Have Yamaha amp RX-V850 Receiver Amplifier, need 18vac display mini bulb which is no longer available, would like help with circuit to use 6v LED to replace it.

 

How much current does this 6 volt LED require? Any specs or a link?

 

Diode and resistor. Select resistor value and wattage to suit.

 

Because the circuit is AC, you do need that second diode to prevent reverse voltage damage. I am going to guess that you will choose a white light LED with a forward voltage drop of about 3.3 volts, for convenience. And I guess that the series diode will have a drop of 0.7 volts. Then I will pick the LED current to be 15 milliamps, and guess that the AC voltage is actually 18.00 volts. So the resistor will need to drop 18-4=14 volts (supply minus LED+diode drop) at 15 mA=0.015A.
Then R=V/I=14/0.015=933 ohms (not a standard value). So use two 470 oh resistors in series, because 470 is a very common value. If that is not bright enough, change one of both resistors to 390 ohms(another standard value.

 

A capacitor of 100nF ceramic above the LED is a good choice, especially if you don't use a white one. This should reduce flicker. In a white LED the response time of the phosphor reduces the flicker.
Another point to consider is, that the calculation of MrBill2 referrers only to one half wave, so current is 15 mA *0.5 = 7.5 mA effectively.
After R1 you can do a bridge rectifier and the small capacitor, this gives low flicker and real 15mA, but the circuit with just one diode is simpler and should do the job.

 

Another option will be to put THREE six volt LEDs in series, but still with the extra diode toavoid excess revers voltage.