500mV (0dB) on the voltmeter?

Can’t understand why they would "arbitrarily" refer to something in the service manual.

In the context of voltage a given number of dB refers to the ratio between two voltages, as I indicated. Therefore "0 dB" must be defined as corresponding to **some** specific voltage. And there is not a single specific voltage that can be defined which will be best suited for all applications. So in that sense whatever voltage 0 dB is assigned to in a particular case will have been chosen "somewhat arbitrarily" (as I put it in my previous post), meaning that it will have been chosen by a particular organization or individual for a particular purpose.

In some contexts 0 dB may be defined as representing 1 volt, in which case voltages may be stated as dBV, the "V" signifying that the number of db is relative to 1 volt. "dBu" is also often used, 0 dBu corresponding to about 0.775 volts. In this case they chose 0 dB as corresponding to 0.5 volts, probably for reasons related to the signal levels used on the cassette tapes.

Also, why not just refer directly to 397 mV?

Some voltmeters provide a scale calibrated in dB, in addition to a scale calibrated in voltage. Especially older analog meters, if I recall correctly. Also, VU meters on tape decks are usually calibrated in dB, so specifying input voltages in dB may make it more convenient to correlate and compare changes in input voltage to changes in VU meter readings.

Regards,
-- Al

Thanks Al. Can’t understand why they would "arbitrarily" refer to something in the service manual. Also, why not just refer directly to 397 mV?

I love this.

I tell my track and field athletes

"Math is hard, so is track, deal with it"

I took a look at the service manual, at hifiengine.com. It appears that they are somewhat arbitrarily defining 0 dB as corresponding to a voltage of 500 mv, with other voltages that are expressed in dB undoubtedly being related to that 500 mv level in accordance with the standard formula for expressing the ratio of two voltages in dB:

dB = 20 x log(V1/V2)

where "log" is the base-10 logarithm.

So for example the reference in the procedure to a voltage level of -20 dB would correspond to a voltage of 50 mv, since:

20 x log(50/500) = -20 db

And the reference you stated to 500 mv -2 dB would mean 2 dB less than 500 mv, which I calculate to be 397.1 mv:

20 x log(397.1/500) = -2.00 db

Hope that helps. Regards,
-- Al