Charge pump voltage drop

[Esppse]

Hey,


I am building a tube driver here.

http://effectslayouts.blogspot.com/2019/09/bk-butler-tube-driver.html?m=1

So far it works only with a 12AU7 i have laying around, but the voltage on my charge pump is dropping.

With the tube driver disconnected, it is 12v off the pump.

When the tube drive is connected, that 12v output drops to 7v.

I used a vero layout somewhere online but can't find it anymore.

It uses a 7660s to bring it to 18v, then drops it to 12v with an L7812. 

The 7660s gets very hot too, not sure if this is normal.

Does anyone know why it does this?

[amz-fx]

The heater of the 12AU7 is drawing way too much current from the charge pump for it to be able to supply the proper voltage.

The tube heater elements in series need 12.6v at 150ma, plus the current that is used in the L7812.

The 7660S charge pump can only supply 50ma max, so it is unable to power the tube and voltage reg.

The more current that you draw from a charge pump, the more the output voltage drops from its maximum unloaded condition.

regards, Jack

[Esppse]

Ah makes sense, I had a feeling it was underpowered.

Are there any 9v to 12v charge pumps that can handle such current? The 9v coming into the charge pump is also daisy chained to other 9v effects so i can't use a 12v adapter.

[Clint Eastwood]

Why not connect the tube heaters parallel to a 6.3 volts supply?

[Keppy]

Many charge pumps have the ability to regulate their output, so you can get 12v from 9v without wasting current in a regulator. That still wouldn't get you all the current you need in this case, but it might help if you can find a more powerful pump. The highest one I know of is the LT1054 at 100mA, but I haven't ever needed to look for something more powerful.

The LT1054 datasheet shows some applications using two charge pumps to get higher output current, so it seems to be possible, but you'd be designing the power supply yourself at that point.

[antonis]

A series pass transistor might be your friend..

[Rob Strand]

Why not connect the tube heaters parallel to a 6.3 volts supply?

If your are starting with a 9V rail then 9V into a 6V3 regulator to power the heaters makes more sense than trying to get a 150mA voltage doubler working.   

The linear regulator is only 70% efficient.   Going from 18V to 12.6V is also 70% *but* when you factor in the additional losses from the doubler and the headaches getting it to work at 150mA the advantages of going down to 6.3V becomes clear - the circuit is much simpler and no noise considerations as everything is linear.   (When the charge pump droops there is an improvement in the efficiency of the 12V reg.)

[Esppse]

Ive mocked up the 18v to 12v from another existing pump. I was planning to run 2 of these in parallel. Both 12V outs to the 12V in on the pedal board.

I added the L7812 and the 100nf cap from this preexisting layout:

http://tagboardeffects.blogspot.com/2014/06/toneczar-openhaus.html



Will this be ok?


I also am considering the other option with the 6v3 regulator, but im not sure where to start on rewiring the tube heaters in parallel.

[Keppy]

Why not connect the tube heaters parallel to a 6.3 volts supply?

If your are starting with a 9V rail then 9V into a 6V3 regulator to power the heaters makes more sense than trying to get a 150mA voltage doubler working.

Yes, but the whole circuit runs off 12v, not just the tube heaters. Still need the doubler. But yeah, adding a separate 6v heater supply might still be the least complicated fix.

im not sure where to start on rewiring the tube heaters in parallel.

For 12v, the heater supplie is connected to pin 4 & pin 5. For 6v, tie pins 4 & 5 together, and connect the supply to that connection and pin 9. See the first two examples below (not sure what the third one is).

[Rob Strand]

With the tube driver disconnected, it is 12v off the pump.

When the tube drive is connected, that 12v output drops to 7v.

I used a vero layout somewhere online but can't find it anymore.

It uses a 7660s to bring it to 18v, then drops it to 12v with an L7812.

Yes, but the whole circuit runs off 12v, not just the tube heaters. Still need the doubler. But yeah, adding a separate 6v heater supply might still be the least complicated fix.

I'm confused about what the OP is actually building!

It's a pain the butt back-tracking through the layout in the first post.
I see the original circuit as:
- Input supply 12V DC
- Heaters supplied by 12V
- LT1054 used to generate -12V
- opamp powered from +/- 12V

LT1054 is not used for heaters.

So is the intention to rejig the whole circuit to run from +9V?

[Esppse]

Why not connect the tube heaters parallel to a 6.3 volts supply?

If your are starting with a 9V rail then 9V into a 6V3 regulator to power the heaters makes more sense than trying to get a 150mA voltage doubler working.

Yes, but the whole circuit runs off 12v, not just the tube heaters. Still need the doubler. But yeah, adding a separate 6v heater supply might still be the least complicated fix.

im not sure where to start on rewiring the tube heaters in parallel.

For 12v, the heater supplie is connected to pin 4 & pin 5. For 6v, tie pins 4 & 5 together, and connect the supply to that connection and pin 9. See the first two examples below (not sure what the third one is).

Ah I see, whats the best way to get 6.3V from 9v? I think there might be other stuff that runs off 12v, which was what I thought might be convenient. My source has to be 9V originally though.

[Esppse]

With the tube driver disconnected, it is 12v off the pump.

When the tube drive is connected, that 12v output drops to 7v.

I used a vero layout somewhere online but can't find it anymore.

It uses a 7660s to bring it to 18v, then drops it to 12v with an L7812.

Yes, but the whole circuit runs off 12v, not just the tube heaters. Still need the doubler. But yeah, adding a separate 6v heater supply might still be the least complicated fix.

I'm confused about what the OP is actually building!

It's a pain the butt back-tracking through the layout in the first post.
I see the original circuit as:
- Input supply 12V DC
- Heaters supplied by 12V
- LT1054 used to generate -12V
- opamp powered from +/- 12V

LT1054 is not used for heaters.

So is the intention to rejig the whole circuit to run from +9V?

Oh sorry yes, 9V is my starting point because of other daisy chained effects in this big box of mine. So I decided to use the pump for 12V. The LT1054 on the tube drive board is for -12V. The other board layout pump is for turning 9v to 12v. Two in parallel should be able to get 200ma, which should be sufficient, correct?

If I run the heaters in parallel, this means i would need both a 12v pump for other stuff in that circuit, and a separate way to get 6.3v right?

[amz-fx]

6.3v heater connection is going to require 300ma. so it would be best to use an L7806 to get the heater voltage directly from the 9v supply.

The more current you draw from a charge pump, the more the voltage will sag from its unloaded condition.

The more current you draw from a charge pump, the higher the noise will be on the voltage output.

The flying capacitor is dumping its charge into the doubled output only 50% of the time, and the other 50% it is getting recharged from the supply.  http://www.muzique.com/news/how-charge-pumps-work/

Charge pumps always struggle to provide large amounts of current, and while the datasheet may claim that it can make 100ma, you have to check to see how that is impacting the output voltage. Breadboard the charge pump and load it down and measure the output voltage to see what you can realistically get from it.

There are better voltage boosters than charge pumps if you need more than 20 or 30 ma.  You can buy one on Ebay for $3 or $4 that can easily provide an amp or two at the full regulated output voltage.

regards, Jack

[Rob Strand]

Oh sorry yes, 9V is my starting point because of other daisy chained effects in this big box of mine. So I decided to use
the pump for 12V. The LT1054 on the tube drive board is for -12V. The other board layout pump is for turning 9v to 12v.

OK got it.  So you are doing a 9V to 12V conversion then feeding that into the original circuit.

Two in parallel should be able to get 200ma, which should be sufficient, correct?

In principle it should work but paralleling stuff never quite doubles because you can't guarantee each side shares 50/50.  It's a bit of a gamble.   You can urge better sharing by adding series resistance the output, see pic below.    A second issue is the two halves won't have synchronized clocks.  You might be able to link one pump to the other.   It's all a bit of an uncharted area.   For a one-off DIY project you can take your chances but for production runs you'd want to do your homework.

Instead of regulators feeding the resistors it would be charge pumps.


One thing I can recommend is to use a largish cap on pin 2 (CAP+) of the LT1054, say 47uF or 100uF.   While the maximum *output current* is defined in kind of an iffy fashion in the datasheet, one thing that is made clear is the peak *switch current* of 300mA.   If you use a larger cap on pin 2 (CAP+) it can help lower the peak switch current.  That will buy you some reliability. [FWIW the 300mA peak current lines up with about 133mA output current, so 100mA output current is a good choice as it adds some safety margin.]

If I run the heaters in parallel, this means i would need both a 12v pump for other stuff in that circuit, and a separate way to get 6.3v right?

The idea was to do what you are doing now to get the 12V to feed the tube driver board.  Only use a single charge pump for the +12V (keep the -12V charge pump on tube driver board).   The next idea is to off-load the heater current from the charge pumps and the 12V linear regulator.  The heaters are powered with a separate power circuit:  The 9V DC in feeds into a 6.3V regulator and the output of that goes to the parallel heaters: for example +6V3 to pin 4 and pin 5 of the 12AU7 and 0V to pin 9.

(I'm slightly reluctant to mention this but if you did off-load the heaters it's possible to do +/-12V with a single charge pump.   It would need a lot of redesign of the power supplies and you would end-up needing a +12V regulator and a -12V regulator.  Not a clear-cut design choice.)

[antonis]

whats the best way to get 6.3V from 9v? My source has to be 9V originally though.

Q1/Q2 can be a single Darlington..

P.S.
A single BJT and 6V8 Zener could be implemented in case of not precise 6.3V demand..

[EBK]

Another way to get the heater voltage is this snippet from the MBP Archibald (another Tube Driver clone):

Although it says 12V, 9V works too.
The LM317 gets crazy hot though.  Give it a heat sink or at least plenty of air space.

[amptramp]

A switching regulator could get you from 9 VDC to 6.3 VDC without the losses of a resistor or linear regulator.

[Clint Eastwood]

I understand your starting point is a 9v power supply in 'a big box' filled with pedals. So there probably would be room in that box for a dedicated 12v, or better 12.6v supply for the tube pedal. I think in the end that is the simpelest solution. Also, what is the power rating of your 9v supply? The tube heaters will consume about 2 watts, add to this another watt or so loss in a regulator, that's probably more power than all other pedals in you box combined.

[Rob Strand]

If you drive the heaters with a voltage source which has no current limiting when the tube powers up the heaters are cold and there is a surge in current which reduces the life of the heaters.   A 3-terminal regulator limits the current because the regulator has a current limit.

The LM317 voltage regulator EBK posted would work since it has current limiting.  Power dissipation = (9-6.3)*0.3 = 0.81W.   The TO-220 package has a thermal resistance of say 50 degC/W so the temperature rise will be 0.81 * 50 = 41 degC.     So it will run hot but you will probably get away with it and it would be wise to locate the regulator away from the tube itself.

The simplest way to power the heaters with with a series resistor.  9V in and 6.3V out at 300mA means the resistor needs to be R = (9-6.3) / 0.3 =  9 ohms.   The resistor will dissipate 0.3^2 * 9 = 0.81W.   The 9 ohm resistor limits then power-on surge current as well.

An issue is if the socket is dirty an one of the heater contacts doesn't connect then the 300mA goes down one of the heaters.   The way around that is easily fixed.  Use an 18 ohm resistor for each heater.  (9 - 6.3) / 18 = 150mA per tube.

Also, if you accidentally plug in 12V then the heater current will be more than 300mA.   The voltage regulators stop that.

Another way is to power the heaters via a current source.   There's no power-up surge at all and it doesn't care what the input voltage is.

There's plenty of posts on the web about these.

Unfortunately a simple current using a LM317 regulator doesn't quite work on 9V when you consider the regulator drop out.  It might work in practice but on paper there's not enough input voltage for the current to regulate properly.   (There's no need for a heatsink because the  regulator and the current sense resistors share the power dissipation.)

[No good]


These types are also simple but you have to be careful of temperature dependency,


[The single current source doesn't solve the dirty socket problem.]

So least parts and headaches is probably EBK's LM317.

[Esppse]

I put some thought into the eBay voltage boosters, it seems they can run quite high current.

https://www.ebay.com/itm/224996442243

I was planning to get 2 of them, run 12v off one (which powers a second 12v overdrive unit, this is why I prefer 12v over 6.3v supply)

I was thinking about the plate voltage on this circuit. The boost said it can go to 40v. Can I run the plate at this voltage safely with everything else staying the same?