Statistics: Posted by Augustica.com — 21 Dec 2016, 10:45
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Additional pairs of output tubes may be used but calculations remain the same. The output stage is fixed biased, but cathode biased amplifiers are usually fully cathode bypassed so treatment would also be the same. We will start with finding the open-loop gain from input to output before applying feedback. For a push-pull amplifier we can consider one half of the output stage when calculating gain.
Strictly speaking, because the pair of EL34 is not perfectly balanced, its gain will be different depending on which of its grids and anodes we consider. Nevertheless, for the sake of simplicity, we will assume the balance is fair and take the gain to one anode as a half the differential gain:
R = R5 = R6 = 100KOhm || 220KOhm = 69KOhm
Ra = 65KOhm
The voltage gain of each EL34 pentode is -GmZpri. The output transformer will usually be specified in terms of its anode-to-anode impedance, which might be 3.5KOhm for a pair of EL34, but we are only considering one half of the output stage. Assuming the output stage operates in class A or AB, the impedance seen by either tube is half the anode-to-anode impedance under small signal conditions.
From EL34 datasheet the Gm is about 10mA/V. Assuming this is a class AB amplifier with a 3.5KOhm output transformer, the magnitude of the gain of one side is:
If feedback is taken from the 8Ohm speaker tap then we find the turns ratio to be:
Thus, the voltage gain of the transformer is 1/20.9 = 0.048. The total open-loop voltage gain of the circuit is:
If we aim for a feedback factor of 10dB, then the gain of the amplifier will be reduced by a factor of:
Now we can find the feedback coefficient B by rearranging the universal feedback equation:
In other words, for every volt of signal appearing at the speaker, 0.1 volts will be fed back to the phase inverter via the potential divider formed by R2 and R10 on the schematic above. Rearranging the formula for a potential divider to find R10 gives:
The nearest standard resistor is 43KOhm. A variable resistor could be used to set the exact value to taste, and presence, resonance and variable feedback controls finally added.
Statistics: Posted by Augustica.com — 29 Sep 2013, 16:06
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The easiest way to select component values is to remember that a tube can be regarded as a signal generator in series with resistor R2 and the same rules that apply to parallel resistors also apply to tubes - the total resistance is less than either resistance alone - and total current increases while total voltage remains the same. If the triodes inside 6H30Pi are connected as shown above, then we can regard the whole arrangement as a single tube with twice the current handling ability. We can easily represent this on the anode characteristics of 6H30Pi by doubling the anode current scale. If we had connected three identical triodes, we would triple the current scale, and if four triodes would be connected in parallel, we would quadruple the current scale.
A single 6H30Pi triode has maximum anode dissipation limit of 4 Watt, thus two in parallel should have a total limit of 8 Watt, since currents are doubled. Keep in mind that because the anode voltage scale and grid curves are unchanged, Mu also remains unchanged. However, for a given change in grid voltage, the change in anode current is now doubled, so Gm is twice its original value. And since
Mu/Gm = Rp
it also follows that Rp (which is R2 in our example) is halved, and this is to be expected since two identical resistances placed in parallel make a total of one half their individual tubes. If we had used three tubes in parallel, we would find Gm tripled and Rp (which is R2 in our example) would be one-third its original value, while Mu would still remain unchanged. Exactly the same principles apply to pentodes.
Note that because the current has doubled, if we wanted the same gain and bias from a pair of parallel valves as a single tube, then we would need to use half the anode load and bias resistances, and the cathode bypass capacitor would need to be doubled in order to maintain the same frequency response.
These design principles are adopted in our Frigate headphone amplifier kit.
Parallel tubes are commonly found in hi-fi amplifiers since they give an improvement in the Signal-to-Noise Ratio (SNR). This happens because the two tubes combined can operate at lower anode current while still achieving the same gain as a single tube. Less anode current means less flicker noise. In fact, it is a general rule of electronics that every time we double the number of amplifying devices in parallel, we get a 3 dB improvement in SNR, in theory at least. In practice the figure may be better or worse than this, depending on how other circuit parameters change.
For example, headphone amplifier kit Frigate, which is using a parallel arrangement, operates at 70% of the current of the single stage, so the total anode noise current of both circuits is the same. But because the parallel stage has a lower anode resistance Ra, its output noise voltage is lower, while its gain is higher. These two elements combine to give an overall improvement in SNR of 4 dB. And even more importantly, the parallel stage will give lower harmonic distortion and more output swing.
Statistics: Posted by Augustica.com — 27 Sep 2013, 10:57
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Having selected our operating point at 85V, we know anode, and hence, cathode current. (In triode anode and cathode currents are equal). We know the value of Vgk we need - it has to be 2V below the potential of the cathode. If the greed is at 0V, the cathode has to be at +2V. If we know the voltage across, and the current through, an unknown resistor, we can find the value of the resistor through Ohm's law.
We know that the cathode voltage is 2.0V, so the cathode bias resistor R3 will have value of 363 Ohm. The closest standard value is 360 Ohm.
Statistics: Posted by Augustica.com — 07 Aug 2013, 13:21
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When we come to calculate the gain, we find that the anode characteristic begins to curve as we reach its intersection with the load line. It is perfectly valid to treat the anode curve as a straight line, and to project this line onto our load line in order to find the gain. Projection on the axis Va of intersection of the load line and the -4V curve gives us 85V. Projection on the axis Va of intersection of the load line and the -8V curve gives us 170V. The gain is:
R5 is chosen either by a detailed perusal of the datasheet, or by observing that, in general, the anode current is a fixed ratio of G2 current. For the EL84 this ratio is approximately 9:1 (48mA divided by 5.5mA). Therefore, if the anode voltage and the G2 voltage are to be the same, the G2 resistor R5 has to be nine times larger than anode resistor R4. We select R5 at 56KOhm.
Now let's calculate value of C4 in order to hold second grid at AC ground potential. Using the curves for the EL84, at Va=130V, Vg=-6V, Ra=5.2KOhm, we calculate resistance of the second grid R2 at approximately 20KOhm. This resistance of 20KOhm is in parallel with R5 of 56KOhm, giving a final resistance of 15KOhm. Placing the value of 15KOhm to the formula for low cut off filter capacitor calculation gives us value of C4 equal to 10uF. F in the formula below represents a cut-off frequency and is selected to be 1 Hz.
Evaluating Gm from the datasheet, by holding anode voltage constant, and measuring the change in anode current for grid voltage, produces value of 11.3 mA/V. For the pentode, the cathode resistance Rk=1/Gm and gives us R3=100 Ohm. In turn, the value of C3 will be equal to 150 uF for a 1 Hz cut-off frequency.
Statistics: Posted by Augustica.com — 05 Aug 2013, 22:42
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Because the output signal of the first tube of the cathode follower is derived from changing Ia flowing through R4, and Ia also flows through R3, there will also be a voltage across R3. In order to restore full gain, we must suppress the feedback voltage produced at the cathode using a decoupling or bypass capacitor C3. The bypass capacitor C3, together with resistor R3, will form a local low-pass filter.
Let's calculate the resistance between the anode of the first tube and the ground. Let's title this resistance Rk.
Mu for 6922 was previously calculated here and is equal to 32.5. And Ra for 6922 was previously calculated here and is equal to 3.25.
In parallel with 360 Ohm cathode bias resistor R3 this gives us a total resistance Rk of 245 Ohm.
To calculate capacitance of the cathode bypass capacitor C3 we will use the following formula where f is the filter's cut off frequency which we select to be equal to 1 Hz.
Accordingly C3 has a value of 650uF. The closest standard value to 650 uF is 680 uF.
Statistics: Posted by Augustica.com — 05 Aug 2013, 13:02
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Cascode operates as follows - the upper tube has anode load resistor R4, however, instead of modulating signal by changing the grid voltage, and holding the cathode voltage constant, we vary cathode voltage and hold the grid voltage constant. The grid of the upper tube is biased to the voltage that is required for linear operation of the upper tube, and is held to AC ground by the capacitor C3.
The lower tube operates as a normal common cathode circuit, except that it has at its anode load the cathode of the upper tube. Because the lower tube has a low value of load resistance, it would generate considerable distortion if it were allowed to swing over a wide range. Conveniently, most of the gain is provided by the upper tube, and distortion of the lower stage should not be a significant problem.
It is appropriate to operate lower tube's anode at 75 Volt. Thus, if the high tention voltage is 285V, the voltage across the tube is 210V as shown on the 6922's diagram below. We can chose an anode load for the upper tube and draw a load line in the usual way. In our example R4 is 100KOhm, and Vg is -2.5V, giving us Va 76.5V. The corresponding anode current is 1.35mA.
If the anode of the lower tube is operated at 75V, and the upper tube has Vg of -2.5V, then the grid of the upper tube must have potential of 72.5V. Because the grid of the upper tube does not draw any current, its voltage determines the operating conditions of the upper tube.
Attempts to investigate the lower stage using anode characteristics of the 6922 is not very helpful. Instead, we will use curves demonstrating correlation between anode current and grid voltage.
We know that Va of the lower tube is at 75V, so we can look along the curve for Va = 75V (blue curve on the graph) until we come to the point where Ia = 1.35mA (upper and lower anode currents are equal), this is the operating point of the lower tube, and it gives us corresponding Vg of about 2.5V. Plotting the point (red dot) with coordinates Va = 75V and Ia = 1.35Ma on the anode characteristics gives Vg of approximately 2.5V. Accordingly, using Ohm's law, we find that R3 is 1.8KOhm.
Statistics: Posted by Augustica.com — 02 Aug 2013, 11:36
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Mu = amplification factor (no units)
Gm = transconductance (mA/V)
Ra = anode resistance (KOhm)
1. Determining Amplification
The amplification factor Mu of 6922 is ratio of the change in anode voltage Va to the change in grid voltage Vg with anode current held constant.
We can measure Mu at the operating point of 6922 by drawing a horizontal line through the operating point and calculating the gain.
Typically, Mu is one of the more stable tube parameters and varies little with anode current.
2. Determining Trasconductance
The transconductance Gm is the ratio of the change in anode current Ia to the change in grid voltage Vg with anode voltage held constant.
To find Gm we draw a vertical line through the operating point and measure the change in anode current.
3. Determining Anode Resistance
Anode resistance Ra is the ratio of the change in anode voltage Va to the change in anode current Ia with grid voltage held constant.
There is a very useful equation which links Mu, Gm, and Ra together:
Of course, we can arrange this equation as necessary to find the third parameter if we know the other two.
Statistics: Posted by Augustica.com — 01 Aug 2013, 08:35
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Typically there are two most important and usually conflicting parameters - maximum voltage swing and linearity. Since we are designing an audio amplification circuit, we are more concerned with linearity than with voltage swing. Triodes mainly produce second harmonic distortion, which is generated by the amplifier having unequal gain on the positive half cycle of the waveform compared to the negative half cycle, and the distortion is directly proportional to amplitude. In order to maximize our amplifier's linearity, we should select an operating point where the distance to the first grid line either side of the operating point is equal. Looking at the 6922's chart below, we see that biasing the anode voltage at 85 Volt by applying -2.0 Volt to the grid will provide for a operating meeting the above mentioned requirements.
We will now determine the dynamic or AC conditions of the circuit to verify that the parameters satisfy our requirements.
1. Voltage Amplification
The first parameter is voltage amplification Av, also called, gain. We determine gain by looking an equal distance either side of the operating point to the first intersection with a grid line, noting the corresponding anode voltage. Referring to the 6922's chart above, if we move from the operating point to the right, we meet the 3 Volt grid line, which intersects at a voltage of 110 Volt, and the 1 Volt line that intersects at 55 Volt. Amplification Av is the change in anode voltage divided by the change in grid voltage and equals to 27.5. The minus sign signifies the fact that the amplifier is inverting, but the minus has no significance for our calculations.
2. Maximum Undistorted Voltage Swing
Again, we look symmetrically either side of the operating point, but this time we look for the first limiting value. In our example, we look to the left and see that at 55 Volt we are approaching, but not reaching, the region of positive grid current. This is our first limit. Looking to the right, we see that there is no limit until we reach the point of 250 Volt. Although, this means that the 6922 ca swing a large voltage positively, it cannot swing as far negatively. Thus, the first limit of 55 Volt is our operating limit. We now can see that the maximum undistorted peak to peak swing at the output is double that of the distance from the bias point to the first limit or 27.5 Volt peak to peak. Taking in consideration that AC voltages are specified as RMS value of a sine wave, we should divide this number by a factor of 2 multiplied by square root of 2. The resulting value is approximately 10 Volt RMS.
3. Output Resistance Ra
To find Ra we return to the anode characteristics for 6922 and draw a tangent to the curve where it touches the operating point and intersects the axis of Ua and the top boarder of the 6922's chart. Then we will measure the gradient of the curve at those points.
The intersection of the tangent line the with axis Ua gives us point with the values Ia = 0 mA and Ua = 65 Volt. The intersection of the tangent line the with the top boarder of the 6922's chart gives us point with the values Ia = 20 mA and Ua = 130 Volt.
Plugging these numbers in the formula gives us anode resistance of 3.25 KOhm.
Proceed to the next chapter.
Statistics: Posted by Augustica.com — 30 Jul 2013, 14:14
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1. Grid Bias Voltage
In the circuit below 6922 is biased by applying a bias voltage to the grid via R2.
Returning back to the 6922 load line depicted below, we find that as anode voltage Va rises, the grid curves are getting closer to each other. This is indication of non-linearity. Non-linearity is particularly profound close to the 250 Volt point. The region around 250 Volt point is called cut-off point. Designing a circuit to operate around the cut-off point is not advisable if a good linearity of an amplifier is required.
2. Maximum Allowable Anode Voltage and Anode Dissipation
Moving in the opposite direction along the load line, we use the 6922 harder and harder, until finally there is no voltage across it. If however, we have a voltage across the 6922, and a current flowing through it, we are dissipating power within the tube. The limit of power that the 6922 can dissipate is known as the maximum anode dissipation and it appears on the 6922 datasheet. The 6922's anode dissipation limit is 1.5 Watt for each anode.
The 6922 datasheet also specifies two additional, and connected, restrictions for the choice of the bias point. They are maximum anode voltage Va and Va(b). Maximum Va for the 6922 is the maximum DC voltage at which the 6922 could be continuously operated and is 250VDC. The Va(b) is the maximum voltage to which 6922's anode is allowed to swing under signal or cold conditions and is about 300 VDC for the 6922. Exceeding the Va and Va(b) limits usually results in destruction of the 6922 tube.
3. Maximum Allowable Cathode Current
The final limitation is the maximum allowable cathode current Ic. The maximum allowable cathode current for the 6922 tube is 100 mA.
Proceed to the next chapter.
Statistics: Posted by Augustica.com — 29 Jul 2013, 09:48
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Let's design a load line of a cathode follower based on 6922 tube. Cathode follower has a voltage gain of slightly less than 1, a low output resistance of about 1 KOhm, a high output resistance of about 500 MOhm, and is a non-inverting circuit.
1. Designing the Gain Stage
The first two design considerations are the 6922's the plate current and plate voltage. The plate current is selected at 8.5mA and plate voltage is 250V. The load line of the 6922 (blue line on the graph below) is defined by two points - the plate current of 8.5mA and the plate voltage of 250V. From the 6922's average plate characteristics graph below, select a linear region as the idle operation point. The point of intersection of the curve representing -2V grid bias line and the load line represents a very linear region. The red dot is depicting the idle operation point of the gain stage. At the selected idle operating point the grid voltage is -2V, the plate voltage is about 85V, and the plate current is at 5.5mA. This voltage limit between heater and cathode has to be taken into consideration. We select the cathode voltage to be at 90V so that the 150V limit is not exceeded.
With these numbers we can calculate the resistors values in the schematics by using Ohm's law.
Let's start with plate resistor. Assuming that the 6922 is open and there is current flow through the tube, all plate voltage will drop across the plate resistor and none across the tube. Using V = IR, we have:
250V = 8.5mA x Rplate
Rplate = 250V / 8.5mA = 30KOhm
Now let's calculate cathode resistor. Since the grid voltage has to be at -2V below cathode's voltage, and the grid is grounded via a 1MOhm resistor, the cathode's voltage has to be at 2V above the ground. For idle current of 5.5mA, the cathode's resistor value is:
Rcathode = 2V / 5.5mA = 363 Ohm
The gain of the Gain Stage is calculated according to the formula :
Gain = (Tube Gain x Rplate)/(Rplate + Plate Resistance + ((Tube Gain +1) * Rcathode)).
The 6922 has plate resistance of 3 KOhm. The 6922 has gain of 33. Plugging these numbers in to the formula above gives Gain Stage gain of 22.
2. Designing the Cathode Follower Stage Load
When designing the Cathode Follower Stage, keep in mind that the 6922's maximum difference between cathode voltage to heater voltage is limited to 150V. This voltage limit between cathode and heater has to be taken into consideration. We select the cathode voltage to be 90V which is significantly below the 150V limit.
A steeper load line will be chosen for the cathode follower stage in order to obtain higher output current.
Rcathode = 250V = 25mA x Rcathode
Rcathode = 250V / 25mA = 10KOhm.
Cathode bias resistor is calculated according to the formula:
Rcb = Bias voltage / Idle current
Rcb = 4V / 8.5mA = 470 ohm
At the idle point, 6922 will lose about 160V as seen from the graph and the cathode voltage at the idle point will be equal to 250V - 160V = 90V. The grid potential will be 4V below the cathode or 86V.
Proceed to the next chapter.
Statistics: Posted by Augustica.com — 26 Jul 2013, 17:18
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1.Triodes
In a triode the relationship between Mu (voltage gain), Gm (transconductance) and Rp (plate resistance) is determined by the Ohm's law:
Mu = Gm x Rp
Gm is measured in the units of mhos or amps per volt. Rp is measured in the units of Ohms or Volts per Amperes. When these two parameters are multiplied, their units cancel, making Mu (voltage gain) unit-less parameter. A Mu of 25 means that voltage of 1 Volt applied to grid of a vacuum tube is amplifier 25 times and increases to 25 Volt on anode of the tube.
For triodes, Mu (voltage gain) is the most important parameter to measure and is the parameter that makes the biggest impact on sound quality of amplifier. Since Mu is what defines voltage gain of a tube, and the voltage gain is what we hear, matching Mu for tubes in the right and left channel is the most important task in a triode based amplifier. For example, let's say we have an amplifier where a triode in the right channel and a triode in the left channel are matched by their Gm (transconductance) values. However, there is no guarantee that these two triodes will have equal voltage gain, because Mu (voltage gain) is the product of multiplication of Gm and Rp. Since Mu is a product of multiplication of Gm and Rp, the triodes in the right and left channel may have unequal values of Mu. And if we have a triode with a Mu of 25 in the right channel of an amplifier and a triode with a MU of 31 in left channel, we will have a 2 dB imbalance in this amplifier. Accordingly, Rp (plate resistance) must also be measured and matched.Thus, for a triode Rp (plate resistance) is the second most important parameter to measure and match as it is directly affects Mu, and also directly affects the output impedance of an amplifier. For these reasons amplifier designers are primarily concerned with Mu, followed by Rp, and hardly ever at all consider Gm. Presently, there is only one tube tester - Amplitrex - that gives us ability to measure and match Rp.
The two examples below illustrate importance of Rp in matching triodes. The first example represents measurements of two well matched triode sections of the same 6922 tube as well as a plot of plate voltage against related plate current for each section of the 6922. The measurements are made on Amplitrex tube tester.
The second example represents measurements of two poorly matched triode sections of the same 6922 tube as well as a plot of plate voltage against related plate current for each section of the 6922.
In the first example sections of the 6922 have Mu of 29.7 and 28.1 respectively. This is a significant mismatch of Mu for 6922 tubes and without consideration of plate resistance would probably render this 6922 as poorly matched. However, the plate resistance of both sections is the same - 2.2 KOhm. The plate resistance offsets the Mu disbalance and produces perfectly matched sections of the 6922.
In the second example sections of the 6922 have Mu of 26.2 and 28.3 respectively. This is also a significant mismatch of Mu for 6922 tubes which is further compounded by difference in the plate resistance of 300 Ohm. As the result, the Mu disbalance and the difference in plate resistances produce profoundly mismatched sections of the 6922. Take a note that the plot represents the correlation of the plate voltages and the related plate currents between 45V and 90V. Once the plate voltage increases to more usual for the 6922 150V - 200V operating range, the mismatch between the two sections will become even more profound.
2.Pentodes
Pentodes are primarily installed in the output (power) stages of amplifiers. Voltage gain of a pentode is defined by pentode's Gm multiplied by the impedance of an output transformer connected to the pentode. For this reason, Gm (transconductance) must be matched for optimum performance of pentode within the same channel and pares of pentodes in the right and left channels. Further, Gm of a pentode (or pares of pentodes) has to stay matched over the range of bias currents of an installation of that particular pentode or pentode pare. Typically, pentodes that are matched for both their bias current and for their Gm (transconductance) tend to stay matched over a wide range of bias currents applicable to power stages of amplifiers.
Statistics: Posted by Augustica.com — 24 Jul 2013, 10:35
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