Power supply filtering question/current draw

[midwayfair]

You are not the first to follow this path. Plagiarize!! Study every small-signal audio amplifier plan you can find. Most of your ??s relate to power distribution and filtering, so look at that part of the plans.

To be fair I was trying to plagiarize! The changes I made and the questions I've been asking were trying to fix a problem I encountered that I apparently shouldn't have.

I think he's building something like this

http://i222.photobucket.com/albums/dd317/awlivoyl/SSEGM3.gif

With the 390R 2W the maximum current should be limited to no more than 1/8th watt (48v/390r=0.123A)

I can't see how the 100r emitter resistor blew unless there's some mods done that we don't know about.

The output buffer on the Hamptone (almost identical to this one) was giving me trouble even when I breadboarded it exactly as shown in the article on 24V. Obviously something is wrong if even a 470R 1/4W on the emitter is lighting on fire running at 48V ... I just can't figure out what that something is. The Hamptone uses a Darlington on the top of the buffer, that's the only real difference. Other people have clearly built things like this in the past without issue.

Is A fundamentally different from B?
A:
48V\
390R
|
Buffer
|
100R
\G

B:
48V\
48R
|
Buffer
|
470R
\G

I mean, those resistances are still in series, aren't they?

I'm just more and more confused because everyone's telling me that I should be drawing way less current than I am measuring. I'm bad at math but I don't think I'm this bad. If a 100R is supposed to work on the emitter, I don't understand why I need several times as much resistance and much higher wattage just to keep from destroying the resistor.

[midwayfair]

I breadboarded the circuit hymenoptera linked to (including using 2SK170). The FET stage + source follower are still hooked up. I didn't have a 390R 2W so I used a 470R 2W. Each FET is set to ~4.6x gain (22K and 4k670 [3k9 bypassed +470R]. The input is grounded through a big film cap. The gate is biased so I can adjust the gate-base connection to half supply.

This is what I measured:
No output buffer transistors:
Supply voltage after the 470R: 45.2 = 5.3mA draw.

With output buffer transistors in:
100R source resistor: supply voltage after the 470R = 19V = 60mA, and my voltages occasionally jump all over the place
470R source resistor: supply voltage after the 470R = 28V = 40mA
1K source resistor: supply voltage after the 470R =  33V = 29mA
2k7 source resistor: supply voltage after the 470R = 39V finally something that looks reasonable.

Changing the 68K for a smaller or larger resistor has almost no effect.

Anything below about 1K on the source of that buffer makes the upper transistor too hot to touch. The lower transistor seems okay at all times, but a 100R is not long for this world (even 1/2W). Even though that or a similar value -- of even smaller values! -- is what's shown in ever example of this buffer I can find.

I've tried almost every BJT transistor I have in there. I've also tried multiple similar buffers with low-value resistors and either a BJT or Darlington on the top. All of them draw massive amounts of current.

I can't see anything wrong with it. I've checked the pinout on every transistor, and I can't believe that every single schematic is making a mistake.

What can I possibly be doing wrong that would result in the buffer drawing 55mA on its own when everyone is telling me that it shouldn't even be drawing half of that?

[hymenoptera]

I am so sorry. I was running a SERIOUS fever and should NOT have been posting. 

Ignore whatever I was saying. I was somehow confusing watts and amps, and I don't know what else. 390 ohms + 100 ohms = 490 ohms. 48v across 490 ohms would be about 100mA. That seems about right for what you're getting. It's a pretty high current stage there, designed I think to drive a transformer.

Still not well enough to figure out what's burning up your resistor though. Just ignore what I was saying about an 1/8th watt yesterday. Carry on. 

[hymenoptera]

MJE800 and BD139 are TO-126 package. Medium-duty big chunks of stuff. Wouldn't recommend this circuit with any TO-92 transistors in those positions! They're gonna get a little warm

[midwayfair]

MJE800 and BD139 are TO-126 package. Medium-duty big chunks of stuff. Wouldn't recommend this circuit with any TO-92 transistors in those positions! They're gonna get a little warm

The big metal cans I used didn't get as hot ... I think they're 2269. I'm still concerned that it's drawing so much more than the Hmaptone article specifies (40mA).

Thanks for the help -- hope you feel better soon.

[PRR]

> as shown in the article on 24V. Obviously something is wrong if even a 470R 1/4W on the emitter is lighting on fire running at 48V

The Hamptone article shows 100 Ohms, not 470 Ohms.

In fact where did 470 Ohms on Q3 emitter come from??

That 100 ohms is not critical and could even be omitted.

But note the article shows it as a HALF-Watt part.

BTW, the article shows Q3 Base resistor as 47K, not 68K. IMHO both may be too small for many builds. I already did math for 68K (below); 47K would cause higher current.

Stop fiddling with non-specified FETs; I'm not sure what is going on there.

If the FET Drain is anywhere near mid-supply, the current in the buffer is determined *only* by the 68K resistor and the hFE of Q3. You can simplify to just these two parts, plus the Q3 Emitter and any B+ resistor.

Whatever current flows in the 68K is multiplied hFE times and sucked in Q3 Collector-emitter path.

Assume Q3 has hFE of 100, and that Vbe of Q3 is very-small compared to total supply voltage.

Then you can, for analysis, approximate Q3 as a hundred 68K in parallel, or 680 Ohms.

(Actually 673 Ohms, hFE+1, but we can't be that accurate.)

With straight 24V, this gives Q3 current of about 24V/680r, or 35.3mA.

With straight 24V plus 47 Ohms from 24V and 100 Ohms under Q3, this gives Q3 current of about 24V/(680r+47r+100r), or 29mA. If the 100r is 470r, 20mA.

BUT: Scott specifies Q3 as "generic NPN". We *could* pluck out a 2N5089 with hFE=1,000. Now the 68K is divided by 1,000 (actually 1,001). Q3 acts like 68 Ohms! In the article circuit, 24V/(68r+100r)= 143mA!!

I have seen many people say this plan runs super-hot with the specific parts they used. I have no doubt it worked for Scott as-drawn with the parts *he* used. However I do not believe the current (and power) is well-determined enough for blind following.

I hesitate to re-design Scott's empirical design because he says every part was determined by "sound". However where there is smoke there is no sound.

IMHO, "68K" should, for most "generic NPNs", for most modern loads, be a lot closer to 330K, not 47K or 68K. That puts you in the general ballpark of 10mA. Total dissipation at 24V supply is around 0.25 Watts. If all parts are 1/4W or 1/3W, and none takes all the heat, then nothing will burn. It should pass sound. Any possible sonic improvement at higher currents (lower the 330K) can be approached cautiously without the bee-beep smoke alarms.

> Is A fundamentally different from B?

Yes. (And I had not seen this "A".) The largish 390r resistor will sag the voltage to the 47K or 330K resistor. This is Negative Feedback. The huge (29mA versus 143mA) variation of current with variation of hFE will be significantly reduced. As a special-case, using 68K, if hFE is 173 then 390r and Q3 split the available supply voltage-- 390r:(68K/175) = 390r:390r. As that plan starts from twice Scott's voltage, a halfway split is reasonable.

OTOH half your precious power goes as waste heat in the 390r.

(And it does nothing about the need to hand-trim for the specific JFET used for Q1.)

[PRR]

A further point-out:

If a circuit were "happy" at 24V, and you run on 48V, the heat in each part may QUADRUPLE.

Not so true for chips with constant-current biasing (though dynamic loads may heat as square of voltage). Very true for resistors, and simple resistor bias schemes.

Also light-bulbs, which is part of the reason we had to replace incandescents often where wall-voltage ran high.

[R.G.]

A further point-out:
If a circuit were "happy" at 24V, and you run on 48V, the heat in each part may QUADRUPLE.

Not so true for chips with constant-current biasing (though dynamic loads may heat as square of voltage). Very true for resistors, and simple resistor bias schemes.

The ugly truth is that even with constant current schemes, the power will double. No matter what combination of stuff is between the poles of the power supply, the power that goes out is always V * I. But you're right - a resistive-responding circuit will heat up four times as much. CC-biased circuits will only double.

Also light-bulbs, which is part of the reason we had to replace incandescents often where wall-voltage ran high.

Light bulbs are my favorite example of several things. One is the difference between temperature and heat/power. The filament on any modern incandescent is run at temperatures where the color of the light is the handiest temperature measurement. White-yellow is about the same temp for all normal bulb filaments. But you can have grain-of-wheat bulbs glowing white-yellow with half a watt going in, the same temp as the filaments in a 250W BIG incandescent bulb.

What's different is the power. The big bulb is designed to get rid of 250W or heat by radiation, convection and conduction. The grain-of-wheat can only dump half a watt before glowing the same temperature.

The subtlety here is that when you heat something, the temperature of the thing will rise until power out, by conduction, convection, and radiation combined is equal to power in. This is a fundamental law of the universe. The grain of wheat bulb is getting only a tiny trickle of power in - but it's been set up to have a really difficult time letting heat out, so the temperature rises until radiation lets it get rid of the total power. The glass envelope and tiny wires can't get rid of more heat, so it has to be by radiation. If you put a heat source in a situation where the heat cannot be conducted, convected, or radiated away, there is literally no limit to the temperature the thing will achieve. No limit. That's a little scary and profound.

It's one reason that spacecraft cooling is one of the most critical things in spacecraft design. Space is a near-ideal insulator. There is no conduction or convection in space. All you can do to get rid of heat is to radiate it away. Space is really cold - about 3-4 absolute. But there's no way to get rid of heat except by radiating, so in the short run, if you can't radiate enough, you will *cook* inside your spaceship, not freeze. If you can't manage to radiate away the heat, your body's metabolism will kill you from overheating, not freezing.

The Stefan-Boltzman law says that energy transferred from a surface is proportional to the fourth power of the temperature. Because resistances have internal powers that go up as the square of the applied voltage, and because tungsten filaments have variable resistances with temperatures, it turns out that incandescent filament life varies inversely with the **twelfth power** of the applied voltage. The math conspires to foce light bulbs to run at their designed voltage - or die.

[LightSoundGeometry]

I have a 48V/30W power supply that can supply 650mA.... It's regulated...

That gives us Assumption #1: the regulator is designed OK and gives minimal voltage change between 0 current and full rated current. In other words, a "stiff" or low impedance supply.

Based on that assumption, no "stealing" would be happening. In the case of the multiple parallel connections (with series elements in each parallel circuit) you are really worried about power supply "sag," - to my mind, I think "stealing" would refer more to what happens when you tap off a power supply in different series spots (like the various filter nodes in a tube amp.)

1) Are circuits 2-4 going to steal voltage from circuit 1? In other words, do the current draws appear in parallel to each other, or to the 48V PSU?

They appear in parallel to each other AT the 48 V PSU.

2) I calculated ~5W above for the 100mA circuit. If I use two 5W resistors, does that actually give me 10W of total dissipation at 20 Ohms in this case? Do I need to worry about the capacitors for purposes of determining the dissipation needed from each individual 10R, or can I still treat them as a series?

Each R in series will carry the total current drawn by that subcircuit's load only, say, 100 mA in your examples. Each one will be required to dissipate .1 x .1 x 10 = .1 W. (P = I^2*R)

If we use another formula to double check, we have a voltage drop of V=I*R, V= .1 x 10 = 1 V per resistor, then P = I*V = .1 x 1 = .1W.
OK.
How did you get 5W?

3) I was wondering if I could get away with using the same filtering line for multiple circuits. But they'll have different current requirements when in use, correct? So I should definitely keep them resistor-isolated?

If you hang all the circuits onto one set of filters, then the voltage drop through the filter will depend on how much current is being drawn at that moment - if  Load A, B, C, & D are connected, or just A & B, etc.

If you give each load a separate filter then the maximum voltage drop when all loads are active will be smaller and will not depend on how many loads are connected (because the regulated power supply will take care of making sure your input to the filter is consistent.)

The caps act as open circuits for DC so we can ignore them completely in this DC analysis.
They act as short circuits for high-enough frequencies of AC and so they will filter the noise and ripple.

Think about the difference between these two drawings and especially what happens to the bottom one when some of the 460 ohm resistors are switched in and out.

started on something similar to this today when introduced to parallel circuits..the inverse reciprocal  law..started on the frequency to. introduced to the formula for finding reactance Xc=1/6.28fc, Frequency Fr=1/6.28√LC, Impedance Z=√r^2+x^2 and susceptance BL=1/6.28fL



absorbing all I can PRR   ..one of these days I might even be able to help others lol

[midwayfair]

In fact where did 470 Ohms on Q3 emitter come from??

I upped the resistance because the half-watt 100R burned up. I didn't have any higher wattage 100Rs, so I used the next closest thing on hand, which was a 470R 2W.

If a circuit were "happy" at 24V, and you run on 48V, the heat in each part may QUADRUPLE.

Gah! Also: Aha.

BTW, the article shows Q3 Base resistor as 47K, not 68K. IMHO both may be too small for many builds. I already did math for 68K (below); 47K would cause higher current.

Assume Q3 has hFE of 100, and that Vbe of Q3 is very-small compared to total supply voltage.

With straight 24V, this gives Q3 current of about 24V/680r, or 35.3mA.

With straight 24V plus 47 Ohms from 24V and 100 Ohms under Q3, this gives Q3 current of about 24V/(680r+47r+100r), or 29mA. If the 100r is 470r, 20mA.

BUT: Scott specifies Q3 as "generic NPN". We *could* pluck out a 2N5089 with hFE=1,000. Now the 68K is divided by 1,000 (actually 1,001). Q3 acts like 68 Ohms! In the article circuit, 24V/(68r+100r)= 143mA!!

I have seen many people say this plan runs super-hot with the specific parts they used. I have no doubt it worked for Scott as-drawn with the parts *he* used. However I do not believe the current (and power) is well-determined enough for blind following.

I hesitate to re-design Scott's empirical design because he says every part was determined by "sound". However where there is smoke there is no sound.

IMHO, "68K" should, for most "generic NPNs", for most modern loads, be a lot closer to 330K, not 47K or 68K. That puts you in the general ballpark of 10mA. Total dissipation at 24V supply is around 0.25 Watts. If all parts are 1/4W or 1/3W, and none takes all the heat, then nothing will burn. It should pass sound. Any possible sonic improvement at higher currents (lower the 330K) can be approached cautiously without the bee-beep smoke alarms.

First: Thank you thank you thank you! This fixed the problem.

I rummaged through the bin and found a marked 2N2270, which I know can take really high voltage (like 80?), and it's a big metal can, and it's low gain (it was in a bin marked "50-55"). I used that for Q3 (lower half of the buffer), and upped the 47K to 330K and then lowered it to 220K. Current draw is now quite low with either. Even with a 100K I measured only ~32mA for the entire circuit. (not much bigger than the 68K ... I guess I should have paid closer attention to the measurements when I changed that resistor, but I was scared about more things blowing up.) I left the 470R on the emitter of Q3 (more on that in a second).

A 47R on the power supply now drops just over a volt, and everything's stable. There's a HUGE amount of drive from the FETs and I can safely drop the source resistance some as well if for some reason I need to.

That 100 ohms is not critical and could even be omitted.

I think it might be ... another one bites the dust ... 

A 100R seems to be living. It's my last half-watt. I left it hooked up for an hour to see if any damage occurs but nothing's getting hot. I'm also not hearing any difference with a 470R, though, so I don't know if there's any reason not to go with a safer part except that it might lose a little bit of drive for the transformer. I'll toss some extra 2W in with my mouser order as well.

[amz-fx]

The article says that the hfe of the Q3 ZTX653 is 100 but that's the minimum value and it will typically run about 200 at the currents involved. If you put this into the equation that PRR posted earlier, you can see that the current goes up by a significant amount, which is probably why people are reporting that it runs hot. Scott also stressed that you need enough current in the output devices for adequate drive so you don't want to increase the 47k value too much (depending on the hfe of Q3 of course).

Another important item to check is the dc voltage of the output at the junction of the two bipolar transistors, immediately before the 10uF poly output capacitor. You want to that to be 12v DC when powered with 24v. If it is not close to 12v then some parts values will need to be tweaked. Post your output DC idle voltage and let's see what you have.

A pair of diodes added on the base of Q3 would make the output stage less dependent on the transistor hfe, and I doubt that it would change the sound at all, though that would be tinkering with Scott's empirical design.

Best regards, Jack

[midwayfair]

The diodes are cathode to ground, right?

What exactly do they do? Is it just to shunt voltage higher than 1.2V to ground? Do they need to be anything in particular, or are 4001s ok? Better yet: is there something I can read that actually describes this type of buffer? I don't even know what it's called to go looking for it.

[amz-fx]

The one on the right is the example, and the LEDs are the load there instead of the other transistor. You need to change the value of the emitter resistor from the original schematic.

If we want 40 ma. then R = 0.7/.04 = 17.5 ohms. Use 18 ohms as it's the closest standard value.

The 47k on the base of the Q3 transistor might need to be dropped in value. 10k should be okay. 1N4148 diodes are fine.

You might want to build it as originally designed and see how you like it first.

regards, Jack

[midwayfair]

Thanks, Jack. Also, the article that image came from is quite good. I read it a couple years ago but clearly didn't internalize it.

http://www.talkingelectronics.com/projects/TheTransistorAmplifier/TheTransistorAmplifier-P1.html

[PRR]

> I upped the resistance because the half-watt 100R burned up.

That would be right-thinking if there were a *constant voltage* across the resistor.

Here there is (I agree not obvious) a nearly *constant current* through the resistor.

4.7X resistance then means 4.7X more heat.

470r will burn-up almost 5 times faster than 100r.

It really IS useful to be sent to a Boot Camp and be required to do a Hundred R-I-V-P problems before breakfast, and then all day, for 6 weeks. That's before the drill-sarge lets you touch a transistor.

And R.G.'s clarification to my post bears repeating:

If the voltage OR current in a system is doubled, Power will Double. If *both* are doubled (a common situation), Power will QUADruple.

If the voltage OR current in a system is doubled, Power will Double. If *both* are doubled (a common situation), Power will QUADruple.


We can make exceptions. Most devices have a "breakdown" where current (and power) suddenly goes to infinity. OTOH we can design "foldback" circuits which cut current way-down if voltage is too-high.

Jack has explained some of the possible issues with Scott's values. Since you are In Trouble, may I suggest *simplifying* to a plan that WILL work and you KNOW what it will do before you smoke-test?

Remove Q3. Put about 5K between the E and C holes. This is a resistor-coupled cathode follower. A very respectable design. If the Q-point (Q2 emitter) is the original ~~12V, the 5K will flow about 2.4mA. Some checking will assure you that no part should heat even 0.030 Watts. If run on 24V and mis-biased to ~~20V at the Q-point, the 5K resistor could hit 0.080 Watts, fine for a 1/4W part. If you run 48V and don't have the Q-point under control, the 5K will throw 0.46W, so a 1/2W (or two 1/4W) part will run for weeks.

The 5K emitter load is not low enough to change gain (0.02dB), and will drive typical modern studio 10K loads quite well, nearly as hard as Scott's plan.

Scott's plan "will" drive 600 Ohm transformers, but a single-ended Big Studio 600r output stage is a stretch. Get it working for 10K loads first.

[tubegeek]

started on something similar to this today when introduced to parallel circuits..the inverse reciprocal  law

The way I teach this, we don't look at the law formula until we can do the calculation ourselves directly from Kirchoff's Current Law.



Make a table:

	R1	R2	R3	R4	Req/Total
E	?	?	?	?	?
R	100	25	100	50	?
I	?	?	?	?	?

Next, I point out that the voltage doesn't matter, the Req will be the same no matter what voltage we choose to calculate with. let's try something easy like 1V.

Calculate the current in each branch, I = E/R:

	R1	R2	R3	R4	Req/Total
E	1	1	1	1	1
R	100	25	100	50	?
I	.01	.04	.01	.02	?

KCL says the total current is equal to the sum of the branch currents:

	R1	R2	R3	R4	Req/Total
E	1	1	1	1	1
R	100	25	100	50	?
I	.01	.04	.01	.02	.08

KCL says Req = V/Itot

	R1	R2	R3	R4	Req/Total
E	1	1	1	1	1
R	100	25	100	50	12.5
I	.01	.04	.01	.02	.08

Then when I'm done showing them that I'll show them the formula and how easily they could have f'ed it up if they didn't use the table to keep it straight at first. (Banging the inverse reciprocal-sum formula straight into a calculator is a guaranteed disaster every time for my students, who are not exactly all math whizzes and whose last encounter with the order of operations was many moons ago in a vague painful memory.)

Then the next thing I do is show them how much easier this particular problem would be taking pairs of equal resistors, replacing them with half the value, and repeating.

PSYCH!

[PRR]

> inverse reciprocal-sum formula straight into a calculator

YES YES YES.

My fingers can't find chords, but somehow they have learned to bang the 1/x key in proper sequence (9 out of 10 times). If I watch myself I get confused.

It was different in days of slide-rule. Not better, the SR is a poor adder. But you HAD to keep your fractions straight, because reciprocal was just reading C and D scales as D and C. When I do math, a mental line runs through it all-- my mental fraction-bar. Many problems go easier upside-down then inverted at the end.

[midwayfair]

I think I'm out of the woods.

With a 100K on the base of the lower transistor and the 2N2270 (hfe in the 50s) in the lower transistor, even a 100R 1/4W just gets warm, so a 1W or 2W should be sufficiently overkill. The upper transistor gets slightly warm (I have to take it out and press it to my lips to tell -- I can't feel it through my fingers).

Also, with the bigger transistors and low gain on the bottom, the 47K works okay on 24V without blowing things up. I think the only reason it wasn't working on 24V before was that I was using much too high gain a transistor for the bottom, thinking that the top one was responsible for all the current.

Current consumption is similar between the two voltages just with doubling that resistor.

I still haven't been able to find anything that actually talks about that type of transistor arrangement, even though I've seen it in several mic preamps. I've seen a bunch of close things -- totem poles, push-pull stages, and complementary power amplifiers, but nothing exactly like this. Does it have a specific name that I just don't know to look for? I know Jack linked to the picture showing the LEDs as the load for a constant current transistor, but I don't see how replacing the LEDs with a transistor emitter follower is an obvious arrangement that wouldn't be explained in detail somewhere like that giant article.

[amz-fx]

You might enjoy this page:

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

Best regards, Jack

[midwayfair]

You might enjoy this page:

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

Best regards, Jack

Thanks!