amplifier output stage suggestions - Page 1
Forum section: electronics / projects

author: akis
author_rank: Frequent Contributor
timestamp: June 11, 2012, 08:57:58 pm
content: I have constructed the attached circuit to be able to drive a +/- 4V sinusoidal voltage into a 4R load from a +/- 6V supply. If I remove some output transistors then there is a bigger voltage drop at the load. With 3 pairs of transistors it just about works.In order to get a more reliable circuit and with less parts I have tried to use power op-amps but had no luck with 200 KHz, the low supply voltage and the high load.Is there a more elegant way to do this?

author: hlavac
author_rank: Frequent Contributor
timestamp: June 12, 2012, 03:47:31 am
content: Maybe look atclass D amplifiers?Although 200kHz may be a bit too much requiring the switching frequency of several MHz...I would not recommend building a discrete one but you may be able to find a chip that will do

author: DrGeoff
author_rank: Frequent Contributor
timestamp: June 12, 2012, 05:18:46 am
content: Have you looked at using mosfets?

author: akis
author_rank: Frequent Contributor
timestamp: June 12, 2012, 06:32:33 am
content: I am not very familiar with mosfets but I seem to recall that they have an "on" resistance (is it called RDon) which is much worse than transistors.

author: DrGeoff
author_rank: Frequent Contributor
timestamp: June 12, 2012, 08:01:53 am
content: That's irrelevant since you are running them in linear mode. Either BJT's or MosFets will dissipate a lot of power and require heatsinking. Mosfets have the advantage of being driven by a voltage rather than a current (transconductance) and present a high impedance to the driver circuitry.

author: ejeffrey
author_rank: Super Contributor
timestamp: June 12, 2012, 08:16:00 am
content: MOSFETs have zero DC drain current, but their much lower transconductance and large gate capacitance may make them harder to drive in this application due to the 200 kHz frequency goal. Their higher turn-on voltage also reduces the maximum swing across the load or requires higher supply voltage.

author: T4P
author_rank: Super Contributor
timestamp: June 12, 2012, 08:21:34 am
content: There are mosfet drivers for that very reason.

author: vk6zgo
author_rank: Super Contributor
timestamp: June 12, 2012, 08:41:40 am
content: Quote from: akis on June 11, 2012, 08:57:58 pmWhat are you actually trying to do?The text reads as if you want to see +/- 4 volts (2.8284 volts RMS),assuming a sine wave, across your load,but the schematic shows you with an input of that level.If the output only needs to give you 2.8284 volts RMS across the load,the amplifier only has to supply2 watts,so it shouldn't be incredibly hard to do,even at 200kHz.Obviously,it needs enough power gain to amplify the input power represented by 2.8284 volts RMSacross the higher impedance of the amps input,to that represented by the same voltage across 4 ohms,if thatiswhat you want to do.If,on the other hand,you need a much higher level,you need to tell us.

author: jimmc
author_rank: Frequent Contributor
timestamp: June 12, 2012, 08:54:41 am
content: Shouldn't all those 0.22ohm resistors be in the emitter leads of the output transistors?They're not doing anything useful where they are now.Jim

author: PChi
author_rank: Frequent Contributor
timestamp: June 12, 2012, 09:47:25 am
content: On possibility at a cost may be the Linear Technology LT1210.I agree with jimmc the 0.22 ohm resistors aren't helping.

author: akis
author_rank: Frequent Contributor
timestamp: June 12, 2012, 05:50:56 pm
content: Quote from: vk6zgo on June 12, 2012, 08:41:40 amThis is just it, as simple as that. I am constrained by the +/- 6V rails. If I increase the supply to, say, +/- 7V the curve looks so much better on the scope.Just to say again, I think I have a solution, in the form of 3 pairs of transistors, but am just wondering if there is another, more elegant (less parts ?) way.

author: akis
author_rank: Frequent Contributor
timestamp: June 12, 2012, 07:24:22 pm
content: Quote from: PChi on June 12, 2012, 09:47:25 amHad a look at this chip and t looks absolutely great, but the simplified schematic shows darlingtons at the output and I doubt it can reach the supply rails better than the complementary pair can under heavy load.

author: akis
author_rank: Frequent Contributor
timestamp: June 12, 2012, 07:28:01 pm
content: Quote from: jimmc on June 12, 2012, 08:54:41 amThose resistors actually belong to the BC337/BC327 driver transistor emitters, if you consider the BC337/MJE15029 as a complementary pair. I then added more MJEs to get less voltage drop under load. I could not wire the BC337's emitter to any of the MJEs (which one would you pick?), so it is wired to the output.

author: hlavac
author_rank: Frequent Contributor
timestamp: June 12, 2012, 08:01:59 pm
content: I think the purpose of the 0.22R resistors is just to provide equal load sharing between the parallel transistors. BJTs are not nice when connected in parallel, the most overloaded one tends to get even more overloaded, enters thermal runaway and burns.MOSFETs are much better in this regard due to their resistive nature and can be connected directly in parallel without these resistors.

author: jimmc
author_rank: Frequent Contributor
timestamp: June 12, 2012, 08:25:45 pm
content: OK if you want feedback to stabilise the quiescent current, connect three resistors (say 2.2ohm) to the collectors of the MJE15029s, common the other end of these resistors and join the emitter of the BC337 to this point. Repeat for the MJE15028s and BC327.As hlavac has said, you won't get good current sharing but since a single device is rated for the full output current it will not be catastrophic. Still not good practice.Jim

author: hlavac
author_rank: Frequent Contributor
timestamp: June 12, 2012, 08:56:03 pm
content: Is power consumption an issue? How much power does the +-6V supply have?

author: hlavac
author_rank: Frequent Contributor
timestamp: June 12, 2012, 09:29:03 pm
content: In other news, the 2u output capacitor seems to be a bit low, it's limiting output amplitude about 10% at 200kHz. Use something like 10u.

author: akis
author_rank: Frequent Contributor
timestamp: June 12, 2012, 09:43:51 pm
content: Quote from: hlavac on June 12, 2012, 08:56:03 pmThe supply will be two batteries of the appropriate AH. Currently using the bench multimeter for the prototype, shows up to 200-300mA on each channel at full load (full load varies, the 4R being an average)

author: jimmc
author_rank: Frequent Contributor
timestamp: June 12, 2012, 09:45:50 pm
content: Well I'm not Pythagorus, but it's surprising what a soak in a warm bath does.I've just realised your problem, the drive current to the BC3x7s is very limited, with the O/P at +4v the maximum base current is only about 500uA [6v-4v-Vbe)/3K3] so you need a minimum current gain of 3000. (Similar for -4v).The input can't provide current as this would reverse bias the associated diode.Simple answer is to remove C4 and replace with two 100n caps to each BC3x7 base, of course junction of caps to input).Devices are now positively driven and you should get away with only one output pair.(Paralleling output devices does reduce the gain drop-off at high currents.)Jim

author: akis
author_rank: Frequent Contributor
timestamp: June 13, 2012, 09:35:48 am
content: Quote from: jimmc on June 12, 2012, 09:45:50 pmI ran it again on the simulation quickly (prototype is downstairs, will get to it later)Your calculation of 500 uA concerns the current flowing down the 3.3K resistor, not the current branching off into the base of the BC337.This current is dependent on the combined hFE of the complementary pair [ BC337-MJE15029 ] which at 1A has a theoretical hFe of [ 400 * 140 = 56,000 ].On the simulator the current is shown as 52.8 uA and the current down the 3.3K resistor is at 375 uA. This works out as a hFe of 1 A / 53 uA = 19,000 roughly.The choice of the 3.3K was set almost empirically on the real prototype to see what "works" best. A 2.2K is much better because at the same measurement points, the current it provides is 557 uA which is 10 times greater than the 52.8 uA branching off into the base of the BC337, so we have better linearity. Unfortunately we are also loading the source quite a bit more.I will test your idea on the prototype and come back, because on the simulator I cannot see the real losses on the various parts of the circuit.

author: jimmc
author_rank: Frequent Contributor
timestamp: June 13, 2012, 08:02:13 pm
content: I'll be very interested to see the results on the prototype.A few thoughts...The HFEyou quote (56000) is the typical value, worst case is closer to 4000.Current gain and gain-bandwidth product drop rapidly as saturation is approached. The devices you have chosen are very good in this respect but I do wonder.The only other thing I can think of is supply line inductance, a couple of good low ESR caps directly from emitter of output transistor to the ground end of the load might help.If you're worried about source loading, how about bootstrapping R13/14 ie split into two halves and connect middle to output via a capacitor.You could then raise their values a bit.The 2uF output cap should be OK but watch the ESR (reactance ~0.4ohm so only about 1% drop with a 4ohm resistive load.)By the way I made a mistake in my last post; min required HFEshould be 2000 (I used 6v/4ohm rather than 4v/4ohm).Jim

author: akis
author_rank: Frequent Contributor
timestamp: June 13, 2012, 10:04:46 pm
content: Quote from: akis on June 12, 2012, 09:43:51 pmYes you are right. I looked again today in my box of caps and I have either electrolytics for anything over 1uF or polyester/film for anything below 1uF. To make up 20uF I simply have to use electrolytics, but these are not good at 200KHz. So just for today I used 2 X 1000uF electrolytics and 2 x 480nF polysters in parallel. I will buy some film 10uF and 20uF.

author: jimmc
author_rank: Frequent Contributor
timestamp: June 13, 2012, 10:14:25 pm
content: One thing I forgot to ask, when you say 'there is a bigger voltage drop at the load' (in your first post) do you mean it starts clipping if you raise the drive to try and get 4v out, or do you mean that with a fixed 4Vpk input the output is undistorted but lower than the 4v you want?I've been assuming the first alternative, if it's the second remember that connecting the emitter of the input transistor to the collector of the output device will mean that you're voltage gain will be reduced to just under 0.95 (4/(4+.22)) ie output of 3.8v for 4v input. Of course when you changed to multiple devices you removed this feedback.If you must have unity gain and still want to stabilise the quiescent current, you'll have to add large caps across each of the 0.22ohm resistors.Jim

author: akis
author_rank: Frequent Contributor
timestamp: June 14, 2012, 06:48:52 am
content: Quote from: jimmc on June 13, 2012, 10:14:25 pmYes the input is fixed at 4Vpk. Ideally I would want to raise the input to even 5Vpk, but because the rails are fixed to +/-6V, there is no way (I know of) to drive a 4R or 2R load that way. The 4Vpk was a sort of compromise. It is also to do with energy conservation since the unit is powered from batteries. The 4R "load" in reality will be a step-up transformer to reach 50V pk, and there will be losses on the tranny too, and the greater the step up the greater the issues arising at the transformer (I have a parallel thread about the trransformer).I did some tests yesterday on the prototype, trying out different configurations. I have already adopted your suggestion of two parallel caps feeding the bases of the transistors. I have reduced the resistors to 0.15R. Many problems go away if I increase the rails. Increasing the rails past 7.5 Volts offers no benefits except the transistors now get awfully hot and run away. On the prototype the transistors are not mounted on the same heatsink, so they can run away.A single complementary pair in its normal configuration works as expected but cannot drive the load (asymmetrical clipping and general mangling of the waveform). This is not shown on simulation. The simulation shows a slightly reduced output, perhaps, but not asymmetrical mangling and other artefacts. Adding more MJEs presents the problem of where to connect the emitter of the BC337. There are two choices, connect it to a random MJE's collector, or connect it straight to the output rail. I got better results with connecting it to the collector of a random MJE. I am attaching a new diagram.A question still remains, is there a better way to do this ? Something less than 6 output transistors, plus a whole bunch of circuitry to do with current limiting which is not even shown here.