One of the most sought after search query related to Tube Amp builds…

I’ll explain the biasing process as applicable to a typical Push Pull Amp, in my case this is the procedure I followed to bias Custom 50.

WARNING:

Tube amps have extremely lethal DC High Volates to the tune of over 500V. Here are some quick safety tips before attempting to do any kind of DIY…

  • Keep One hand at the back or away from the circuit
  • Use only one hand to do stuff inside the amp head
  • Wear Rubber sole/Plastic Sole shoes
  • Stand on a thick mattress
  • Check your hands for dryness
  • If possible ask someone to stand at a distance while you are working live and instruct her/him about emergency procedures (Cutting off supply)
  • Use plastic pointers to make room or displace wiring for working
  • Test your meter leads for any exposed area
  • Never attempt to make any connection or soldering live
  • Always discharge the capacitor terminals through a small resistance to ground after turning off the power (In some amps, turning mains switch OFF with standby switch in ON condition discharges the capacitors automatically in a couple of minutes)
  • Feel comfortable and do NOT work if your heart is racing…

Procedure

Install Adjustable Bias

A typical Bias adjust schematic is described iun the diagram. You need to install a pot to vary the negative voltage (Bias).

Bias Adjust Credit: Lenard Audio Institute

The cathode resistor method

This is the method that is best for hobby techs and do-it-yourselfers. While not as accurate as the output transformer shunt method (detailed below, after the cathode resistor procedure) it is far and away the safest of the two methods, and can be successfully done with medium- and even low-quality test equipment. It is performed by reading the cathode current through each power tube; the accuracy is lower because the cathode current is composed of the plate current *plus* the screen current. Plate current can be identical on two tubes (tubes are matched by plate current readings) while one tube is drawing more screen current; with this method, the readings will appear to indicate a mis-match when such is not actually the case. Since the cathode current will always be higher than the actual plate current, the readings obtained with this method will tend to make you set the tubes a little colder than your calculations will indicate that they are. This promotes slightly more conservative operation, which is beneficial to tube life. Note that these instructions assume that your amplifier is biased by applying a negative voltage to the control grids; cathode-biased amplifiers cannot be adjusted other than by changing the value of the cathode resistor(s) so this method does not apply to them. BE AWARE THAT THE ACCURACY OF THE RESULTS YOU OBTAIN FROM *ANY* BIASING METHOD WILL BE DIRECTLY AFFECTED BY THE QUALITY OF YOUR TEST EQUIPMENT, AND YOUR SKILL IN USING IT. If any part of the following instructions doesn’t make sense to you, seek help from someone with more experience.

CREDIT: Regis Coyne

DIAGRAM

Install Cathode Resitors

Procedure

  1. Install cathode resistors to ground, Resistor should ideally be, 1 Ohm, 1%, 2 W or more
  2. Connect the end of the resistor which is connected to the cathode to the bias monitoring post as its safer and convenient to have them. You’ll need to check bias whenever you change tubes or when they get older.
  3. Next, you need to measure voltage drop across the red coloured resistors (1 ohm), and since the resistance value is 1 Ohm, the voltage will be equal to the current.
  4. Measure the Plate voltage. Take care as it has HIGH DC Voltage ~ 500 V DC
  5. Measure the dissipation, (Plate voltage) X (Cathode Current, measured above)
  6. You should adust the negative bias using the bias adjust pot and will have to adjust the current so that the dissipation calculated above should be around 70% of rated dissipation (i.e. wattage, for ex.: EL34s have 25W rated power, so the dissipation should be around 18W)

Illustrative Example:

Plate Voltage = 490 V

Cathode Resistance Voltage drop = 40 mV => Current = 40 mA

Dissipation = 40 mA X 490 V = 19.6 W

Cheers!

:-)