fisher 400 class A, A/B.
- Thread starter freeman5
- Start date
Register to hide this ad
Free -- As Sony mentioned, the 400 employs a Class AB1 output stage. With such a design, the output stage operates for the first few watts in Class A mode before switching to Class B, as you surmised. In a properly designed Class AB amplifier however, the amount of Class A power output produced is rather small, relative to the total power output capability. Usually, no more than 20% of the total power output of a Class AB amplifier will be produced in strict class A operation. In the case of the Fisher 400, and assuming that the output stage quiescent current is set for 32 ma per tube, then the early 400 receivers will produce about 2.75 watts RMS in class A before switching to Class B operation, while the later versions will produce as much as 4.25 watts before Class B operation commences. As folks reduce the quiescent current of the output stage, even less Class A power is produced. Class A operation in a push-pull design is defined as both output tubes conducting current at all times during the complete input cycle. At the point, where one or both tubes begins to "cut-off" all current flow during some portion of the input cycle, then Class B operation has commenced. Therefore, any Class AB amplifier can operate in strict Class A mode -- just keep the volume down!
The numeric suffix "1" after the class letters indicates that the output tube grids are never driven positive during any portion of the input signal cycle. A suffix of "2" indicates that the output tube grids are in fact driven positive (with respect to bias voltage), allowing greater amounts of power output to be developed, for otherwise similar "1" conditions, where positive grid drive does not occur.
I hope this helps!
Dave
The numeric suffix "1" after the class letters indicates that the output tube grids are never driven positive during any portion of the input signal cycle. A suffix of "2" indicates that the output tube grids are in fact driven positive (with respect to bias voltage), allowing greater amounts of power output to be developed, for otherwise similar "1" conditions, where positive grid drive does not occur.
I hope this helps!
Dave
freeman5
Member
Thank you for the valuable information and for the speedy response:thmbsp: So if I understand this, each of the 4 output tubes are drawing 32 ma of current at idle (not amplify any signal). Then as I rotate the voulme up, the current draw increases until I hit the 4.25 watt point? Or is the current staying at 32ma and the voltage increasing in which case P=IxE, P=4.25, I=32ma, so E would be 132 volts? Then from this point on up the output is class B.
The current remains essentially constant until the output power reaches 4W or so. This is because the audio signal causes the current in one tube to go up during a portion of the waveform, while the other tube's current goes down by the same amount. Once the audio level is increased enough, the DC current begins to rise. This is the point where AB operation begins - one tube is dropping to zero current during part of the waveform, while the other one increases. Since the current can't go any lower than zero, while the opposite tube is conducting MORE, the average current increases.
Free -- Tom's answer is spot on as always. To expound a little further, when any tube in a push-pull or single ended design operates in Class A, the AVERAGE current draw of the tube or tubes does not change under perfect conditions, regardless of power output produced (up to the onset of clipping). This is true of each tube within a push-pull pair as well while operating in Class A conditions.
Now, the peak instantaneous current draw certainly rises with the application of signal -- otherwise, no power output would be produced. However, it is quickly followed by a mirror image dip in instantaneous current draw (assuming a sine wave signal is applied), for a net average current change of zero over the complete cycle. This happens in single ended designs, or in both tubes of a push-pull design while operating in Class A. Additionally, in a push-pull design, since both tubes are driven 180 degrees out of phase with each other, not only is the net current change within either tube zero, but then each tube is further acting to cancel the other tube out to reinforce the net zero current change as well.
So, if your Fisher is a later unit, then output stage current draw remains constant from zero output, up until (about) 4.25 watts of power output, beyond which average current draw starts in increase quite rapidly. I'm not sure what you were trying to determine with the formulas you provided, or what voltage you are trying to solve for. Therefore, let me give you a quick run through on how the 4.25 watt figure was determined, which as a back of napkin calculation, has proven to be rather accurate in reality:
1. The OPT primary impedance in your 400 is 10,200 ohms plate to plate.
2. Each tube in a push-pull pair effectively operates into 1/4 of this impedance, or 2,550 ohms.
3. With each tube in a push-pull pair set for a quiescent plate current of 32 ma, it means that the maximum plate current that either tube can draw is 64 ma before its mate can no longer compensate for the current increase in the tube being positively excited. At 64 ma of peak (instantaneous) current draw in one tube, its mate is then completely "cut off", such that any further increase in current draw in the excited tube then starts to raise average current draw of the stage (considering both tubes), with class B operation commencing.
4. A 64 ma swing across a 2550 ohm load equals a 163.2 peak volt swing in each half of the primary winding.
5. Since each half of the primary winding is effectively in series with each other, across the full secondary winding, this equals 326.4 peak volts.
6. Convert this to RMS volts by dividing by 1.414, which equals 230.8 volts RMS.
7. Since output transformers are only about 90% voltage efficient, the value in #6 must be multiplied by 90%, producing 207.8 vac RMS, normalized for transformation losses.
8. This value can now be used to determine power output developed at that voltage level by using the OPT primary impedance as the load value, but represents real world power produced from the secondary winding, because the losses have been accounted for. 207.8 volts squared, and divided by 10,200 ohms equals 4.23 watts RMS.
This is all based on ideal conditions which do not exist in reality, due to non-linearities of the grid characteristic in all tubes near the cut off region. However, this exercise is accurate enough, with real world measurements on my own later version 400 returning a power output of 3.90 watts RMS at the point that class B operation commenced.
I hope this helps!
Dave
Now, the peak instantaneous current draw certainly rises with the application of signal -- otherwise, no power output would be produced. However, it is quickly followed by a mirror image dip in instantaneous current draw (assuming a sine wave signal is applied), for a net average current change of zero over the complete cycle. This happens in single ended designs, or in both tubes of a push-pull design while operating in Class A. Additionally, in a push-pull design, since both tubes are driven 180 degrees out of phase with each other, not only is the net current change within either tube zero, but then each tube is further acting to cancel the other tube out to reinforce the net zero current change as well.
So, if your Fisher is a later unit, then output stage current draw remains constant from zero output, up until (about) 4.25 watts of power output, beyond which average current draw starts in increase quite rapidly. I'm not sure what you were trying to determine with the formulas you provided, or what voltage you are trying to solve for. Therefore, let me give you a quick run through on how the 4.25 watt figure was determined, which as a back of napkin calculation, has proven to be rather accurate in reality:
1. The OPT primary impedance in your 400 is 10,200 ohms plate to plate.
2. Each tube in a push-pull pair effectively operates into 1/4 of this impedance, or 2,550 ohms.
3. With each tube in a push-pull pair set for a quiescent plate current of 32 ma, it means that the maximum plate current that either tube can draw is 64 ma before its mate can no longer compensate for the current increase in the tube being positively excited. At 64 ma of peak (instantaneous) current draw in one tube, its mate is then completely "cut off", such that any further increase in current draw in the excited tube then starts to raise average current draw of the stage (considering both tubes), with class B operation commencing.
4. A 64 ma swing across a 2550 ohm load equals a 163.2 peak volt swing in each half of the primary winding.
5. Since each half of the primary winding is effectively in series with each other, across the full secondary winding, this equals 326.4 peak volts.
6. Convert this to RMS volts by dividing by 1.414, which equals 230.8 volts RMS.
7. Since output transformers are only about 90% voltage efficient, the value in #6 must be multiplied by 90%, producing 207.8 vac RMS, normalized for transformation losses.
8. This value can now be used to determine power output developed at that voltage level by using the OPT primary impedance as the load value, but represents real world power produced from the secondary winding, because the losses have been accounted for. 207.8 volts squared, and divided by 10,200 ohms equals 4.23 watts RMS.
This is all based on ideal conditions which do not exist in reality, due to non-linearities of the grid characteristic in all tubes near the cut off region. However, this exercise is accurate enough, with real world measurements on my own later version 400 returning a power output of 3.90 watts RMS at the point that class B operation commenced.
I hope this helps!
Dave
freeman5
Member
Thank you Dave and Thank you Tom. You have given me a wealth of information here and it's going to take me a bit to digest it all. Its been 15 years since electronics school and I want to take this opportunity to delve back in. My 400 needs some attention and as soon as I get a new scope I am going to tear in and hopefully get a decent grasp of this circuit in the process. Do either of you know of maybe a website that would offer a sort of "tube amplifier 101"? Thanks a million, chris.
Grbluen
Super Member
I'd also be interested in the tube amp 101.
Similar threads
