Groundplane in poweramplifier PCB design

Haven't built a poweramp in years, but I found *my* golden rule was to have a *significant* starpoint ground system, all grounded at the Big Capacitors in the power supply. Big Thick Tinned Copper Wire.

Short is beautiful (never thought I would say that)

martin

My amp consists of four mono boards, with one power supply. The starpoint is 4 filter caps (two per rail), connected to eachother by a piece of unetched PCB. In the centre is the bolt, which acts as a starpoint. Every amp and speaker is connected to it by heavy gauge wire.

What I mean by that is, I have a proper starground setup. My question is about the circuitry present on one amplifier PCB. There are a few grounds there: one or two for the input stage and one for the decouplers/byassing. The question is, should I connect all that to one big ground plane, or use a "sub-stargrounding" setup on the PCB as well, routing every ground to the 0V pin seperately?

After thinking about it some more, seperate tracks to the 0V pin seems like the best idea, to keep the noise from the decouplers and bypass capacitors out of the ground of the input stage.

I get lots of mail every day from people who don't agree with you...

UGH.... Basically you can theorise until the sun cools down. You start off small, build boards, play around with earthing, until it works to your satisfaction.

You probably wont get it correct first time. But I found that having a thick flexible cable that I could use to short between various ground points and chassis, while measuring the O/P hum with a microvolt meter.

Trial and error...Theory is wrong, Practice sucks(TM)

martin

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No!... No ground planes. In audio circuitry you have to be very carful where current flows to avoid ground loops and current induced feed back from unsuspecting points. A ground plane complicates everything and you can't define where currents flow, where troublsome microvolt drops occur or where leakage flux induces hum. Use a star ground as others have mentioned. It's ok to use a ground plane as a shield connected only at ONE place to the star but be sure there are no other connections to it and no current flows in it. Bob

This is indeed what I've concluded after some thinking.

Your suggestion of having a groundplane as shield, connected to the 0V pin is a good idea. I'll see if I can do that.

On Sunday 16 April 2006 01:02, martin griffith wrote:>>

To be clear, I was talking about Viagra spam here... (I hope I don't set off any spamfilters...)

The thing is, this already is the second time. I have a board in use now, which has problems. In the mean time, I've learned a lot more, and the new one looks a lot better. But, theory _is_ involved there.

Agreed - I have layed out a few high power amps for a good analogue engineer and the single point in the preamp was always used - even though it went against my (RF) grain to add all these long thin parallel tracks. Guarding of sensitive inputs with single point earthed tracks was common. Any spare areas of board (not that there was much) was filled with earth taken back to the local star but not employed as a plane as such.

Geo

There can also be thermal reasons for putting copper in. You can't ignore these.

You want the extra copper as shielding. If you can arrange to effectively have a thick shorted turn around the whole circuit, this will help to keep AC magnetic fields from going through the PCB.

Also remember that every bipolar transistor is an AM radio just waiting to happen. The guy next dooor will buy a 1KW CB tomorrow.

There are other places in the design where you want the trace to be short and thick. These are the places to look at. It is likely that you will find that you want those traces so wide that they end up really being a plane. ie:

Ascii art:

pin trace pin O-------------------O

Becomes:

If you have to, you can initially make the plane as a different net than GND. It will flow around the GND traces. Then you can join it up using whatever trick your layout tool allows.

I assume that this amplifier has feedback. Where are you picking off the two sides of the speaker for feedback? These connections can also be trouble makers.

For me, that's not an issue. The driver transistors will be fitted with internal heatsinks, and the output devices are connected through brackets to external heatsinks. Everything that needs cooling is taken care of this way.

Sounds like a good idea. Will do.

Will it matter (much) BTW if that track has a gap in it?

I have this for the supply tracks, and speaker output. And, to a lesser degree also some ground tracks, mainly for those used for decoupling and bypassing.

What do you mean, the two sides of the speaker? The voltage feedback for the long tailed pair input stage is taken care of on the PCB, if that's what you mean.

Ah, so in radio it is better to use a ground plane? I wondered about that, because I have seen radio PCBs which indeed had a ground plane.

Don't forget the thermal stabilization for the bias. The device(s) used for this need to be at the same temperature as the power transistors. Sudden changes in the output power can lead to sudden changes in device temperatures so those components need to be on the heat sink if you aren't using PCB copper to conduct the heat.

Yes, it matters about any gap. Think of it this way:

When a magnetic line force passes through a conductor it makes a tiny voltage right at the part it goes through. Any current that this voltage may cause is always such that it opposes motion of that line of force.

Now imagine a ring of copper, with 101 lines of force going down through the center of it. If any of these lines of force try to get away, they will cause a current in the copper ring that slows their departure. If you try to slip another one in, a current will be created to slow that down too.

The result of this is that the magnetic field can't change as quickly as otherwise. This is how the copper reduces the high frequency changes in the field.

Ideally, the power amplifier puts a controlled voltage onto the speaker. In real life, you are putting the voltage onto the terminals the speakers get wired to.

Lets say, your power amplifier has a pair of connections for the input signal on one side and another pair for the speaker on the other. (This would be a simple mono amp) You want the voltage on the speaker to be some number times the voltage on the input. You don't want any of the voltage drops in the internal wiring to get into the picture. Using an op-amp symbol in some ASCII art, I think makes the idea clear:

R1 R2 (in-) -+--/\\/\\/----+---/\\/\\/---- ! ! ! [?] ! ! ! ---!-\\ ! GND ! >-----+-- (Out+) ---!+/ ! ! (in+) ---/\\/\\------+-----/\\/\\---+-- (Out-) R1 R2 ! [?] ! GND

The two mystery components represent the wiring. If the R1's are the same and the R2's are the same, a voltage drop in the wiring won't change what gets applied to the speaker.

It's not a darlington pair setup, it's a compound pair. In such a setup, the bias servo needs to be in thermal contact with one of the driver transistors (not the class-a driver of course). If you connect it to the main heatsink, the quiscent current will only get higher as the temperature increases.

OK, I'll see if I can make some adjustments to make the circle complete.

I'm still not sure what you mean. The only voltage feedback this amp has, as far as I know, is that it's output is fed back to the long tailed pair through a resistor.

This is the amp I'm making BTW:

In article , Wiebe Cazemier wrote: [... shifted out of order ...]

The core of this circuit is a power op-amp.

The base of Q1 is the non-inverting input. The base of Q2 is the inverting input.

The voltage divider (C3 + R4) / (C3 + R4 + R5) sets the gain of the circuit. In the middle of its band it has a voltage gain of:

(R4 + R5)/R4 = 23

Consider the schematic's R5 connection to the OUT terminal. The way it is drawn, the R5 connection is to C5. This is based on the theory that this node is all at the same voltage. This isn't really true. The R5 connection should be as close to the OUT terminal as practical.

If I was designing this, I'd try to make the ratio of R1/R2 equal to the ratio of R4/R5.

Now consider the R2's ground connection. It is drawn as being near the input jack. In fact you want it nearer the speakeer's return connection.

Actually you want it to be sort of the average of the two driver transistors. If the amplifier is used to product a signal with a lot of even harmonics in it, there may be a slight temperature difference between the two.

I'm renumbering this to match the schematic. I think you'll see the point.

Indeed, 27 dB.

I have. The R5 connection is taken from the exact middle between the collector resistors, as is the speaker terminal. I don't take it from the path to which C5 is connected. C5 has a seperate track.

The latter part goes against what I know about line input connections. For example, in my current state of design, I have one SPK_OUT on the PCB, and the return goes to the PSU's 0V. When you connect the input ground to the 0V of the PSU, instead of close to the input stage on the PCB, it will oscillate. I have tried this, and indeed it did.

My PCB has a 0V pin, which is used as a starpoint for all the sections which require 0V. That means there is one for the input stage as well. However, the input ground is not fed to to this star seperately. Instead, it's connected to the input stage (near R2) and from there, one connection goes to the 0V PCB pin. That's how it should be, as far as I know.

I'm beginning to. But, this setup doesn't hold true if I adhere to what I mentioned about the input ground connection above. I cannot have the speaker return and input ground at the same points. It will oscillate, and/or speaker return current pollutes the input ground.

Yeah, OK. But in practice, this is not really necessary. In fact, I have a first version of this amp running with the servo not in contact with anything. It doesn't suffer from thermal runaway.

Wiebe, I just had a short look at the webpage. Even if the author says it is an excellent amp, it doesn't necessarily need to be so. This construction invites any instability and there are a few very weak points. To bring good performance, the input pair has to be operated with equal currents, which is almost impossible here. There will always be some offset voltage at the output. When you trimm it away it will be again there when the temperature has changed. This is due to the current source made with Q4. which relies only on its Vbe. It is also dependent on the beta of the transistor and the forward voltage of the green LED. C3 will be always operated with changing polarity, this calls for trouble too. The worst is when there is an overload on the input and all the current goes through Q1 and saturates Q4, especially without a load. The output goes high and C5 will be pulled down by R10. Thus the current through Q9 increases, and the voltage drop across it. The moment Q6 starts conducting it turns on Q8 and the current will increase until the fuses blow, but by then it is too late. No current limit. In this kind of topology it is impossible to incorporate a simple current limit, because when you pull down the base in the usual way, the other side starts conducting and shorts the supply rails. So it is not the gnd plane, it is the principle of operation which results in self-destruction, whatever you do. There are many more details that can be improved, but IMHO it wouldn't make sense.

I'm not talking about moving the ground of the input connection. Only the ground side of the resistor is moved to the speaker's return.

Again look carefully at what I suggested. Notice that the R2 ground is the only thing moved. I've also assumed that R5/R4 now equals R2/R1. Also, I've assumed that the signal source has low impedance. If it doesn't the values have to change to include it's effects.

You often have to run the amplifier at about 1/2 power at a low frequency to get it to run away.

I disagree with this and will explain my reasoning in some detail for the OP.

The tempco of the LED almost exactly matches that of the Vbe of Q4. So long as they are at the same temperature, the current will not vary much. Both silicon transistors and LEDs have a tempco of Vbe of nearly 2mV per degree C.

If D1 has about 3V on it, there is still more than 1mA of current flowing in R8 and about 4 in R7. I admit I haven't looked up the spec. on Q3 but I don't its HFE will be under 10.

I don't see it as likely that the voltage on C3 can ever reverse. How did you conclude that?

Note that the (R1+R2)C1 time constant is shorter than the the C10 time constant. The input would have to swing a long way to keep Q1 on like this.

The current in Q9 roughly doubles. Its VI curve will only bias Q6 on enough to double the idle current of the output stage. This shouldn't be enough to cause damage. I'd worry a lot more about a reactive load.

Try this idea .. Simplified circuit Ascii ART:

! !/ -----+---+-------! ! ! !\\e Q5+Q7 Q9 \\! ! ! !-------+ ! e/! NPN ! ! ! Q98 [R13] ! ! ! ! ! +------------ load ! ! ! ! ! [R14] ! e\\! PNP ! ! !-------+ ! /! ! ! ! Q99 !/e ---+---+-------! Q6+Q8 !\\ !

Assume that Q5+Q7 are passing enough in R13 to make 2 diode drops of current. Both Q98 and Q99 will be biased on. Q99 will be super-saturated so its base current rating has to be enough.

Now the remaining problem is that Q4 is not current limited. It needs a resistor in its emitter and R6 needs to be a larger value + a diode. You can make it so that the collector current of Q4 is nearly a constant times the collector current of Q1.

The whole R9,R10 + C5 network could be replaced by another constant current circuit using D1 voltage.

"Ken Smith" a écrit dans le message de news:e2494i$ssq$ snipped-for-privacy@blue.rahul.net...

Or put a current limiter in its collector, like a JFET current source.

I've also done this for a supply, which transposed here gives:

| |< -| Q4 |\\ | +----- | | | .-. | | | | | |4K7 | '-' \\| | |---+ | \\| .-----+ | |------' .-. |

In article , Fred Bartoli wrote: [....] I think you made an error. I've marked the connection with "XXX" that I don't think should be there.