Blues Harp stompbox A/B/off, please weigh in with any ideas

[Skruffyhound]

Great! this

The | | marks around "-5V" mean Absolute Value. Throw-away the sign.

was pretty important to know. Thank you.

I've learnt more this week than in the last year I think.

Now I have to find out what R1 should be in the diagram
There's more of it here http://www.diystompboxes.com/smfforum/index.php?topic=100797.0;topicseen

Your drawing was good

[PRR]

> what R1 should be

You know your V. You can find I on the load-chip's (filter?) spec-sheet. B is 100 for practical purpose (see transistor sheet's minimum hFE at approximate current and aim lower).

[Skruffyhound]

Great Paul



2N2222 says ^^^at 10mA and 10V Hfe= 75
                         at 150mA and 10V Hfe = 100-300
                         Lets say 100  

I think my circuit is about a 50-60mA load and I'm running at lets say 12V
N5532 supply current at 12V = about 6.4 mA

TL072IP supply current at12V = about 1.34 mA per amplifier. So 2.68mA ?

6111 supply current at 12V = about 0.5 mA per plate. So about 1mA - not sure about this because that doesn't include the cathode. (the heaters are not part of this circuit) This is less than I previously thought

2 superbright Led's at 12V = Vin(12V)-VF(3.2V)/R(4.7K)= I (11.99uA) - so about 24uA  Although if I limit the current by half then : P=I^2R, 0.012 x 0.012 x 4700= 0.6768 Watts Thats a bit much, might have to give them more current.
                                             (12-3.2)÷0.02 = 440                                                                                                                                           0.02 x 0.02 X 440 = 0.176 back to quarter watt resistors and 40mA current

So altogether 50mA

(Vout(11.60)xBeta(100))÷Iout(0.05)=R=23.200Ω           23k     er.... does that sound reasonable?

The obfuscated data sheet wants this:

RX should be chosen to provide enough base drive to the external transistor so that it is saturated under nominal output voltage and maximum output current conditions

[PRR]

I don't see why superbright LEDs on a stage as status indicators would need even 2mA each.

Even rounding-up I don't see 15mA there. Perhaps 25mA if all your chips run to the high end of tolerance.

OTOH, the transistor must be "saturated". That means heavy over-drive. So if you did need 60mA, and if a '2222 can be as low as 75 (not to mention the 35 at -55 deg C spec because you won't be playing that cold), then you need nearly 1mA into the Base.

And the specs are for 10V drop. The idea here is "NO" drop. In practice, less than a Volt. The hFE will fall some at very low voltage. They don't like to tell you that. In the Fairchild lit Fig 12 shows the trend, but not all the way to zero volts. Fig 2 shows low Vce _but_ assumes you drive with Beta of 10. So 1mA into Base for 10mA load, 5mA for 50mA. It is likely you don't have to push it that hard. It is likely that very few modern '2222s have low hFE. But that's how you'd figure it if you were making a million units.

Yeah, use 22K and see what happens. If the Collector won't pull very close to Emitter, try 10K and see if it gets better.

I'm annoyed because this is a Known Issue and they could have built-in 4 extra transistors instead of "fixing it in the application sheet". Aside from the relative cost of transistors versus paper, this is untold hours of engineering salary over the product's lifetime. And a sure way to make the engineer look at other chips for the next product.

[Skruffyhound]

I'm going on this http://www.banzaimusic.com/LED-5mm-blue-ultrabright-8000mcd.html info, because these are the LED's I have. As you could see though when I trimmed the current down to 12mA I suddenly had to dissipate over half a watt.
Nobody uses 1 Watt resistors around here. I know I'm running at 12V, but anyway this suggests to me that they should be around 20mA. I'll take some measurements. I don't know where the 20mA spec comes from.
I could also just stick a little old timey red LED's in, they have a 2mA spec.

This made me think though. Some fairly big part of the load will be switched on and off. I may need to recalculate this resistor when I figure out the switching. There's no need to include the LED's in this load on the chip (I don't think).
Interesting how this project evolves as I get more insight.

It is annoying that they didn't just stick this transistor in the chip, I can only think that it must be detrimental for one of the many other functions (voltage doubler etc.)
It is trying to be all things for all people this chip, which I guess just leads to compromise.

[PRR]

Don't understand.

> had to dissipate over half a watt.

12V at 20mA 0.020A is 0.240 Watts. Total in R and LED.

Given say 3V in LED, the resistor is 9V 0.020A or 0.180 Watts.

1/4 will do. 1/2W may live forever.

And 20mA in that LED gives 8000mcd, 8 Candellas, a LOT of light.

That's the maximum. _Any_ LED may be run at lower current; and often is, because we don't need that much light.

> old timey red LED's in, they have a 2mA spec

All that I can recall are rated 10mA-30mA (usually 20mA) MAX. Again, often run at a small fraction of rated power because who needs that much?

[Skruffyhound]

Ok, I've somehow been thinking about this the wrong way round, the Power calculation above is confusing me.
If I say  (Vs - Vf) ÷ If = R    (Supply Voltage - LED forward voltage drop) ÷ LED forward current = R       (12 - 3.6) ÷ 0.010 = 840Ω

It would be better with a graph on the datasheet  because the Vf is stated at a fixed If of 20mA, and the If is what I want to change. Maybe that's what is confusing me. I split the difference to give Vf of 3.6

Then :    12V - (voltage drop from the LED) - (voltage drop from the resistor) = 0
            (12V -  3.6) - 8.4 = 0

P = IV    0.01 x 8.4 = 0.084 W
R = 840 (or thereabouts) dissipates 0.084 W.
I use 1/4 watt resistors, no problem. My resistor was a guess (3 posts above), I thought I could remember using 4K7 resistors, but it was just too big.

Thanks Paul

[PRR]

> Vf is stated at a fixed If of 20mA

In solid-state diodes, including LEDs, the forward voltage "hardly changes" over a w-i-d-e range of currents.

(That's why they make OK clippers/fuzzes.)

That 3.6V at 20mA is probably above 3.0V at 1mA.

So it's easy. (And I think a spreadsheet is just a distraction.) You got 12V, and you got a 3V diode. You need to waste-off 12-3= 9V.

OK, spreadsheet:

9V @ 20mA = 450 ohms
9V @ 2mA = 4500 ohms
9V @ 1mA = 9000 ohms
9V @ 0.20mA = 45000 ohms

Yeah, if you compute the *exact* answer for 12V and 3.6V and 450 ohms you get 18.7mA (because the approximation "3.0V" was off). But 20mA, 18mA, who cares?

Anyway, you reach for your 450 ohm resistor and find 470 ohm nearest common value. Now it's only 17.9mA, and who cares?

Likewise the "1mA" approximation of 9K leads to a 10K standard-value, And say the LED is really 3.4V near 1mA. You really get 0.86mA. That's plenty close to 1mA. You don't know that you "need 1mA", that was just a handy guess to get "much less than FULL brightness" (which you rarely need).

The LED dissipation is limited by the rated 20mA MAXimum current. Apparently 3.6V*0.020A= 0.072 Watts (72mW) is all it can take. And if you stay down to 20mA or less, you don't have to compute LED dissipation.

You, as you say, need resistor dissipation. In this case, you start from 12V and at high current the LED won't be under 3V. So 9V in the resistor. V^2/R= power. 9V*9V/450= 81/450= 0.18 Watts.

And at lower currents? The voltage is very-nearly the same, the current is much less, we expect the dissipation will be less. Assume the same 9V across a 4700 ohm resistor. 0.017 Watts. Get very paranoid and 'assume' the LED drop might be zero. The full 12V across 4,700 ohms is 0.03 Watts.

[Skruffyhound]

I can't replace the photobucket pics
this would be the correct version