Field effect transistors are increasingly used in digital audio today. Its principles, advantages, and common sense have been discussed in some reference books and newspapers, and will not be repeated here. This article illustrates the rational application of field effect transistors through two points that are easily overlooked by enthusiasts, especially beginners. At present, the field effect transistors used in the audio field include junction transistors (JFET) and insulated gate field effect transistors (MOsFET), and the latter are divided into LDMOS, VMOS and UHC and IGBT that have appeared in recent years, and are still developing With perfection.
1. JFET lacks complementary tubes within matching tolerances. The maturity of today's bipolar transistor manufacturing process has reduced the pairing error of NPN and PNP complementary transistors to a level generally accepted by the majority of professional manufacturers and audiophiles. In contrast, the selection of the field effect tube is much more difficult, and the JFET used as the input stage of the amplifier lacks complementary pairs that meet the requirements (this is determined by the current manufacturing level). The attached table lists the main characteristic data comparison of Toshiba's twin field effect transistor (DualFET) K389 / Jl09. As can be seen from the attached table, the differences between K389 and J109 are V C C and NF, of which C and C are two values, and the difference between N and P grooves is as large as 5 times. I once bought 8 pairs of K389 / J109, but the results of the test before installation were quite disappointing: â‘ The so-called twin tube is just the same performance of the two tubes in the same shell, while the 8 pairs of tubes purchased at the same time are N groove The difference between them is quite large, the difference between the N channel and the P channel is greater; â‘¡ The Idss, gm, and Vgs of K389 and J109 are different, and the actual waveform test is also asymmetric. Finally, the author can only select two tubes with an error of 3.8% from K389 as a single-sided differential input stage. (In the past, when using bipolar twin tubes, it was always difficult to control the errors of the same polarity tube at 1%, the matching error of the heteropolar tube will not be greater than 3%). Through the above data comparison and actual testing, the following enlightenment can be obtained: When the JFET is used in the complementary input stage, the discreteness of V and I will cause a large deviation of the static operating point of the circuit, thereby changing the stability of the circuit Poor; the inherent differences in gm, Cis, Cis affect the dynamic indicators of the upper and lower waveform symmetry and transient response speed of the entire push-pull stage. On this point, some well-known foreign manufacturers have long reached a consensus. For example, Tianlong, Marantz and other products often see the field effect differential input stage made by K389, but it is always difficult to see the shadow of J109. / J109 was originally "La Lang Pei".
Compared with JFET, the MOS tube's withstand voltage, power consumption and transconductance are easily made higher. In addition, the parts of the amplifier other than the input stage (such as the driving stage and the output stage), their complementary pairing requirements can be relatively relaxed, and some pairing defects can also be overcome by careful design of the circuit. Therefore, the application of MOS tube in the final stage of the power amplifier is not a big problem. The problem of MOS tubes used in the power amplifier output stage is not complementary pairing, the key is low efficiency.
2. It is well known that the efficiency of the MOSFET output pole is greater than the loss of the output stage of the lowest MOS tube than the bipolar transistor. Usually in the same circuit, in order to obtain the same output power as the bipolar tube, the method used is to increase the power supply voltage by ± 5V to compensate for the loss of the MOS tube. However, the actual production proves that it is far more than that simple.
The parameters of common field effect tubes for audio are as follows:
You can learn from the attached table the main characteristic data of several representative MOS tubes. Now take Hitachi's old LDMOS tube K135 / J50 as an example. The gate-source turn-on voltage threshold of K135 / J50 is from 0.15 to 1.45V. Measured when Io = 10mA VO. 25V, and when I ~ = 100mA typical value, V increases to 0,6 ~ 0,85V. It can be seen that the voltage control characteristics of the field effect transistor determine that the gate-source loss voltage rises with the increase of the drain current I (relative to this, the V of the bipolar transistor is almost constant at 0.7V). The internal loss of the M OS tube mainly depends on the size of the drain-source on-resistance R ,. The parameter RDs 1 is not directly given in the parameters of K1 35 / J50, but through the drain-source conduction voltage VDs (sat): 12V and ID 7A two data, using the formula RDs (oN) = UDs) / ID can be calculated The R of K135 / J50 is about 1.7 Q. This is equivalent to putting a 1.7 Q resistor in series with the load. For a standard 8 Q load, the power loss is close to 20%. If the loudspeaker impedance drop at low frequencies and the negative temperature voltage-current characteristics of the field effect tube (ie, the current decreases when the temperature rises, that is, R increases at this time) are considered, then the actual internal loss of the MOS tube will be greater. In contrast, the situation with bipolar transistors is very different. For example, Toshiba's A1265 / C3182, when Ic = 7A, V, = 2V. If the output is a two-stage follower, then add the last stage loss (<1V), the total V. (sat) <3V. Compared with K135 / J50's VDs (sat) = 12V, which is better is naturally self-evident.
(2) The output current of the NOS power tube is as mentioned above. Under the same circuit conditions, only increasing the power supply voltage by about ± 5V does not enable the MOS power output stage to obtain the same power as the bipolar tube. We tend to underestimate the dynamic loss when the MOS tube is actually working. Set a post-stage of 100W. When the load is 8 ohms, the power supply voltage for the bipolar transistor is V_Gc (8P. × RL) + 2 × [VcE (sat) + I () × RE. For example, IcM = 5A, vc. 1.5V, RE = 0.22 Q, then VC 85.2V (± 43V). For the MOS tube, V? = 99.2V (± 50V) can also be calculated. Of course, this is theoretical and the value at maximum power. In practice, the no-load voltage when using an unregulated power supply is obviously higher.
In fact, the AC secondary side power supply value of the 100W / 8Q bipolar transistor power amplifier is about AC 33V × 2. When AC38V × 2 is used to supply power to the MOS power amplifier according to the usual practice, P can only reach 70 ～ 80W, and the high R of Hitachi MOS tube depends on the power amplifier NFB network to improve the overall internal resistance, so the sense of hearing at the actual large output is lacking (It's even worse when the amplifier is designed to have no feedback). So there are various misunderstandings about MOS tube, such as poor linearity when MOS tube is high current is one of them.
In fact, compared with bipolar transistors, MOS tubes have excellent high-frequency characteristics and distortion is dominated by even harmonics. Because there is no second breakdown, Hitachi has given the application current of MOS tubes such as K135 / J50 to be close to Recommended limit. The linear current of the VMOS tube can reach tens of amps, and the pulse current of the UHC-MOS tube is as high as more than 300A. So why is the MOS audio tube misunderstood as a large current linear difference? The root cause is still not enough understanding of the power loss caused by the internal resistance of the MOS tube and the corresponding countermeasures. Although the internal resistance of VMOS and UHC-MOS tubes is very small, the V under high current is as high as 5V or more, so it should be paid attention to.
3. Efficiency improvement of MOSFET output stage
The loss of the MOS power output stage is always greater than that of the bipolar transistor, which is determined by its inherent characteristics. The improvement mentioned here should actually be how to reduce the power loss of the MOSFET output stage.
①If you want to obtain the same output power as the bipolar transistor under the same circuit, the power supply voltage of the MOS output stage should be higher than ± 10V than when using the bipolar transistor. When it is necessary to reduce the loss of the MOS tube, the voltage and current levels can be used to supply power separately. At this time, the current stage and voltage stage Nl: k, the output stage of the bipolar transistor is higher than ± 5V and ± 10V.
â‘¡Use multi-tube parallel output stage. The parallel connection of multiple tubes is to reduce the equivalent on-state resistance of the MOS power tube, not to improve the poor linearity of the MOS tube at high current. The parallel connection of multiple MOS tubes can not only increase the current driving force, but also greatly reduce the power loss, and improve the open loop. When the K135 / J50 4 groups are connected in parallel, the equivalent internal resistance drops to 1/4 of the single tube, which is 0.4. Below Q, the power loss for a load of 8 ohms is correspondingly reduced from 20% to 5%). In addition, the error of the parallel parameter of the MOS tube can be properly relaxed than that of the bipolar tube, that is, the error of the parallel tube is slightly larger and it will not deteriorate the hearing.
③ Adopt common source output stage, namely the collector output form of bipolar tube. This output mode is more suitable for VMOS tube, because the current of VMOS tube is large and the internal resistance is small. When the circuit design is reasonable, it can take into account good efficiency, low distortion and low output impedance. The power supply voltage at this time may be higher than the bipolar circuit by ± 3 to ± 5V.
The author made a common source output stage amplifier with op amp OPA604 ~ DVMOS tube IRF540 / 9540 ~ 1, the specifications are v is ± 40V, Po is 60W / 8Q, 100W / 4Q. The actual audition results show that the driving force is not weak, and the calorific value is not necessarily greater than that of the bipolar tube. As an audiophile, I certainly hope that in the near future, I will fully meet the requirements of today's bipolar pairing and even exceed the bipolar mutual ~ FET, high voltage UHC-MOS complementary power tube. This is not an illusion for today's rapidly changing, but one thing, I hope that the price of those "supplements" will not be too high.
1. JFET lacks complementary tubes within matching tolerances. The maturity of today's bipolar transistor manufacturing process has reduced the pairing error of NPN and PNP complementary transistors to a level generally accepted by the majority of professional manufacturers and audiophiles. In contrast, the selection of the field effect tube is much more difficult, and the JFET used as the input stage of the amplifier lacks complementary pairs that meet the requirements (this is determined by the current manufacturing level). The attached table lists the main characteristic data comparison of Toshiba's twin field effect transistor (DualFET) K389 / Jl09. As can be seen from the attached table, the differences between K389 and J109 are V C C and NF, of which C and C are two values, and the difference between N and P grooves is as large as 5 times. I once bought 8 pairs of K389 / J109, but the results of the test before installation were quite disappointing: â‘ The so-called twin tube is just the same performance of the two tubes in the same shell, while the 8 pairs of tubes purchased at the same time are N groove The difference between them is quite large, the difference between the N channel and the P channel is greater; â‘¡ The Idss, gm, and Vgs of K389 and J109 are different, and the actual waveform test is also asymmetric. Finally, the author can only select two tubes with an error of 3.8% from K389 as a single-sided differential input stage. (In the past, when using bipolar twin tubes, it was always difficult to control the errors of the same polarity tube at 1%, the matching error of the heteropolar tube will not be greater than 3%). Through the above data comparison and actual testing, the following enlightenment can be obtained: When the JFET is used in the complementary input stage, the discreteness of V and I will cause a large deviation of the static operating point of the circuit, thereby changing the stability of the circuit Poor; the inherent differences in gm, Cis, Cis affect the dynamic indicators of the upper and lower waveform symmetry and transient response speed of the entire push-pull stage. On this point, some well-known foreign manufacturers have long reached a consensus. For example, Tianlong, Marantz and other products often see the field effect differential input stage made by K389, but it is always difficult to see the shadow of J109. / J109 was originally "La Lang Pei".
Compared with JFET, the MOS tube's withstand voltage, power consumption and transconductance are easily made higher. In addition, the parts of the amplifier other than the input stage (such as the driving stage and the output stage), their complementary pairing requirements can be relatively relaxed, and some pairing defects can also be overcome by careful design of the circuit. Therefore, the application of MOS tube in the final stage of the power amplifier is not a big problem. The problem of MOS tubes used in the power amplifier output stage is not complementary pairing, the key is low efficiency.
2. It is well known that the efficiency of the MOSFET output pole is greater than the loss of the output stage of the lowest MOS tube than the bipolar transistor. Usually in the same circuit, in order to obtain the same output power as the bipolar tube, the method used is to increase the power supply voltage by ± 5V to compensate for the loss of the MOS tube. However, the actual production proves that it is far more than that simple.
The parameters of common field effect tubes for audio are as follows:
You can learn from the attached table the main characteristic data of several representative MOS tubes. Now take Hitachi's old LDMOS tube K135 / J50 as an example. The gate-source turn-on voltage threshold of K135 / J50 is from 0.15 to 1.45V. Measured when Io = 10mA VO. 25V, and when I ~ = 100mA typical value, V increases to 0,6 ~ 0,85V. It can be seen that the voltage control characteristics of the field effect transistor determine that the gate-source loss voltage rises with the increase of the drain current I (relative to this, the V of the bipolar transistor is almost constant at 0.7V). The internal loss of the M OS tube mainly depends on the size of the drain-source on-resistance R ,. The parameter RDs 1 is not directly given in the parameters of K1 35 / J50, but through the drain-source conduction voltage VDs (sat): 12V and ID 7A two data, using the formula RDs (oN) = UDs) / ID can be calculated The R of K135 / J50 is about 1.7 Q. This is equivalent to putting a 1.7 Q resistor in series with the load. For a standard 8 Q load, the power loss is close to 20%. If the loudspeaker impedance drop at low frequencies and the negative temperature voltage-current characteristics of the field effect tube (ie, the current decreases when the temperature rises, that is, R increases at this time) are considered, then the actual internal loss of the MOS tube will be greater. In contrast, the situation with bipolar transistors is very different. For example, Toshiba's A1265 / C3182, when Ic = 7A, V, = 2V. If the output is a two-stage follower, then add the last stage loss (<1V), the total V. (sat) <3V. Compared with K135 / J50's VDs (sat) = 12V, which is better is naturally self-evident.
(2) The output current of the NOS power tube is as mentioned above. Under the same circuit conditions, only increasing the power supply voltage by about ± 5V does not enable the MOS power output stage to obtain the same power as the bipolar tube. We tend to underestimate the dynamic loss when the MOS tube is actually working. Set a post-stage of 100W. When the load is 8 ohms, the power supply voltage for the bipolar transistor is V_Gc (8P. × RL) + 2 × [VcE (sat) + I () × RE. For example, IcM = 5A, vc. 1.5V, RE = 0.22 Q, then VC 85.2V (± 43V). For the MOS tube, V? = 99.2V (± 50V) can also be calculated. Of course, this is theoretical and the value at maximum power. In practice, the no-load voltage when using an unregulated power supply is obviously higher.
In fact, the AC secondary side power supply value of the 100W / 8Q bipolar transistor power amplifier is about AC 33V × 2. When AC38V × 2 is used to supply power to the MOS power amplifier according to the usual practice, P can only reach 70 ～ 80W, and the high R of Hitachi MOS tube depends on the power amplifier NFB network to improve the overall internal resistance, so the sense of hearing at the actual large output is lacking (It's even worse when the amplifier is designed to have no feedback). So there are various misunderstandings about MOS tube, such as poor linearity when MOS tube is high current is one of them.
In fact, compared with bipolar transistors, MOS tubes have excellent high-frequency characteristics and distortion is dominated by even harmonics. Because there is no second breakdown, Hitachi has given the application current of MOS tubes such as K135 / J50 to be close to Recommended limit. The linear current of the VMOS tube can reach tens of amps, and the pulse current of the UHC-MOS tube is as high as more than 300A. So why is the MOS audio tube misunderstood as a large current linear difference? The root cause is still not enough understanding of the power loss caused by the internal resistance of the MOS tube and the corresponding countermeasures. Although the internal resistance of VMOS and UHC-MOS tubes is very small, the V under high current is as high as 5V or more, so it should be paid attention to.
3. Efficiency improvement of MOSFET output stage
The loss of the MOS power output stage is always greater than that of the bipolar transistor, which is determined by its inherent characteristics. The improvement mentioned here should actually be how to reduce the power loss of the MOSFET output stage.
①If you want to obtain the same output power as the bipolar transistor under the same circuit, the power supply voltage of the MOS output stage should be higher than ± 10V than when using the bipolar transistor. When it is necessary to reduce the loss of the MOS tube, the voltage and current levels can be used to supply power separately. At this time, the current stage and voltage stage Nl: k, the output stage of the bipolar transistor is higher than ± 5V and ± 10V.
â‘¡Use multi-tube parallel output stage. The parallel connection of multiple tubes is to reduce the equivalent on-state resistance of the MOS power tube, not to improve the poor linearity of the MOS tube at high current. The parallel connection of multiple MOS tubes can not only increase the current driving force, but also greatly reduce the power loss, and improve the open loop. When the K135 / J50 4 groups are connected in parallel, the equivalent internal resistance drops to 1/4 of the single tube, which is 0.4. Below Q, the power loss for a load of 8 ohms is correspondingly reduced from 20% to 5%). In addition, the error of the parallel parameter of the MOS tube can be properly relaxed than that of the bipolar tube, that is, the error of the parallel tube is slightly larger and it will not deteriorate the hearing.
③ Adopt common source output stage, namely the collector output form of bipolar tube. This output mode is more suitable for VMOS tube, because the current of VMOS tube is large and the internal resistance is small. When the circuit design is reasonable, it can take into account good efficiency, low distortion and low output impedance. The power supply voltage at this time may be higher than the bipolar circuit by ± 3 to ± 5V.
The author made a common source output stage amplifier with op amp OPA604 ~ DVMOS tube IRF540 / 9540 ~ 1, the specifications are v is ± 40V, Po is 60W / 8Q, 100W / 4Q. The actual audition results show that the driving force is not weak, and the calorific value is not necessarily greater than that of the bipolar tube. As an audiophile, I certainly hope that in the near future, I will fully meet the requirements of today's bipolar pairing and even exceed the bipolar mutual ~ FET, high voltage UHC-MOS complementary power tube. This is not an illusion for today's rapidly changing, but one thing, I hope that the price of those "supplements" will not be too high.
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