With respect to Nobu Shishido. KT88-based Push-Pull Amplifier PDF

Title With respect to Nobu Shishido. KT88-based Push-Pull Amplifier
Author Khoi Anh
Course Operating Systems
Institution HCMC University of Technology
Pages 45
File Size 4.6 MB
File Type PDF
Total Downloads 65
Total Views 158

Summary

I research some electrical device.Computer Operators support the users of their employer’s computer systems by working to improve and maintain important computer processes. Their responsibilities may vary depending on their employer’s size and technical infrastructure but usually include the followi...


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1/25/2021

With respect to Nobu Shishido. KT88-based Push-Pull Amplifier

With respect to Nobu Shishido. KT88-based Push-Pull Amplifier. The scheme of a push-pull amplifier (the input stage and driver which made based on this scheme, developed by genius of sound - Nobu Shishido (WAVAC audio lab)), was developed by famous tube designer - Michael Bron, ideas and sensible comments which helped me to implement this project, for which I am very thankful to him!

Content Schematic of the Amplifier Schematic of the Amplifier Module Schematic of the Power Supply Module Design of the Power Supply Design of the Anode Rectifier Static Load operation mode of the Anode Rectifier Design of the Choke Filter Calculation of the Anode Resistors The Anode Voltage Delay circuit Design of Rectifier for fixed grid bias Design of the output stage of the Amplifier Design of the input stage of the Amplifier Design of the driver stage of the Amplifier Parts List of the Amplifier Mechanical Parts Electromechanical Parts Electronic Parts Mechanical Circuits Photo Gallery Appendix The standard range of diameters of drills on metal (mm) How to read screw sizes in inch Correspondence between the U.S. and the European diameters of wire

Schematic of the Amplifier The Amplifier consists of two modules: the module of Amplifier is shown in Fig. 1 and the module of Power Supply is shown in Fig. 2.

Schematic of the Amplifier Module

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Fig. 1. Schematic of the Amplifier Module

Schematic of the Power Supply Module

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Fig. 2. Schematic of the Power Supply Module

Design of the Module of Power Supply Design of the Anode Rectifier (Section "A") The basis of this module - power transformer МЕ–225 manufacturd by ISO Tango.

Fig. 3. Power Transformer МЕ–225 with the following parameters:

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Primary winding voltage (RMS) U1AC = 230V Secondary winding voltages (RMS) U2AC = 400V–360V–0–100V–360V–400V (for power supply the anode circuits the taps 360V is used). The rated current of the anode winding through the 400V tap - I2AC = 0.225A. The power consumption of the transformer (calculated based on the secondary windings): P2 = 2 x 5.0V х 3.3A + 6.3V х 3.3A + 10V x 3.3A + 400V x 0.225A = 177VA Calculation of power consumption of the anode and filament circuits Anode Winding Resting Current of the output tubes: 2 х 65mА = 130mА Resting Current of the driver tube: 27mА Resting Current of the input tubes: 3.8mА Current of the bias of filament of the "top" tube of the input stage: 2.5mА Total Resting Current in the Anode Winding: 130 + 27 + 3.8 + 2.5 = 163.3mА (164mA). Voltage on the Anode Winding: U2AC = 360V Power Consumption of the Anode Winding: I2AC х U2AC = 0.164A х 360V = 59VA. Filament Windings Filament Current Filament Current Filament Current Filament Current transformer)

of of of of

the GZ34 Rectifier: 1.9A (both – 3.8А) the КТ88: 1.6А (both – 3.2А) the EL38: 1.4А the 6J5G: 0.3А (the calculation takes into account only "top" tube of the SRPP stage, because "bottom" tube filaments is powered by a separate

Total Filament Current: 3.8А + 3.2А + 1.4А + 0.3А = 8.7А. Power Consumption of the Filament Winding: 5.0V х 3.8А + 6.3V x 3.2А + 6.3V x (1.4А + 0.3А) = 19 + 20.6 + 10.7 = 50.3VA. Total Power Consumption of the Secondary Windings: Р2 = 59VA + 50.3VA = 109.3VA. The Power Trancformer Connections Filament Windings of 0–5V 3.3А connected in parallel for supply the filament of both rectifier tubes. Filament Winding of 0–5.0V–6.3V 3.3A with 6.3V tape is used for supply the filament of the "top" triode of the input stage and for driver tube. One of the outputs of this winding is connected to the voltage divider, so that half of the anode voltage of the input stage (DC bias) raises the potential of filament with a view to remove the potential difference between the cathode and the filament. Filament Winding 0–6.3V–10.0V 3.3A with 6.3V tape is used supply the filament of the output (Push-Pull) stage. Due to the DC bias is not supplied to the filament of the "bottom" tube of the SRPP, then the filament of the "bottom" tube, as well as anode voltage delay circuit is powered from a separate filament transformer T2 - 266JB6, manufactured by Hammond.

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Measured resistance of one half of the anode winding of the transformer = 41.3Ω (400V tape) and 37.2Ω (360V tape), the second half of the anode winding – 43.3Ω (400V tape) and 39Ω (360V tape), so the average value of resistance of the half of anode winding of the transformer RТР2 = 42.3Ω (400V tape) или 38.1Ω (360V tape) can be taken for further calculations. The transformation ratio (ratio of the number of turns of the primary winding to secondary windings or ratio of the voltage on the primary winding to the secondary winding) of the anode winding (360V tape): nР = UА / U2АС = 230V / 360V = 0.64. The measured resistance of the primary winding of the transformer RТР1 = 4.4Ω. Resistance of the transformer from the secondary (anode) winding: RТР = RТР2 + RТР1 / nР = 48.8Ω.

Static Load operation mode of the Anode Rectifier If there is no input audio signal (idle mode), the amplifier is the static load for the rectifier with total current of anode circuit IР = 164mА and total current of filament circuit IF = 8.7А.

Fig. 4. Circuit of the rectifier operating on the static load Voltage drop on the anode winding of the transformer Anode circuit current IР = 164mА, flowing through the anode winding of the transformer with active resistance 48.8Ω, lead to voltage drop across this winding: 0.164А х 48.8Ω = 8V. Therefore, the voltage UР, supplied to the anodes of the rectifier, will be: U2АС – 8V = 352V. Voltage drop on the rectifier.

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Because two rectifiers are connected in parallel, through an one diode will flow only half the current, i.e. 164 mА / 2 = 82mА. For GZ34 tube (see. Fig 5) for current 0.082А voltage drop on one diode is 13.5V.

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Fig. 5. Plate Characteristics GZ34 (by Philips Data Handbook) from frank.pocnet site) Thus the total voltage drop on the active resistance of the anode winding of the transformer and rectifiers is: ΔU = 8V + 13.5V = 21.5V. Direct voltage applied to the anodes of the rectifier in idle mode: UP0 = √2 х U2AC = √2 х 360V = 509V. A first capacitor of the filter in idle mode is charged prior to this voltage. Operating voltage of the first capacitor of the filter should be about 10% more than this voltage, i.e. 509 + (509 х 0.1) = 560V (600V). Since the anode winding and the first capacitor of the filter are connected to the rectifier in series, then when the negative half cycle of the voltage applied to the anode (rectifier is closed) the cathode of the rectifier is under a positive voltage of the first capacitor of the filter. Thus, a voltage between the anode and the cathode of the closed rectifier is double of the amplitude of the anode winding (Peak Inverse Voltage): UPEAK = 2 х UP0 = 2 х 509 = 1018V. Voltage on the cathode of the rectifier. UК = √2 x (U2AC – ΔU) = √2 x (360V – 21.5V) = 479V. The amplitude of the voltage ripple across the capacitor С1 (47μF): UC1~ = IOUT / (2 x fC x C) = 0.164 / (2 x 50 x 47e–6) = 35V (p–p). DC voltage on the capacitor С1: UС1 = UК – UC1~/2 = 479 – 35/2 = 461V. The load of the rectifier can be assumed as: RН = UOUT / IOUT = 461 / 0.164 = 2811Ω. (with taking into account the 40Ω resistance of the choke, the resistance of the rectifier load becomes equal to 2851Ω).

Design of the Choke Filter (Section "B") To further reduce of the ripple of the voltage, the choke filter (see. Fig. 6) is used. The filter is based on the choke LC–3–350D manufacturd by ISO Tango. with following parameters: L = 3H. IRATED = 350мА IMAX = 450мА R = 40Ω

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Fig. 6. The Choke Filter Since the choke has an active resistance, the voltage at the filter output (UC2) is less than the input voltage (UС1) by the IР х 40Ω. For 164mА load the voltage drop becomes equal to 6.6V, so the voltage on the capacitor С2 at 164mА load will be 454.4V. Filtration factor of the circuit КF = 4 х π2 х f2 x L x C2, where f – frequency of pulsation of the filtered voltage (for a full-wave rectifier circuit the frequency of pulsation is 100Hz). L – inductance of the choke, H. С – capacity of the capacipor, following the choke (С2), F. shows by how many times the ripple voltage at the output of the filter is less the ripple voltage at the input of the filter, i.e. КF = UC1~ / UC2~. Thus, for the capacitor С2 = 470μF, the КF= 4 х π2 х 1002 x 3 x 470e–6 = 556.6 and ripple voltage at the output of the filter UC2~ = UC1~ / КF = 35 / 556.6 = 0.063Vp–p. The operating voltage of the capacitor at the output of the choke in accordance of the slight ripple voltage may be selected by approximately 5% greater than the output voltage of the filter, i.e. = 454.4V + 0.05 х 454.4V = 477V (it is possible to use a capacitor with a standard operating voltage - 550V). Further filtering of ripple can be achieved by using the rejector, consisting of a choke L1 and capacitor C3, connected by parallel to the choke (parallel resonance circuit). The circuit has a maximum resistance on the 100 Hz resonant frequency. The circuit can be calculated on the basis of the following rule. The condition of the resonance of currents: YC = YL (where Y is conductivity), so ωC = 1/ωL, so ω = 1/√(LC). Given, that ω = 2πf, the f (100 Hz) = 1/(2π√(LC)). For the 3 H choke, the shunt capacitor value will be equal: CS = 1/(L x (2 x π x f)2) = 1/(3 x ((2π x 100)2)) = 0.844μF (a standard value is 0.82μF). The minimum value of the current flowing through the Choke: IMIN = 2 x √2 x UC2 / (6 x π2 x f x L) = 2 x √2 x 461V / (6 х π2 х 100 x 3) = 73mА. when value of the load current is less than this minimum value, then the smoothing capacitor, connected after the choke, will be charged to the peak value of the voltage at the cathode of the rectifier under load, i.e.. up to 479V.

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Estimated value of the anode voltage of the output stage: UB1 = 452V at IB1 = 130mА. Estimated value of the anode voltage of driver stage UB2 = 320V at IB3 = 27mА thus, the value of the driver tube anode resistor is equal to (UB1 – UB2) / (27mА + 4mА + 3mА) = 3.9кΩ. Power dissipation of this resistor: (UB1 – UB2) х (27mА + 4mА + 3mА) = 4.5W Estimated value of the anode voltage of the input stage UB3 = 250V at IB3 = 4mА thus, the value of the input stage anode resistor is equal to (UB2 – UB3) / (4mА + 3mА) = 10кΩ. Power dissipation of this resistor (UB2 – UB3) х (4mА + 3mА) = 0.5W A current, flowing through a DC bias voltage divider is: I = 3мА, thus, the value of the resistance of this divider is equal to UB3 / 3mА = 83кΩ.

Calculation of the delay circuit of the anode voltage (Section "С") Response time of the circuit τ = C x (R1 x R2 / (R1 + R2)). So, with the values С = 100μF, R1 = 470кΩ, R2 = 680кΩ, the τ is 28 seconds.

Design of the grid fixed bias rectifier (Section "D") The grid bias range is: UBIAS = {–35 ... –70}V, i.e. voltage drop across the regulating resistor will be 30V. AC input voltage of the rectifier: U~ = 100V. DC output of the rectifier: U= = √2 х 100V – U drop on diode = 141V – 1.0V = 140V. The DC Filter resistor: RF = 10кΩ. Common current of two voltage dividers: I0 = 6mА, so voltage drop on the DC Filter resistor: UR = 10кΩ x 6mА = 60V. Thus, the voltage applied to the two dividers: U0 = √2 x 100V – Udiode – UR = 141 – 1.0 – 60 = 80V, and resistance of one divider: R = U0 / (I0 / 2) = 80V / 3mА = 27кΩ. A current, flowing through each divider: I1 = I2 = 6mА / 2 = 3mA. A bottom resistor of the divider is chosen from the condition of the lower limit value of the bias voltage -35V: 35V / 3mА = 11.7кΩ (a standard value of 12кΩ is used, the lower limit value of the bias voltage will be –36V). The potentiometer of the divider should provide a change in voltage from 36V to 70V, so the voltage drop across the potentiometer will be 70V - 36V = 34V, at a current of 3mA, the voltage drop determines its resistance equal to 34V / 3mA = 11.3kΩ. (a potentiometer of 10кΩ is used, and the range of grid bias voltage adjustment is 10кΩ х 3mА = 30V). A top resistor of the divider is: 27кΩ – (12кΩ + 10кΩ) = 5кΩ (a standard value of 5.1кΩ is used). The DC filter resistor RF power dissipation is: 10кΩ х 6mА2 = 0.36W.

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Design of the Output Stage of the Amplifier Since the output stage utilized a ultralinear connection to the specified output transformer XE-60-5, manufactured by ISO Tango then the calculation reduces to determining the idle current and power dissipation of the stage.

Рис. 7. Operation mode of the KT88 tube in push-pull output stage (the tube datasheet (by The General Electric CO. LTD of England) can be downloaded from frank.pocnet site) First point of the loadline IА (UА = 0) = EА / RА, где RА is determined based on the resistance RА–А of the output transformer XE–60–5 (5кΩ), recalculated for one tube: RА = RА–А / 4 = 1.250кΩ. Then IА (UА = 0) = 452 / 1.250 = 362mА. Second point of the loadline UА(IА = 0) = EА = 452V.

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The point "Р" is defined at the intersection of the loadline with the characteristic at UС = 0, wherein IА max = 328mA, UА min = 42V. The idle current of the tube IА0 = ~(1/3 ... 1/5) IА max / 2 = 65mА (Point "Т") is located at the intersection of the loadline with the characteristic at UС approximately equal -43V, this is a voltage offset of the tube in idle mode. Point "Т" determines a plate voltage in the idle mode: UА0 = 370V, relative to the idle current of the tube IА0. Resistance in the anode circuit of two tubes: RА–А = 22 x (UА0 – UА min) / (IА max – IА0) = 4 x (370 – 42) / (0.328 – 0.065) = 5кΩ. Power dissipation of the anode: PA = UА0 x IА0 < PA max = 370 х 0.065 = 24W < 40W. The maximum power given to the load by two tubes with the efficiency of the ultralinear connection ~60%: P~ = (IА max x (UА0 – UА min) x η) / 2 = (0.328 x (370 – 42) x 0.60) / 2 = 32W. The amplitude of the AC component of the anode current of the tube: ImА = (IА max – IА0) / 2 = (328 – 65) / 2 = 132mА. Max. effective value of of the anode current of the tube at maximum power: IА0 max = (IА max + 2 x IА0) / 4 = (328 + 2 x 65) / 4 = 115mА. Max effective value of of the anode current flowing through output transformer Imax = 2 х IА0 max = 230mА.

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Fig. 8. Creation of IА (UG) characteristics of tube KT88 of the Push-Pull output stage (the tube datasheet (by The General Electric CO. LTD of England) can be downloaded from frank.pocnet site) The peculiarity of this output stage is the presence of feedback, which is provided from the output transformer to the cathodes of the output tubes (i.e. "super-triode" connection). Details on this scheme is described in "Universal system and output transformer for valve amplifiers" brochure, see Menno van der Veen site.

Design of the Input Stage of the Amplifier The input stage is created based on Shunt Regulated Push-Pull (SRPP) design.

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Fig. 9. Schematic of the input stage

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Fig. 10. Plate characteristics of the 6J5G tube (the tube datasheet (by RCA) can be downloaded from frank.pocnet site) For a given idle current 4mA, flowing through the bottom tube, the grid voltage of the tube will be 4V, so resistance of the self-bias in the cathode of the bottom tube (as well as of the top tube) will be 4V/4mA = 1kΩ.

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Amplification factor of this stage, assuming that as a "top" and "bottom" are used identical tubes, and the cathode resistor of the "bottom" tube is shunted by a capacitor: А = μ х (rА2 + RК2 х (μ + 1)) / (rА1 + rА2 + RК2 х (μ + 1)) = 20 х (8000 + 1000 х (20 + 1)) / (8000 + 8000 + 1000 х (20 + 1)) = 15.7. Where: rА1 – internal resistance of the "bottom" tube rА2 – internal resistance of the "top" tube RK2 – resistance of the self-bias of the "top" tube μ – amplification factor (gain) of the tube The amplifier is designed for a nominal input voltage of the audio signal ~1.0VP–P, so at this level of the input signal, the output voltage of this stage will be 1.0 х 15.7 = 15.7VP–P. Since the connection between the input stage and the driver stage is direct, so the value of the grid voltage of the driver tube will be: UК + 15.7/2 = 125+7.85 = 133V.

Design of the Driver Stage of the Amplifier As noted earlier, the voltage of bias UК of the driver tube (voltage drop on the cathode resistor) should be at least 133V. With selected current, flowing through anode of the tube А0 = 27mА, cathode resistor of the tube RК will be 133/27 = 5кΩ. Power dissipation of the cathode resistor PRК = UК х IА0 = 133V x 0.027mА = 3.6W.

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