Power supply filtering question/current draw

[tubegeek]

You might enjoy this page:

http://sound.westhost.com/project83.htm

Best regards, Jack

If Rod Elliott ever selects anything I design to put up on his site, I'm gonna do a great big victory lap and feel very satisfied with myself!

[PRR]

> haven't been able to find anything that actually talks about that type of transistor arrangement

There are many "similar" circuits. I can't say I have seen this exact plan, or particularly in a published "DIY" project.

As has been said, the output stage current depends entirely on Q3 actual hFE. hFE is very loosely controlled on a given type. I have seen 2N-numbers with hFE "40-800". The article does say "generic NPN", which doesn't narrow it down. Further, what matters is the hFE *at the operating point*. The collector voltage here is likely to be higher than the spec-sheet or test-set test condition. hFE also varies with current, and we don't know the current until we know the hFE. So the output stage current is not predictable with reasonable certainty.

If a rough-guess suggested 0.1mA, we might go on and see how it works. Even if "0.1mA" was 20:1 in error (we hit 2mA), nothing would burn.

However the intent here *seems* to aim near 25mA. If it really is 25mA, 10V*25mA is 250mW, safe-enough for small transistors. However with unspecified transistor it could easily be 90mA. This causes large drop in the 100r, which will dissipate 0.8 Watts, and possibly 0.6 Watts in Q2 or Q3.

And as-said, all much worse with a higher supply voltage.

It is not clear why he used this bias plan.

The clear intent is for Q3 to be a quasi-constant current load. There are better (predicable design) schemes for this. Jack's drawing in reply #32 gives two fine plans. The one on the right is ample and simpler. As your supply voltage is higher, the "2k2" would be larger, yet smaller than 47K (since we need to feed both Q3 B and the diodes). If you kept the 100r emitter resistor, current would be 7mA (for ANY transistor!) which is ample for all but the heaviest loads. Four diodes with 100r would pass 18mA which will drive 600 Ohm loads.

However first tests can be done even simpler, with 5K resistor in place of Q3 and chums.

[midwayfair]

The clear intent is for Q3 to be a quasi-constant current load. There are better (predicable design) schemes for this. Jack's drawing in reply #32 gives two fine plans. The one on the right is ample and simpler. As your supply voltage is higher, the "2k2" would be larger, yet smaller than 47K (since we need to feed both Q3 B and the diodes). If you kept the 100r emitter resistor, current would be 7mA (for ANY transistor!) which is ample for all but the heaviest loads.

Yeah, I already amended my schematic and layout for Jack's plan. I just had your suggestions running on the breadboard for a couple days.

Four diodes with 100r would pass 18mA which will drive 600 Ohm loads.

Does only the diode voltage drop matter? Can I use an LED and a single diode instead of four diodes? Better to just change the source resistor?

[PRR]

> just change the source resistor?

I dunno. It doesn't HAVE to be there; the circuit works without it. I am wondering if Scott found that it increased Q3 collector resistance in a favorable way.

Yes, some green(?) LED may be a good voltage equivalent for four simple Si diodes. (It has been a long time since I questioned LED voltage against color.)

[midwayfair]

Yes, some green(?) LED may be a good voltage equivalent for four simple Si diodes. (It has been a long time since I questioned LED voltage against color.)

Hmm ... actually, now that I think about it and remembered to check the datasheets, this is probably a bad idea if I have to shoot for 18mA to drive a 600:600 transformer. Max current on a 3mm LED is only 20mA. Max on 1N4148s is 300mA. I'll stick with the Si diodes -- there's plenty of room, and I can always use an LED and some jumpers if I don't need to drive a transformer (e.g. op amp output).

>I dunno. It doesn't HAVE to be there

The Westhost and the picture Jack posted from the Talking Electronics article BOTH indicate that it controls current. Jacks' picture says 0.7/R is current. If R is 0 ...

[PRR]

> Max current on a 3mm LED is only

The LED gives a control voltage, NOT the output current!

I suggested a resistor value higher than 2K (because your voltage is higher), but lower than 47K (because that is just-enough without the LED). Say 12K. Takes around 2mA. LED won't blow-up.

Check: we probably can't want even 50mA in Q3 (heat). Taking a lowest-case hFE of 50, we need 1mA base current. With 2mA in the resistor, we can have that 1mA base current and still have 1mA in the LED. If Q3 is plucked as hFE=500, we have 0.1mA base current and 1.9mA LED current. It seems to do the right thing for nearly "any" hFE of Q3 that we find.

[PRR]

me> I can't say I have seen this exact plan, or particularly in a published "DIY" project.

That statement bothered me.

I finally remembered the very-similar bias plan: the original JLH power-amp.

http://oi62.tinypic.com/jpeutx.jpg

This is NOT a different way to do Scott's plan. Rather it can be understood as a plan which worked good for the originator but caused trouble in the hands of many builders. AND due to hFE variability with fixed resistor bias.

I sketched-out the bias action, and have attached JLH's later comments about constructor troubles.

Overall NFB forces "Ve" to be about half supply. Tr2 Base sits a bit higher, in this case about 13V *below* "+27V". The 13v across R1 plus R2 flows about 20mA. This then *splits* two ways, to Tr2 Base and (via Tr3) to Tr1 Base. Tr1 and Tr2 flow essentially the same current. If Tr1 and Tr2 have very similar hFE, then they have very-similar base currents, so the 20mA splits nearly half/half. 10mA to each Base.

Scott's plan does not have the half-split, and runs much lower power, but otherwise works the same. Current in 47K is multiplied by very uncertain hFE to be output bias current.

While the chart says you want 1.2Amps in Tr1 TR2, in fact the current will BE the 10mA times the hFE. JLH's results suggest his MJ480 had hFE near 120 (at 1.2 Amps). The data-sheet says MJ480's hFE (at 1A) can be 30 to 200! With 10mA in, the output stage current could be 0.3A or 2A! (Actually not so broad because 1A is on the falling side of the hFE/Ic curve; still large variation is possible.)

Speculation: MJ480 was a single-source part, probably all made in the same vat, and the production hFE may have been a narrower spread than the data-sheet promises. Moreover I speculate that in 1969, "all" the MJ480 in England came from a single batch, so JLH could have bought hundreds and not had much spread of idle current. This project was VERY popular, which implies that Motorola had to rush another crate or two of MJ480s, from another batch, and now builders were getting quite different currents.

His comments about "collection of odd resistors" is relevant, though only half the situation. The need to trim, and the *process* for trimming, is not clearly explained.

Rod has an annotated copy of the first article.
http://sound.westhost.com/jll_hood.htm

While JLH was developing this very "simple" power amplifier, others (D. Meyer, J. Bongiorno, D. Jensen, Dyna, Heath, many more) were developing circuits where every current and voltage was *nailed* (+/-5%) to an exact value by basic design. It's like building a house with logs and a rubber tape-measure, or building with planed timber and a good level. The less-accurate way works, unless it gets so far out of kilter that it falls down. Result may depend on builder's "knack" (or lack of knack). Since extra electronic parts got cheaper than a good level, the fad has been for more designed-in circuit control.