0

Do this with no power applied.
\[R_A\]
\[ R_B \]
A small difference between \( R_A \) and \( R_B \) is normal because the inner and outer halves of the winding use different lengths of wire.
Example values:
\[ R_A = 70 \ \Omega \]
\[ R_B = 72 \ \Omega \]
You will use these resistances when calculating current.
Now you will power the amp and measure the DC voltage drop across each half of the OT primary.
Because multimeters are not perfect and the HT voltage can bounce slightly, the most reliable way to get the voltage drop is to measure it directly between the center tap and each plate lead, rather than measuring both to ground and subtracting.
For side A:
\[ V_{\text{dropA}} \]
For side B:
\[ V_{\text{dropB}} \]
These are the actual DC drops across each half of the primary at idle, with no subtraction error. This improves accuracy, especially when B+ moves a little as you adjust bias.
It is also useful to note plate voltages to ground for later dissipation calculations:
\[ V_{PA} \]
\[ V_{PB} \]
Ohm’s law in this context is:
\[ I = \frac{V}{R} \]
So the total idle current in each half of the OT primary is:
\[ I_A = \frac{V_{\text{dropA}}}{R_A} \]
\[ I_B = \frac{V_{\text{dropB}}}{R_B} \]
Here:
For a 2‑tube push‑pull amp (one tube per side), suppose:
\[ R_A = 70 \ \Omega, \quad R_B = 72 \ \Omega \]
and you measure:
\[ V_{\text{dropA}} = 5 \ \text{V}, \quad V_{\text{dropB}} = 4 \ \text{V} \]
Then:
\[ I_A = \frac{5 \ \text{V}}{70 \ \Omega} \approx 0.071 \ \text{A} = 71 \ \text{mA} \]
\[ I_B = \frac{4 \ \text{V}}{72 \ \Omega} \approx 0.056 \ \text{A} = 56 \ \text{mA} \]
In a 2‑tube amp, these are the idle currents of each tube (ignoring small screen current).
How you divide that current depends on how many tubes share each half of the primary.
One tube on each side:
\[ I_{\text{tubeA}} = I_A \]
\[ I_{\text{tubeB}} = I_B \]
So each tube’s current is exactly the side current.
Two tubes in parallel on each side.
Assuming the tubes on a side are reasonably matched, you can approximate per‑tube current as:
\[ I_{\text{tubeA}} \approx \frac{I_A}{2} \]
\[ I_{\text{tubeB}} \approx \frac{I_B}{2} \]
Example:
\[ I_A = 120 \ \text{mA}, \quad I_B = 116 \ \text{mA} \]
Then:
\[ I_{\text{tubeA}} \approx \frac{120 \ \text{mA}}{2} = 60 \ \text{mA} \]
\[ I_{\text{tubeB}} \approx \frac{116 \ \text{mA}}{2} = 58 \ \text{mA} \]
Three tubes on each side.
If tubes are reasonably matched:
\[ I_{\text{tubeA}} \approx \frac{I_A}{3} \]
\[ I_{\text{tubeB}} \approx \frac{I_B}{3} \]
A big difference between the per‑tube currents on each side suggests tube mismatch or bias‑balance issues.
Tube datasheets give maximum plate dissipation \( P_{\text{max}} \) in watts (for example, EL34 ≈ 25 W, 6L6GC ≈ 30 W).
Plate dissipation at idle is:
\[ P = V_{\text{plate}} \times I_{\text{plate}} \]
For biasing, it’s common to approximate plate current as the measured current from this method, knowing that a small portion actually goes to the screen.
For each tube:
\[ P_{\text{tube}} \approx V_{\text{plate}} \times I_{\text{tube}} \]
If side A has plate‑to‑ground voltage \( V_{PA} \):
\[ P_{\text{tubeA}} \approx V_{PA} \times I_{\text{tubeA}} \]
Similarly, for side B with plate voltage \( V_{PB} \):
\[ P_{\text{tubeB}} \approx V_{PB} \times I_{\text{tubeB}} \]
For fixed‑bias Class AB guitar amps, a common rule is to idle around 60–70% of maximum plate dissipation:
\[ P_{\text{target}} \approx 0.6 \text{ to } 0.7 \times P_{\text{max}} \]
Given \( V_{\text{plate}} \), the corresponding target current is:
\[ I_{\text{target}} = \frac{P_{\text{target}}}{V_{\text{plate}}} \]
Example for a 6L6GC (\( P_{\text{max}} \approx 30 \ \text{W} \)) at 450 V and 70%:
\[ P_{\text{target}} = 0.7 \times 30 \ \text{W} = 21 \ \text{W} \]
\[ I_{\text{target}} = \frac{21 \ \text{W}}{450 \ \text{V}} \approx 0.047 \ \text{A} = 47 \ \text{mA} \]
Biasing cooler (50–60%) increases tube life; biasing hotter (toward 70%) reduces crossover distortion but shortens tube life.
Now you adjust the bias pot and iterate measurements.
As you bias hotter (higher idle current), the HT / B+ voltage usually drops because the supply sags under load. When you bias cooler, HT tends to rise.
Because both \( V_{\text{plate}} \) and \( I_{\text{tube}} \) change, you must:
Let the amp idle for several more minutes and confirm that the bias stays in range.
The old “shunt” method puts the meter in series between B+ and a plate lead:
The transformer‑resistance method:
For fixed‑bias guitar amps, this balance of accuracy and safety makes it one of the best everyday bias techniques.
To bias a fixed‑bias push‑pull amp using the output‑transformer resistance method:
Hope you enjoyed the ride. Thanks for reading, and stay safe on the bench.
Marko, Slightly Technical









We use cookies to make our site work (essential). With your consent, we use optional analytics cookies to understand traffic and improve content. You can change your choice anytime via “Cookie Settings” in the footer.
Your browser sends a Global Privacy Control (GPC) signal. We respected it by defaulting analytics to off.