Hi guys!
I have a maybe stupid question but still I would really apreciate if someone would answer it to me. When I am setting an operating point of the power tube, as a plate resistor should I be counting with the dc resistance of the primary side of the output transformer or it's impedance? In my case the dc resistance is 170R but the impedance is 5k. Thank you.
I have a maybe stupid question but still I would really apreciate if someone would answer it to me. When I am setting an operating point of the power tube, as a plate resistor should I be counting with the dc resistance of the primary side of the output transformer or it's impedance? In my case the dc resistance is 170R but the impedance is 5k. Thank you.
This should help. It sounds like you are making a single ended power amplifier.
The plate AC voltage swings up and down from the quiescent B+ voltage.
http://www.valvewizard.co.uk/se.html
The plate AC voltage swings up and down from the quiescent B+ voltage.
http://www.valvewizard.co.uk/se.html
The quiescent operating point of a single ended triode output stage's plate (let's use a triode to simplify):
1. B+ voltage at the top of the output transformer primary
2. Plate Current
3. DCR of the output primary
Example:
B+ = 300V
Plate current = 60 mA (0.060A)
DCR of the primary = 170 Ohms
300V - (0.060A x 170) = 300V - 10.2V = 289.8VDC at the plate.
Suppose you use a 300B tube, and have the filament ends returned to ground through two 25 Ohm resistors,
and have -60V bias supply returned to the bottom of the grid return resistor (no the grid end of that resistor).
Then you will be close to 60mA, and the plate will be close to 289.8V.
30mA (0.030A) through a 25 Ohm resistor will reduce the plate to filament voltage by 0.030 x 25 = 0.75V in each resistor.
289.8 - 0.75V = 289.05V plate to filament.
Now, apply + 5V peak and - 5 Volts signal to the 300B grid, and the plate current will vary about + 27.5mA peak and - 27.5mA peak. (0.0275A).
The output transformer's primary impedance is 5k Ohms.
27.5mA x 5,000 Ohms = 137.5V peak. The plate voltage will vary + and - 137.5V (above and below its quiescent voltage).
Power, rms = ((Peak Volts squared) / Impedance)/2
137.5 squared 5000/2 = 1.89 Watts rms
Of course the Watts is low, because we only used +/- 5V peak signal.
If we want to use the full power of the tube in that circuit, we need the signal to drive the grid with more signal . . .
+60V to - 60V peak signal from the coupling cap to the grid and grid return resistor.
1. B+ voltage at the top of the output transformer primary
2. Plate Current
3. DCR of the output primary
Example:
B+ = 300V
Plate current = 60 mA (0.060A)
DCR of the primary = 170 Ohms
300V - (0.060A x 170) = 300V - 10.2V = 289.8VDC at the plate.
Suppose you use a 300B tube, and have the filament ends returned to ground through two 25 Ohm resistors,
and have -60V bias supply returned to the bottom of the grid return resistor (no the grid end of that resistor).
Then you will be close to 60mA, and the plate will be close to 289.8V.
30mA (0.030A) through a 25 Ohm resistor will reduce the plate to filament voltage by 0.030 x 25 = 0.75V in each resistor.
289.8 - 0.75V = 289.05V plate to filament.
Now, apply + 5V peak and - 5 Volts signal to the 300B grid, and the plate current will vary about + 27.5mA peak and - 27.5mA peak. (0.0275A).
The output transformer's primary impedance is 5k Ohms.
27.5mA x 5,000 Ohms = 137.5V peak. The plate voltage will vary + and - 137.5V (above and below its quiescent voltage).
Power, rms = ((Peak Volts squared) / Impedance)/2
137.5 squared 5000/2 = 1.89 Watts rms
Of course the Watts is low, because we only used +/- 5V peak signal.
If we want to use the full power of the tube in that circuit, we need the signal to drive the grid with more signal . . .
+60V to - 60V peak signal from the coupling cap to the grid and grid return resistor.
Transformers do what they say on the tin - they transform voltage, current and impedance. They have no intrinsic impedance of their own but, due to practical limitations, they have an impedance range over which they work the best over the audio frequency range. So the transformer you have works best at 5K. So, if your transformer is a 5K:8 type then an 8 ohm load will be transformed by it so that looking into the primary winding it appears as 5K. But remember, transformers only work on ac, so at dc all you will see looking into the primary is the 170 ohms DCR and this is what you use when setting the bias in the way described in the post above.
Then for working out the power, you use the ac impedance reflected from the primary which in this case should be 5K
Cheers
Ian
Then for working out the power, you use the ac impedance reflected from the primary which in this case should be 5K
Cheers
Ian
The secondary impedance is fed back into the transformer primary relative to the number of turns ratio squared.
Hence leaving output unloaded causes havoc on primary.
Hence leaving output unloaded causes havoc on primary.
Actually answering your question, you consider the impedance, so 5k
Resistance IS a part of impedance and to be very very precise you "should" add it to the total, so 5170 ohm load but value is so small that you just ignore it.
Yes, but the OP wants to set the dc operating point so the impedance is not used for that.
Cheers
Ian
"OP wants to set the dc operating point so the impedance is not used for that."
Anode operating point is probably irrelevant with a transformer.
The grid/cathode end is more important for setting conditions.
Anode operating point is probably irrelevant with a transformer.
The grid/cathode end is more important for setting conditions.
Theoretically you could measure the DC voltage drop across the transformer primary to set idling anode current. The danger is that if you accidentally get a large AC signal while doing that poof goes your meter. By measuring across the cathode resistor any AC signal won’t get large enough to do damage. Especially if it’s bypassed.
Sorry but no.
Being a Class A amplifier, idle current must be set to Vswing/Z or Δ Voltage/Impedance , emphasis on Impedance.
Δ voltage is the swing available from idle plate voltage to saturation voltage.
Setting current to that value maximizes symmetrical voltage swing , so maximizes Power/Headrom.
Compared to that, DCR has a very minor influence: it will lower idle plate voltage a little.
Actually, if the output transformer is measured properly, you want to measure the complete impedance of the primary.
First, Load the 8 Ohm secondary with an 8 Ohm resistor.
Now, measure the "5k" primary impedance at 1kHz.
You will be measuring the complex impedance of the primary turns in series with the primary DCR.
The vacuum tube will see the same complete impedance of the primary turns in series with the primary DCR.
Output transformers are much more complex than that, especially at other frequencies.
Primary inductance, primary DCR, primary distributed capacitance, leakage inductance from primary to secondary, secondary inductance, secondary DCR, primary capacitance to secondary and laminations, secondary capacitance to primary and laminations, and perhaps I forgot a few (I probably did).
I am reminded of the output transformer measurements I did with a $50,000 vector network analyzer, and $10,000 Cal Kit.
I measured impedance, insertion loss, inductance, leakage inductance, distributed capacitance, un-loaded resonance, etc.
I wish I had that equipment at my disposal, I no longer do.
Please remember, if you calculate the power that the tube puts into an output transformer, you will get less power at the output tap.
5k total impedance primary with 250 Ohms DCR, and 8 Ohm secondary with 0.4 Ohms DCR.
The output transformer insertion loss will be approximately 1 dB at mid frequencies.
With 10 Watts in from the tube, output transformer's 1dB insertion loss, you will get about 7.94 Watts out.
A properly designed amplifier like that will still sound good.
Enjoy listening!
First, Load the 8 Ohm secondary with an 8 Ohm resistor.
Now, measure the "5k" primary impedance at 1kHz.
You will be measuring the complex impedance of the primary turns in series with the primary DCR.
The vacuum tube will see the same complete impedance of the primary turns in series with the primary DCR.
Output transformers are much more complex than that, especially at other frequencies.
Primary inductance, primary DCR, primary distributed capacitance, leakage inductance from primary to secondary, secondary inductance, secondary DCR, primary capacitance to secondary and laminations, secondary capacitance to primary and laminations, and perhaps I forgot a few (I probably did).
I am reminded of the output transformer measurements I did with a $50,000 vector network analyzer, and $10,000 Cal Kit.
I measured impedance, insertion loss, inductance, leakage inductance, distributed capacitance, un-loaded resonance, etc.
I wish I had that equipment at my disposal, I no longer do.
Please remember, if you calculate the power that the tube puts into an output transformer, you will get less power at the output tap.
5k total impedance primary with 250 Ohms DCR, and 8 Ohm secondary with 0.4 Ohms DCR.
The output transformer insertion loss will be approximately 1 dB at mid frequencies.
With 10 Watts in from the tube, output transformer's 1dB insertion loss, you will get about 7.94 Watts out.
A properly designed amplifier like that will still sound good.
Enjoy listening!
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I think we are talking at cross purposes. The optimum quiescent current is indeed calculated from the transformer primary impedance. But, setting the bias to that value for the tube in question requires finding the intersection of that plate current with the anode voltage (which will be as near as dammit B+). For this the transformer impedance is not required and the dcr only if you want to be pedantic.
Cheers
Ian
Thanks for confirming my point.
Worrying about DC resistance is indeed pedantic, unless so gross that it significantly drops +B
Not the usual case.
Take care, friend.
I have measured many single output transformers of several different manufacturers.
I do not have access to the really expensive ones.
But the ones I did measure often have DCR that is 5% to 10% of the primary impedance.
Take a 5k primary that has 250 Ohms DCR (5%).
Take a 300B, with rp = 700 Ohms
Set the bias so that the plate current is 60mA.
0.060 x 250 Ohms = 15V drop from B+, due to the primary DCR.
With 15V less B+, and rp = 700 Ohms, the plate current would drop by 15V / 700 Ohms = 21 mA
You predicted 60mA, but now you have 60mA - 21 mA = 39 mA.
You do care about the DCR.
To get that 21 mA back (to get 60 mA), you have to reduce the grid bias by about 4V; change the bias from -60V to -56V.
I would rather increase the B+ by 15V, instead of reducing the bias by 4V.
20 log (39V / 60V) = 3.74 dB loss of dynamic range (3.74 dB loss of output power).
Did I describe the effect correctly?
Thanks!
I do not have access to the really expensive ones.
But the ones I did measure often have DCR that is 5% to 10% of the primary impedance.
Take a 5k primary that has 250 Ohms DCR (5%).
Take a 300B, with rp = 700 Ohms
Set the bias so that the plate current is 60mA.
0.060 x 250 Ohms = 15V drop from B+, due to the primary DCR.
With 15V less B+, and rp = 700 Ohms, the plate current would drop by 15V / 700 Ohms = 21 mA
You predicted 60mA, but now you have 60mA - 21 mA = 39 mA.
You do care about the DCR.
To get that 21 mA back (to get 60 mA), you have to reduce the grid bias by about 4V; change the bias from -60V to -56V.
I would rather increase the B+ by 15V, instead of reducing the bias by 4V.
20 log (39V / 60V) = 3.74 dB loss of dynamic range (3.74 dB loss of output power).
Did I describe the effect correctly?
Thanks!
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Consider the DC resistance of the transformer as a loss. It is something unwanted but unavoidable. So is the leakage inductance. (Remember the ideal transformer in the textbooks?)
So far so good.
Now you are comparing apples to onions, not the same thing, by far.
IF you have 15V less of +B, that´s it, Rp is irrelevant there.
You will NOT drop "more" because of it.
Does it mean plate will saturate down to Ground if driven hard?
No, that is described by the saturation voltage (which I mentioned), a different thing.
You bias to whatever voltage is needed to get expected idle current, nothing new or special there.
Just curious how would you do that ... using a regulated Power Supply?
Your Math is really hard to follow ... if it means anything at all.
Sorry.
Sad to say, it only adds confusion.
. . . I was objecting to an earlier statement in the thread that said the DCR did not count.
1. Set the 300B plate to 300V; -61V bias; and it will have 60mA of plate current (the most often printed quiescent state in the many 300A and 300B historical Western Electric data sheets. It states that the plate resistance, rp, at that quiescent condition is 700 Ohms.
Now, put 250 Ohms in series from the 300V to the Plate (the DCR of the 5k primary, added into the circuit).
At 60mA, the 250 Ohms would drop 15 Volts.
Of course, when the 300B plate current drops to less than 60mA, the new voltage across the 250 Ohms DCR would be less than 15V, but there will be some drop in plate voltage, it will no longer be 300V.
With the lower plate voltage, the only ways to get 60mA plate current is:
Either to reduce the bias from -61V.
Or to raise the B+ at the top of the output transformer primary, to make up for the drop across the DCR.
Or to use a transformer that has less than 250 Ohms DCR in the primary.
So, if you raise the B+ from 300V to 315V, the desired plate current of 60mA will be restored.
2. Dynamic range:
A 300B has a u of 3.85.
If the bias is -39V, we can swing the grid up by 39V (0V grid to filament). That is as far as we can swing it before drawing grid current, and we are still in Class A1.
A sine wave that swings up by 39V, it also swings down by 39V. 2 x -39V (-78V grid to cathode). Of course the 300B will be at or near cutoff.
The peak swing is 39V.
If the bias is -61V, we can swing the grid up by 61V (0V grid to filament). That is as far as we can swing it before drawing grid current, and we are still in Class A1.
A sine wave that swings up by 61V, it also swings down by 61V. 2 x -61V (-122V grid to cathode). Of course the 300B will be at or near cutoff.
The peak swing is 61V.
Since the u is fairly constant, we can estimate the dynamic range by simply calculating the ratios of the 2 bias voltages, and the ratios of the peak signal voltage (which is equal to the bias voltage).
20 x Log (39V / 61V) = 3.885 dB less power with 39V peak swing, versus 61V peak swing.
All else being equal, which bias is likely to have more power out in Class A1?
3. How do I raise the B+ by 15V.
Easy, I know that I want 60mA plate current, and that I have an output transformer primary with a DCR of 250 Ohms, so I design the B+ to be 315V.
Engineering is often done up front, not after the fact, unless you planned poorly, or you are modifying something that already exists.
One way . . . Plan the output tube you are going to use, the quiescent state you want to use for the output tube, and the output transformer you are gong to use. With that method, only after that do you even start to design the B+ circuitry.
If you design the circuit to work the way you want, with 315V, 300V, or 285V, then OK, you do not need to worry about DCR.
And then, when planning the B+, it is also a good idea to know the power transformer's primary DCR and secondary DCR,
versus the peak currents, so you can predict the voltage to the rectifiers.
All things being equal (they never are), for the same DC current out, a capacitor input filter has much larger peak currents, than a choke input filter has. That means the DCR becomes more important for a cap input filter than for a choke input filter.
If the primary is 120V, and the secondary is 240V, the primary DCR is reflected to the secondary at a 2:1 ratio.
I hope those are a little clearer.
1. Set the 300B plate to 300V; -61V bias; and it will have 60mA of plate current (the most often printed quiescent state in the many 300A and 300B historical Western Electric data sheets. It states that the plate resistance, rp, at that quiescent condition is 700 Ohms.
Now, put 250 Ohms in series from the 300V to the Plate (the DCR of the 5k primary, added into the circuit).
At 60mA, the 250 Ohms would drop 15 Volts.
Of course, when the 300B plate current drops to less than 60mA, the new voltage across the 250 Ohms DCR would be less than 15V, but there will be some drop in plate voltage, it will no longer be 300V.
With the lower plate voltage, the only ways to get 60mA plate current is:
Either to reduce the bias from -61V.
Or to raise the B+ at the top of the output transformer primary, to make up for the drop across the DCR.
Or to use a transformer that has less than 250 Ohms DCR in the primary.
So, if you raise the B+ from 300V to 315V, the desired plate current of 60mA will be restored.
2. Dynamic range:
A 300B has a u of 3.85.
If the bias is -39V, we can swing the grid up by 39V (0V grid to filament). That is as far as we can swing it before drawing grid current, and we are still in Class A1.
A sine wave that swings up by 39V, it also swings down by 39V. 2 x -39V (-78V grid to cathode). Of course the 300B will be at or near cutoff.
The peak swing is 39V.
If the bias is -61V, we can swing the grid up by 61V (0V grid to filament). That is as far as we can swing it before drawing grid current, and we are still in Class A1.
A sine wave that swings up by 61V, it also swings down by 61V. 2 x -61V (-122V grid to cathode). Of course the 300B will be at or near cutoff.
The peak swing is 61V.
Since the u is fairly constant, we can estimate the dynamic range by simply calculating the ratios of the 2 bias voltages, and the ratios of the peak signal voltage (which is equal to the bias voltage).
20 x Log (39V / 61V) = 3.885 dB less power with 39V peak swing, versus 61V peak swing.
All else being equal, which bias is likely to have more power out in Class A1?
3. How do I raise the B+ by 15V.
Easy, I know that I want 60mA plate current, and that I have an output transformer primary with a DCR of 250 Ohms, so I design the B+ to be 315V.
Engineering is often done up front, not after the fact, unless you planned poorly, or you are modifying something that already exists.
One way . . . Plan the output tube you are going to use, the quiescent state you want to use for the output tube, and the output transformer you are gong to use. With that method, only after that do you even start to design the B+ circuitry.
If you design the circuit to work the way you want, with 315V, 300V, or 285V, then OK, you do not need to worry about DCR.
And then, when planning the B+, it is also a good idea to know the power transformer's primary DCR and secondary DCR,
versus the peak currents, so you can predict the voltage to the rectifiers.
All things being equal (they never are), for the same DC current out, a capacitor input filter has much larger peak currents, than a choke input filter has. That means the DCR becomes more important for a cap input filter than for a choke input filter.
If the primary is 120V, and the secondary is 240V, the primary DCR is reflected to the secondary at a 2:1 ratio.
I hope those are a little clearer.
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Alternatively, given a B+ of 300V, you work out the desired quiescent current to be 60mA. Then knowing the dcr of the transformer you calculate the actual anode voltage at 60mA . The intersection of this anode voltage and 60mA on the datasheet gives you your bias point.
Cheers
Ian
Cheers
Ian
ruffrecords,
Of course you can vary the bias voltage to get 60mA from a 300B.
And, keep the output in A1, no grid current.
Start with 300V B+ (fixed), and the output transformer has 100 Ohms DCR, and you set the 300B grid bias until you get 60mA plate current, then the plate voltage will be lower than 300V. 100 Ohms x 0.060A = 6V
300V - 6V = 294V on the plate.
With a WE 300B that is dead-on "perfect": 300V plate to filament, -61V grid to filament bias, and 60mA current . . .
The grid bias will be slightly less than -61V so that with only 294V plate, the current will still be 60mA.
Let's go a little further, and use 500 Ohm DCR, 60mA, 30V drop, 270V plate.
To get 60mA, we have to drop the bias several volts less than -61V (perhaps to -53V, perhaps not).
Whatever new bias voltage we come up with to have 60mA, we will have less peak volts from the driver until we go into A2, grid current.
I call that less dynamic range of power from the 300B (unless you want to use less than 5k Ohm primary, and more than 60mA plate current.
The plate to filament voltage, plate current, and grid bias are all related, pull on any one of those 3, and at least one other of those 3 has to change. All locked together.
Of course you can vary the bias voltage to get 60mA from a 300B.
And, keep the output in A1, no grid current.
Start with 300V B+ (fixed), and the output transformer has 100 Ohms DCR, and you set the 300B grid bias until you get 60mA plate current, then the plate voltage will be lower than 300V. 100 Ohms x 0.060A = 6V
300V - 6V = 294V on the plate.
With a WE 300B that is dead-on "perfect": 300V plate to filament, -61V grid to filament bias, and 60mA current . . .
The grid bias will be slightly less than -61V so that with only 294V plate, the current will still be 60mA.
Let's go a little further, and use 500 Ohm DCR, 60mA, 30V drop, 270V plate.
To get 60mA, we have to drop the bias several volts less than -61V (perhaps to -53V, perhaps not).
Whatever new bias voltage we come up with to have 60mA, we will have less peak volts from the driver until we go into A2, grid current.
I call that less dynamic range of power from the 300B (unless you want to use less than 5k Ohm primary, and more than 60mA plate current.
The plate to filament voltage, plate current, and grid bias are all related, pull on any one of those 3, and at least one other of those 3 has to change. All locked together.
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