Why does this push pull common source FET amp use push pull gate drive? - Page 1
Forum section: electronics / projects

author: Deactivated-1
author_rank: Frequent Contributor
timestamp: August 17, 2023, 12:57:30 am
content: Why did they make a full push pull amp stage for the gate drive? Why not use something simple like a class A single ended gate driver, since no current flows trough the gate right?Also why have that weird connection for the "unused" op amp? Why not connect that 150k to ground instead of the output of that opamp? And connect the "unused" op amp as a buffer and ground the input.Is it to cancel noise or is something else going on there with that setup?[ Attachment Invalid Or Does Not Exist ]FYI The bias supplies are 6.8V and -6.8V from zeeners biased with 10k from VCC+ and VCC-. Which in a fault scenario gets pulled to ~1.2V of each other.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: August 17, 2023, 01:11:46 am
content: Quote from: ELS122 on August 17, 2023, 12:57:30 amThose 5532 opamps had relatively high bias currents.That looks like a bias-offset circuit - it needs to assume both opamps bias currents match.

author: moffy
author_rank: Super Contributor
timestamp: August 17, 2023, 06:34:39 am
content: Quote from: ELS122 on August 17, 2023, 12:57:30 amNo DC current except for some leakage but gates look like capacitors, sometimes reasonably large, so to drive AC into capacitors you need a push pull drive. The gate drive does look a bit weird though, and I agree with PCB.wiz that the extraneous input opamp looks like it is trying to compensate for offsets created by bias current of the 5532.

author: Kleinstein
author_rank: Super Contributor
timestamp: August 17, 2023, 07:08:51 am
content: Using a dual OP-amp to drive the power stage is odd. The 2nd OP-amp needs to be set a neutral state as he buffer configuration.The capacitive load from the gates neede the push pull stage. A class A stage would need to run with quite some current to avoid slow rate limitations.

author: David Hess
author_rank: Super Contributor
timestamp: August 17, 2023, 07:51:02 am
content: U1a implements input bias current compensation for U1b. Audio operational amplifiers typically do not include input bias current cancellation because it increases input current noise while not typically being needed when AC coupling can remove the effects of offset voltage. Here U1b also controls the DC balance of the power output stage.The class-AB drivers can more easily pump charge in and out of the parallel MOSFET output stage. Because the MOSFET output stage is common-source instead of common-drain, the reverse transfer capacitance of the MOSFETs is multiplied by the output voltage gain and pumps considerable charge back into the gate. A similar bipolar output stage would also benefit from class-AB drivers for the same reason.I was surprised you did not ask about what U1c is doing. U1 drives the output stage via its supply pins. The output of U1b is a current feedback input from the output of the power stage. The voltage gain of the current feedback multiplies the gain-bandwidth product and slew rate of U1b effectively decompensating it by exactly the amount needed for highest performance. I am surprised U1c (and U1d?) was not placed in parallel with U1b to provide even more drive.This is the sort of design I might do, but with a common drain output if I used MOSFETs and a boosted voltage for the driver circuits.

author: David Hess
author_rank: Super Contributor
timestamp: August 17, 2023, 09:58:14 pm
content: Quote from: ELS122 on August 17, 2023, 07:19:48 pmIt is a pretty unusual abuse of how an operational amplifier is usually configured. Fairchild included in in their 741 application notes in the early 1970s. I do not know why, but Tektronix used it in their 7A29 vertical amplifier which was released in 1979(?).

author: moffy
author_rank: Super Contributor
timestamp: August 17, 2023, 11:32:47 pm
content: The configuration of using the current out of the opamp power supply pins as the effective output has a real drawback in that you cannot decouple the opamp as the decoupling capacitors would divert higher frequency currents. As a result any reasonably high BW opamp can easily become unstable. A better approach is for the opamp with feedback to drive a push pull pair as the output and the collectors become the current outputs that you monitor, that way you can safely decouple the opamp without compromising the frequency response, and you have a more stable amplifier.

author: David Hess
author_rank: Super Contributor
timestamp: August 18, 2023, 10:05:40 pm
content: Quote from: moffy on August 17, 2023, 11:32:47 pmI am going to do it in an upcoming circuit because I want to drive a power output to one supply rail.In all of the examples I have seen, the power supply pins drive a low impedance like the emitter or base of a bipolar transistor. It would be fun to see how fast the operational amplifier can be without problems in this configuration.QuoteThat was how my audio amplifier design worked, but the operational amplifier output drove the emitters of the push-pull pair instead of the bases. Now I would likely use a diamond buffer, which is what current feedback amplifiers do.Linear Technology showed the same idea but with an LT1010 buffer, as shown below.

author: moffy
author_rank: Super Contributor
timestamp: August 18, 2023, 11:13:04 pm
content: If I had to guess I would say this is a Jim Williams design, a nice use of the LT1010 buffer.

author: David Hess
author_rank: Super Contributor
timestamp: August 18, 2023, 11:45:45 pm
content: Quote from: moffy on August 18, 2023, 11:13:04 pmIt is from Linear Technology application note 18 so Jim Williams, but the idea was hardly new. The LT1010 can be directly replaced with a diamond buffer.

author: David Hess
author_rank: Super Contributor
timestamp: August 19, 2023, 12:46:46 am
content: Quote from: ELS122 on August 18, 2023, 11:48:51 pmCommon base Q55 and Q56 control the voltage seen at the supply terminals of the operational amplifier while passing the currents through to the class-ab drivers. The example below from Walter Jung shows the same configuration with Q5 and Q6 and the zener diodes limiting the supply voltage to 6.8 - Vbe for each supply pin.As an aside moffy, the AD8610 is a 25 MHz part.The Vbe multiplier is not, although it fulfills the same purpose. That circuit generates an adjustable current to set the quiescent current of the output stage. It is not shown, but maybe Q28 is suppose to be mounted on the same heat sink as the output transistors?

author: magic
author_rank: Super Contributor
timestamp: August 20, 2023, 10:18:24 pm
content: Quote from: David Hess on August 19, 2023, 12:46:46 amThis "not a Vbe multiplier" is of course the garden variety two-transistor current sink. It passes current about equal to Vbe/(R55+TRM2), which does indeed add some negative tempco to the voltage across R138/R125 which biases gates of the output devices.The whole thing appears to be included to put more idle bias on the MOSFETs than from the 5532's quiescent current alone, and the trimpot adjusts it.It's an interesting circuit, which I think should be free of the stupid transconductance doubling problem that plagues traditional output stages. It would seem to work best in "hot" class AB bias, and neither side should be allowed to turn off.Quote from: David Hess on August 18, 2023, 10:05:40 pmThis, like countless similar variants found in other IC app notes, is no good for low distortion because it fails to keep the external power transistors biased, resulting in wild output impedance variations when parts of the circuit turn on and off. It would need a quiescent current control scheme, like the amplifier above.

author: David Hess
author_rank: Super Contributor
timestamp: August 21, 2023, 12:25:32 am
content: Quote from: magic on August 20, 2023, 10:18:24 pmI agree, but have you tested the LT1010 example? The output impedance is somewhat constrained by the current feedback network.The Vbe of the common emitter output transistors could be biased on to 0.33 volts as shown by the LT1010's quiescent current. The diamond buffer variation relies on emitter degeneration and transistor matching but would also benefit from trimming the idle current.

author: magic
author_rank: Super Contributor
timestamp: August 21, 2023, 05:51:16 am
content: Well, I did some quick math.I think we can agree that unloaded output resistance is simply (a bit more than) the current feedback resistor, 100Ω here.I think we can agree that under load this resistor still stays in circuit and Q1/Q2 current adds to it, reducing impedance. This starts to look bad already.In order to merely halve output impedance under load, current gain of the 33Ω/Q1/0.18Ω circuit can't exceed unity - rather impractical.These external transistors simply can't be allowed to work in class C under any circumstances.In reality, the LT1010 example ought to have single digit output resistance under heavy load:Q1 base input resistance is β·(0.18Ω + 1/gm), maybe similar in magnitude to 33Ω.Ignoring 33Ω, LT1010 supply current is multiplied by β and output resistance of the whole circuit is approx 100Ω/β.But current "wasted" in the 33Ω makes it about twice worse.

author: magic
author_rank: Super Contributor
timestamp: August 21, 2023, 08:09:04 am
content: Quote from: ELS122 on August 20, 2023, 11:19:17 pmIt's a kind of circuit where I think the more bias you throw at it, the better.The buffer or opamp output pin substantially follows the VAS voltage. To that pin is connected a resistor to ground (Dragster) or to the final output node (LT1010 application example). Either way, change in VAS voltage results in a proportional change in buffer output current and a proportional change in the sum of its supply currents.For overall linearity, we now need to ensure that current gain from each supply pin to the final output node is:- equal on the low and high side, and never drops down to zero (one half can't shut down)- independent of load current (transconductance of output devices can't vary too much, which is ensured by high bias)Notably, high bias doesn't result in transconductance doubling, because there is only one feedback loop through the buffer (which acts as an accurate phase splitter for class B or AB output stage) rather than two paralleled feedback loops like in a complementary Sziklai/Darlington OPS which will never transition cleanly from one loop to another.

author: magic
author_rank: Super Contributor
timestamp: August 23, 2023, 10:58:11 am
content: Quote from: magic on August 21, 2023, 08:09:04 amQuote from: magic on August 21, 2023, 08:09:04 amI take this back.The buffer has to work in class B in order to drive the common source/emitter output devices in class B.Therefore, its output node has the same crossover problem as an ordinary complementary EF output stage.Now the buffer has to be "optimal class B biased" or receive some local feedback correction or whatnot.Its distortion is still the same, it is only "obscured" by the increased output impedance (due to the resistor).editHowever, the resistor isolates driver (buffer) emitters from the load, so maybe VAS Miller compensation could be picked off from there without instability?That should provide some attenuation of transconductance doubling, leaving only power device current gain variation uncorrected.