does have to be dissipated is that that flows through the load, which is now the headphones and the resistive network.
Thanks I will use a dividing network. When you say dumping factor is that the same as damping factor.
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