pentaboost ?'s

[LightSoundGeometry]

got a few questions about rick holts 5672 tube boost - does it matter how the tube heater sees the 1.25 volts ? or do I need the IC regulator ? I dont have one and was going to make a voltage divider to get my heater volts.

and the heat sink ..where is that attached? I can see one in pictures but not enough to see whats going on. what part is heating up and needs to be dissipated ? 

[bluebunny]

You can make the 1.25V any way you like.  Rick often uses a fat resistor to drop voltage.  One way or another, dropping 9V or 12V to 1.25V will dissipate a load of heat.  Either through a regulator (with a heatsink) or through a resistor.  If you go the resistor route, make sure you to the math and pick one with a suitable W rating.  And plan on adding ventilation to your enclosure.

[LightSoundGeometry]

makes sense now, the lost voltage energy has to go someplace and resistance burns energy through heat dissipation.

Its quite obvious I have not ripped apart a lot of old electronics in the name of science , so where and how to place the sink if I do that? can you heat sink a resistors leg by chance or will that cause extra resistance?


I think I am now making the transition from pedal devices to larger voltages and tubes ...my goal in school and with electronics is to make tube amplifiers for audio and guitar.

I have the charge pump on the max1044 cool as a cucumber but this will be my first time stepping down voltage where the energy is lost someplace eg a heatsink ..I have a lot to experiment with ...shold I work on my tube projects or learn to use darlington pairs for effects...or do homework lol ..

[PRR]

> a voltage divider to get my heater volts.

A resistor divider?

To be blunt: that is stupid.

What happens when you put a LOAD on a voltage divider? The division ratio sags. How do you make it not sag much? Pick divider resistors 10X smaller than load resistance. What is the (undesired) power in the divider compared to (desired) power in the load? Huge!

Yes, if you need a nano-Amp to bias a sensitive input, a divider is the bees-knees.

But say you got 9V, and want 1V 0.1 Amps in a heater. The 1V 0.1A heater is a 10 Ohm load. We make the bottom divider resistor 10X smaller, 1 Ohm. The top divider resistor looks like 8 Ohms. Total current looks like 9V/9r or 1 Amp. Total power is 9 Watts. Compared to 1V 0.1A or 0.1 Watts in heater. 8.9 Watts waste! 1% efficiency! And the 1A load would be heavy on many wall-warts. AND we have not done the 1/10X approximation corrections: we really get 0.9183V, not 1V, so will need to re-tweak values.

The heater IS a resistor. A little funky, but when it hots-up enuff to work good, it is resistor-enough. (far more predictable than the 0.1nA to 50nA of an opamp input.) So make that part of this "divider". 9V supply to 1V 0.1A, you need to lose 8V 0.1A, that's an 80 Ohm resistor. And just 0.9 Watts total, not ~9W. 11% efficiency; you can't run heavy equipment that sucky, but fine for small-stuff.

To be fair: we got through the last century on lamps of 1% efficiency. And compact wide-range loudspeakers are doomed to 1% efficiency.

I don't see an advantage in a regulator. Tube heaters are well specified. Resistor dropper is the same heat with less trouble, and slightly reduces start-up stress.

[LightSoundGeometry]

> a voltage divider to get my heater volts.

A resistor divider?

To be blunt: that is stupid.

What happens when you put a LOAD on a voltage divider? The division ratio sags. How do you make it not sag much? Pick divider resistors 10X smaller than load resistance. What is the (undesired) power in the divider compared to (desired) power in the load? Huge!

Yes, if you need a nano-Amp to bias a sensitive input, a divider is the bees-knees.

But say you got 9V, and want 1V 0.1 Amps in a heater. The 1V 0.1A heater is a 10 Ohm load. We make the bottom divider resistor 10X smaller, 1 Ohm. The top divider resistor looks like 8 Ohms. Total current looks like 9V/9r or 1 Amp. Total power is 9 Watts. Compared to 1V 0.1A or 0.1 Watts in heater. 8.9 Watts waste! 1% efficiency! And the 1A load would be heavy on many wall-warts. AND we have not done the 1/10X approximation corrections: we really get 0.9183V, not 1V, so will need to re-tweak values.

The heater IS a resistor. A little funky, but when it hots-up enuff to work good, it is resistor-enough. (far more predictable than the 0.1nA to 50nA of an opamp input.) So make that part of this "divider". 9V supply to 1V 0.1A, you need to lose 8V 0.1A, that's an 80 Ohm resistor. And just 0.9 Watts total, not ~9W. 11% efficiency; you can't run heavy equipment that sucky, but fine for small-stuff.

To be fair: we got through the last century on lamps of 1% efficiency. And compact wide-range loudspeakers are doomed to 1% efficiency.

I don't see an advantage in a regulator. Tube heaters are well specified. Resistor dropper is the same heat with less trouble, and slightly reduces start-up stress.

I will run a few tests and simulations just to see what happens - looking at the TO220 and the 4 fin heat sinks that are made for them, will probably get a few.

http://www.datasheetq.com/LM317-doc-Linear

I was going to to do a resistor divider, not good it sounds from what you are mentioning until the end and you then you lean towards the resistor dropper, lol. the stress you are taling about on the power supply concerns me..

http://www.ti.com/lit/ds/symlink/lm317.pdf

looks like the physics inside the regulator itself is a resistor network with an input and a output. I need to red the sheet some more as I am not fully understanding the entire circuit just yet. At a glance, I can see it shows several ways to regulate I and E, one example being figure 9.3.3 a current limiter circuit.

Now I know what the circular cutout on the regulators is designed for


modified - 1

looking further into the data sheet, I am not understanding exactly what the IC is doing being you have to build a circuit around it to get the results..trying to find a google image of the actual physics inside the IC, like the show the insides of a cpu etc

[PRR]

> regulator itself is a resistor network with

Not seeing the tree for the weeds.

A voltage regulator is a reference voltage, an error-amplifier comparing a fraction of the output to the reference (all weeds), and a big emitter follower which does the Hard Work (the tree).

For efficiency estimates, disregard the small stuff. Assume a magic pass-device which will waste-off whatever voltage you want wasted-off. If the supply and load are highly variable, this needs the error-amp to handle the different conditions. But if your input is reasonably constant (such as domestic wall power), and the load is a hot wire which does not change (or replaced with same-type), there's no "regulation" to be done, just a dumb resistor.

The regulator will also draw added power for its reference and brain. Typically 3mA to 0.05mA. This is small on a 100mA load, but certainly not "more efficient" as I have seem claimed.

> image of the actual physics inside the IC

It is a '741-like amplifier plus a TIP120 power transistor larger than the error-amp. Getting both on one chip without cooking the amp was a minor revolution of the industry.

FWIW: not always emitter follower or Darlington, there are many ways to shave the cat.

[LightSoundGeometry]

> regulator itself is a resistor network with

Not seeing the tree for the weeds.

A voltage regulator is a reference voltage, an error-amplifier comparing a fraction of the output to the reference (all weeds), and a big emitter follower which does the Hard Work (the tree).

For efficiency estimates, disregard the small stuff. Assume a magic pass-device which will waste-off whatever voltage you want wasted-off. If the supply and load are highly variable, this needs the error-amp to handle the different conditions. But if your input is reasonably constant (such as domestic wall power), and the load is a hot wire which does not change (or replaced with same-type), there's no "regulation" to be done, just a dumb resistor.

The regulator will also draw added power for its reference and brain. Typically 3mA to 0.05mA. This is small on a 100mA load, but certainly not "more efficient" as I have seem claimed.

> image of the actual physics inside the IC

It is a '741-like amplifier plus a TIP120 power transistor larger than the error-amp. Getting both on one chip without cooking the amp was a minor revolution of the industry.

FWIW: not always emitter follower or Darlington, there are many ways to shave the cat.

hopefully I can get an hour or two free tonight to experiment ..I ordered the heat sinks and to222 lm317t but I noticed I had a l7805cv on hand to breadboard with and experiment with. I am wanting to set up the resistor divider to see how hot is cn get as well lol

[amptramp]

There are also buck regulators that use the energy stored in an inductor to drop 9 volts to a lower voltage with less loss than a resistor.  It is a switching regulator which means you can regulate the voltage at the output.  The mathematics is here:

https://en.wikipedia.org/wiki/Buck_converter

Most modern designs are clocked but there are some that use a comparator to switch the buck transistor on when voltage is low and off when it is high.  The frequency is variable and since a regulator is a negative resistance element, any input filtering must be heavy on the C and as light as possible on the L.  The negative resistance comes from the fact that as input voltage declines, input current must rise to maintain a constant output power.

[LightSoundGeometry]

There are also buck regulators that use the energy stored in an inductor to drop 9 volts to a lower voltage with less loss than a resistor.  It is a switching regulator which means you can regulate the voltage at the output.  The mathematics is here:

https://en.wikipedia.org/wiki/Buck_converter

Most modern designs are clocked but there are some that use a comparator to switch the buck transistor on when voltage is low and off when it is high.  The frequency is variable and since a regulator is a negative resistance element, any input filtering must be heavy on the C and as light as possible on the L.  The negative resistance comes from the fact that as input voltage declines, input current must rise to maintain a constant output power.

I am new to tubes..I know in building fuzz boxes I am not even concerned with current except the leakage from the device itself. ..maybe for a LED draining a battery ..

not sure how a tube takes a rise in current or if it even matters.,,thinking the heater sees the low volts but a huge spike in I ..I guess thats why they make breadboards lol.

i havent had time to work on anything ..running errands and working etc ..no time to just play around with stuff like I want to ...hopefully tonight I can get an hour or so to mess with stuff

[amptramp]

The 5672 takes 50 mA @ 1.25 volts on the filament.  This very low for a tube filament but 50 mA is pretty much all you can get with any longevity out of a 9 volt battery and you still have to supply 67.5 volts for the plate and screen with 3.25 mA for the plate and 1.1 mA for the screen.  A buck regulator will essentially trade voltage for current and act like a DC stepdown transformer so the 50 mA @ 1.25 volts would be derived from 9 volts @ 6.944 mA at 100% efficiency.  Even if the supply was only 50% efficient, you would still have less than 14 mA drain.

A boost converter for the B+ supply would take 32.625 mA to generate the 4.35 mA drain at 67.5 volts if it was operating at 100% efficiency but fortunately the high voltages make it easy to get about 80% efficiency so you would get 40.78 mA drain from the 9 volt battery.

But the 5672 is a power amplifier pentode putting out a mighty 65 mW of output.  Since it only has a transconductance of 650 µmhos, you may need a bit of experimentation to see if you can get sufficient gain and what the actual current drains are with the values you have chosen.  I will be interested in seeing what you come up with.

[LightSoundGeometry]

The 5672 takes 50 mA @ 1.25 volts on the filament.  This very low for a tube filament but 50 mA is pretty much all you can get with any longevity out of a 9 volt battery and you still have to supply 67.5 volts for the plate and screen with 3.25 mA for the plate and 1.1 mA for the screen.  A buck regulator will essentially trade voltage for current and act like a DC stepdown transformer so the 50 mA @ 1.25 volts would be derived from 9 volts @ 6.944 mA at 100% efficiency.  Even if the supply was only 50% efficient, you would still have less than 14 mA drain.

A boost converter for the B+ supply would take 32.625 mA to generate the 4.35 mA drain at 67.5 volts if it was operating at 100% efficiency but fortunately the high voltages make it easy to get about 80% efficiency so you would get 40.78 mA drain from the 9 volt battery.

But the 5672 is a power amplifier pentode putting out a mighty 65 mW of output.  Since it only has a transconductance of 650 µmhos, you may need a bit of experimentation to see if you can get sufficient gain and what the actual current drains are with the values you have chosen.  I will be interested in seeing what you come up with.

I better take a new look at the schematic lolol ..I thought B+ was 1.25 and the rest of the tube was seeing V+ at 12 volts ..I am learning so much just trying to get started on this circuit ..might be in over my head !

the buck converter seems like a great idea and alternative to the IC ..and like paul was saying the IC regulates the internal errors which led me to thinking its analog tube and not a ttl/cmos and could probably see a slight variance in voltage?

I was reading, a little bit, and I think the buck converter does the same kind of adjustments for a steady output.

[bluebunny]

This article will tell you all about "B" (and "A" and "C" too).  And here's a sneak peek of a picture from that article:

[LightSoundGeometry]

This article will tell you all about "B" (and "A" and "C" too).  And here's a sneak peek of a picture from that article:

we did a brief overview of a class c & d amp which the c amp had a revers bias and a second power source off the base with the emitter straight to ground. right on wiki ...let me google that for you lol

modify 1 - if and when I get this all figured out, I will know it inside and out, be able to come up with my own and not copy other people so much

you must copy the masters to be a master young grasshopper


modify 2 - looks like a class C froma glance but where is the common load/out ? I am blind!

[PRR]

A B C battery is NOT the same as Class A B C.

[boppy100]

why not use a AA battery for the filament.  alkaline AA's will give 2000 to 2500 mAh or  40-50 hours at 50 mA

[LightSoundGeometry]

why not use a AA battery for the filament.  alkaline AA's will give 2000 to 2500 mAh or  40-50 hours at 50 mA

will try it .. getting ready to run a Voltage divider through a current divider and see what happens lol ..I have 3 tubes so what the hell

probably going to run a sim in NImultisim first to get the .05amps in my current divider,  or see what my spice lite does since I have never used spice yet.

need 1.25 Vf and .05 If - to play around until my heat sink and regulator comes in mail ..headed to walmart for batteries 

http://data.energizer.com/PDFs/E91.pdf

rated 1.5v

has to be the zinc carbon type I think

https://en.wikipedia.org/wiki/AA_battery









with a AA battery but it has pin 3 at 1.5 volts

nothing ..I cant find the error , maybe AA batt doesnt work in this? new eyes might see what I am missing

[LightSoundGeometry]

anyone have a new set of eyes on the tube circuit for me? I cant find an error but its not working. using rick holts schematic posted on this forum

the AA battery Ib = 1.5v/10m is .15x10^-6 ..maybe If is too low? looks like .150uA

2n3906:

C -.9
b 11.66
e 12.0

5672 pin 1 - 5:

-.9
5.65
1.4
-.5
.001

I can take more shot of the BB if needed



possible all four of my tubes are bad?

[bluebunny]

Adjust the 100K "trimmer", as per the schematic instructions.  Your transistor is switched off, so there's nothing getting to the tube plate (pin 1).  The collector and plate should be much higher than -0.9V.

[LightSoundGeometry]

Adjust the 100K "trimmer", as per the schematic instructions.  Your transistor is switched off, so there's nothing getting to the tube plate (pin 1).  The collector and plate should be much higher than -0.9V.

I have adjusted the pot ..and I have put in different values from 5k to 1m.

I just got home , so I will try a new transistor and work around pin 1 for higher voltage.

is there anyway to test a tube with a DMM? arent tubes inherently tough as nails?

[LightSoundGeometry]

well the 317's and sinks are in , does anyone know the purpose of the bushing? and the sinks do not come with a fastening screw?

one of these four tubes should fire up if the tubes are good and teh circuit works ..it looks really easy from a glance but we will see