CD4069 vs 40106

[Paul Marossy]

Can you use a CD4069 hex inverter to step up voltage from 9V to 50-60V in the same way as can be done with a 40106 chip? I suppose the absence of the Schmitt Trigger might be a deal breaker?

[danfrank]

I don't know if this helps you but a CD4584 is the same thing as a CD40106, hex Schmidt trigger

[danfrank]

I would plug a 4069 in your circuit and try it out. The 40106 has a Schmidt trigger build it which to my understanding is kind of an "anti-bounce" built in so there's no place with the input signal where the 40106 could go either way. So it probably depends on what is driving the 40106. I'd just sub the 4069 and see if it gets the results you want.
What are you doing? Making a bias osc for tape?

Schmitt , Schmidt... Basically I don't give a Schmitt how it's spelled. Lol!

[Rob Strand]

If you are using the 40106 as a Schmitt-Trigger based RC oscillator then you will need to change the oscillator to a two inverter based RC oscillator.    It's not uncommon to parallel the 4069's to get more output.  The text book pattern would be two inverters for the oscillator and four for the boosted output.    However you might be able to gain an extra boost inverter by having one inverter in the oscillator as is an paralleling  five inverters for the boosted output.   



No doubt it will better to make the boost output of the the gates more than the other.   As shown might work best but it's best to try both under load.

You can often leave off R1 but it's best to keep it.

[Mark Hammer]

PAiA used hex invertors as charge-pumps for several of their tube-based projects.  Here, a 4049 is used to produce 50V for a phantom-powered mic preamp.

[Rob Strand]

PAiA used hex invertors as charge-pumps for several of their tube-based projects.  Here, a 4049 is used to produce 50V for a phantom-powered mic preamp.

So that one uses three inverter RC oscillator and only has three inverters in parallel for the boosted output.

[R.G.]

I still have some old scars from similar work before. 
Any Schmitt trigger input gate can make a reliable one-gate oscillator. Use a cap to ground on the input and a resistor from output to input for feedback. This is an exact analog of the integrator-comparator style of LFO made from opamps. The R-C is the integrator, the Schmitt input is the comparator. You will have to tinker the RC time constant a bit to get a specific frequency. Or at least I always did.
The 4584 is not exactly the same as a 40106 for this. Although it >is< a hex Schmitt input gate, the thresholds are closer to the middle, so for the same RC time constant, it oscillates at a higher frequency. That was one of the scars. 
Other than that, you'll want to look at the output current specs on the gates. C0ckcroft-Walton capacitive multipliers multiply voltage by using a lot more current at the pump input end than goes out the output end of the multiplier. If there's any significant current flowing on the output, you'll need input currents a bit more than the multiplier number times the output current. Maybe more. So using gates with higher output currents and paralleling them is a good move. Or at least if your output voltage sags badly under load, this is where to start looking.

[Rob Strand]

The 4584 is not exactly the same as a 40106 for this. Although it >is< a hex Schmitt input gate, the thresholds are closer to the middle, so for the same RC time constant, it oscillates at a higher frequency. That was one of the scars.

Everyone from the CMOS chip era ends up getting stung at some point.   The tolerances on those things is all over the place which doesn't help the cause.  It even varies from brand to brand so it's hard to design something without the risk of the timing being off.  (There's also the NANDs with Schmitt inputs, IIRC 4093, and all the 74HC/74C stuff).

I suppose we shouldn't forget the NE555 and CMOS variants as an option for a charge pump.

[Paul Marossy]

I probably should've gave a point of reference... a 40106 is used in this manner in a circuit called a valv-e-tizer https://diy.thcustom.com/valv-e-tizer-1-1-diy/ and I was curious if the CD4069 could work similarly.

[Rob Strand]

I probably should've gave a point of reference... a 40106 is used in this manner in a circuit called a valv-e-tizer https://diy.thcustom.com/valv-e-tizer-1-1-diy/ and I was curious if the CD4069 could work similarly.

The drive capabilities of the CD4069 is quite close to the 40106 so you are unlikely to have any issue there.

Where the practicalities come in the the existing circuit uses the first stage as the Schmitt RC oscillator and the voltage multiplier uses up all six of the inverters.  FYI, the voltage multiplier is a Dickson type instead of a common Co-ckcroft-Walton.   The issue is you don't have any "free" inverters to make-up a 2 inverter or 3 inverter RC oscillator.     However, given the inverters in the multiplier are wired in cascade you might be able wrap the RC network for the oscillator around the first two (or three) inverters.  You probably should prototype it under load to make sure the oscillator doesn't fart out.   The other way out is to add an oscillator to drive the first inverter.

The main point is the first stage oscillator won't work when changed from a Schmitt inverter to a conventional inverter.

[noisette]

If we are talking about a circuit like this:
http://zwz.cz/Obrazky/110V%20voltage%20multiplier%20with%2040106.png
my understanding is it consists of two parts,
1. oscillator
2. inverter chain that performs voltage multiplication

The inverter chain needs an ac input so the oscillator "converts" dc into ac, so any oscillator will be fine
if it has enough current output capability, which could be easiest to find out by trial...

Would that be correct?

[Paul Marossy]

I probably should've gave a point of reference... a 40106 is used in this manner in a circuit called a valv-e-tizer https://diy.thcustom.com/valv-e-tizer-1-1-diy/ and I was curious if the CD4069 could work similarly.

The drive capabilities of the CD4069 is quite close to the 40106 so you are unlikely to have any issue there.

Where the practicalities come in the the existing circuit uses the first stage as the Schmitt RC oscillator and the voltage multiplier uses up all six of the inverters.  FYI, the voltage multiplier is a Dickson type instead of a common Co-ckcroft-Walton.   The issue is you don't have any "free" inverters to make-up a 2 inverter or 3 inverter RC oscillator.     However, given the inverters in the multiplier are wired in cascade you might be able wrap the RC network for the oscillator around the first two (or three) inverters.  You probably should prototype it under load to make sure the oscillator doesn't fart out.   The other way out is to add an oscillator to drive the first inverter.

The main point is the first stage oscillator won't work when changed from a Schmitt inverter to a conventional inverter.

I see. That makes sense.

[anotherjim]

It's an interesting circuit. At first sight, it looks like the inverter might be in danger from excessive voltage, but it should only see the chips power supply range.
The current capacity looks to be like it's entirely dependant on the charge stored in the capacitors. Each inverter is constantly switching, so can't be relied upon as a current source directly to a load anyway. The only job it has to do is hold the negative end of a cap at 0v while a diode passes charge to the positive plate from the previous stage. When it switches to +9v, the voltage on the cap positive is jacked up and increased by +9v and passes that on to the following stage.

When CMOS logic switches, the speed of the transition going from high to low or low to high matters in terms of waste power consumption. In between high or low states, both of the complementary MOS transistors are partly conducting letting current pass thru the chip without doing any useful work. With a capacitor on the input, the transition speed at the inputs will be slow causing the CMOS to spend more time in between states -  more wasted current. A Schmitt trigger input will speed it back up and save current.

The existence of a hex Schmitt inverter chip giving an oscillator and several charge pump stages is a happy accident.
It isn't perfect. CMOS has a relatively high output resistance so all of the stages have an RC time constant slowing things down. If the oscillator runs too fast the cap voltage may not have time to reach the threshold of the Schmitt trigger inputs before the next state change and it will freeze the following stages, so the cap value will be chosen small enough in order to fully charge or discharge in time.

You can make a single inverter oscillator without a Schmitt trigger. It's all that many microcontroller chips have to work their external clock oscillator. You need a ceramic resonator or crystal and a couple of caps. However, the output has slow transition edges and another inverter is needed to speed it up.

[Paul Marossy]

It's an interesting circuit. At first sight, it looks like the inverter might be in danger from excessive voltage, but it should only see the chips power supply range.
The current capacity looks to be like it's entirely dependant on the charge stored in the capacitors. Each inverter is constantly switching, so can't be relied upon as a current source directly to a load anyway. The only job it has to do is hold the negative end of a cap at 0v while a diode passes charge to the positive plate from the previous stage. When it switches to +9v, the voltage on the cap positive is jacked up and increased by +9v and passes that on to the following stage.

When CMOS logic switches, the speed of the transition going from high to low or low to high matters in terms of waste power consumption. In between high or low states, both of the complementary MOS transistors are partly conducting letting current pass thru the chip without doing any useful work. With a capacitor on the input, the transition speed at the inputs will be slow causing the CMOS to spend more time in between states -  more wasted current. A Schmitt trigger input will speed it back up and save current.

The existence of a hex Schmitt inverter chip giving an oscillator and several charge pump stages is a happy accident.
It isn't perfect. CMOS has a relatively high output resistance so all of the stages have an RC time constant slowing things down. If the oscillator runs too fast the cap voltage may not have time to reach the threshold of the Schmitt trigger inputs before the next state change and it will freeze the following stages, so the cap value will be chosen small enough in order to fully charge or discharge in time.

You can make a single inverter oscillator without a Schmitt trigger. It's all that many microcontroller chips have to work their external clock oscillator. You need a ceramic resonator or crystal and a couple of caps. However, the output has slow transition edges and another inverter is needed to speed it up.

Yeah I was fascinated by one IC chip bumping up 9V to around 50V. I don't recall ever seeing that before and wanted to know more about the mechanics of the thing. These posts have been a great enlightenment on the subject!

[anotherjim]

Actually, for a single inverter crystal oscillator, you would have to go for a 32KHz watch crystal. The is also 40kHz to suit ultrasonic transducers but it's rarely stocked even where 40kHz transducers are available. Most other options are I think too high in frequency. A 455Khz resonator is the lowest I can find.

[Rob Strand]

When CMOS logic switches, the speed of the transition going from high to low or low to high matters in terms of waste power consumption. In between high or low states, both of the complementary MOS transistors are partly conducting letting current pass thru the chip without doing any useful work. With a capacitor on the input, the transition speed at the inputs will be slow causing the CMOS to spend more time in between states -  more wasted current. A Schmitt trigger input will speed it back up and save current.

That's a good reason to check it on a prototype.  I suppose if you don't compare it against the original it's hard to judge the loss in efficiency, unless it behaves very badly.

If you are keen you can make Schmitt triggers out of every second part of inverters like the two inverters at the top right,
(you would want to use much larger resistors)


It's probably all going down the wrong path.  There's some elegance to the original design.

You can always fall-back onto something like this,

http://recordinghacks.com/images/mic_extras/cad/E300-schematic.png

The first stage oscillator is replaced with a two-inverter RC oscillator (as posted previously) then instead of driving the diode/cap network off a single inverter you parallel all the remaining inverters then feed that into the diode/cap network.    Cap values chosen to suit switching frequency.

if you wanted to play one against the other you would have to build and test the options.

Here's something,
https://gyraf.dk/schematics/Voltage_multipliers_with_CMOS_gates.pdf

I had a skim over it and they use both Schmitt and standard inverters.   They are also recommending using B devices - which would reduce Jim's contention.  Most of the 4069's are unbuffered U's.

[Paul Marossy]

So I built a Valv-E-Tizer for fun because I was intrigued by the 60V from 9V via 40106 thing. I have one unexpected outcome. Does anyone know why the output voltage of a 40106 (around 62V) would drop down to about 45V when you add a 6V voltage regulator to the power supply? I'm a bit perplexed by why that would happen.

40106 data sheet says "DC input current, any one input ±10 mA" So does that mean if we are using six sections of the IC chip that it requires 60mA of power? I have a hunch that my 9V 300mA wall wart doesn't have enough current?

[Rob Strand]

So I built a Valv-E-Tizer for fun because I was intrigued by the 60V from 9V via 40106 thing. I have one unexpected outcome. Does anyone know why the output voltage of a 40106 (around 62V) would drop down to about 45V when you add a 6V voltage regulator to the power supply? I'm a bit perplexed by why that would happen.

It's not clear where you are putting the regulator.

If the 40106 converter was powered by 9V and now it's powered by 6V you will get a voltage drop 62 * (6/9) = 42V.

If you are just placing a 6V across the 9V supply, and the 40106 is still powered from 9V, and the voltage drops it sounds like a problem.  Measure the 9V rail and see if it is dropping.   For a large drop I'd suspect a wiring fault on the 6V circuit.   (At a stretch the 6V regulator might oscillating an pulling a lot of current.)

40106 data sheet says "DC input current, any one input ±10 mA" So does that mean if we are using six sections of the IC chip that it requires 60mA of power? I have a hunch that my 9V 300mA wall wart doesn't have enough current?

Those are absolute ratings.   The meaning is the input protection diodes inside the 40106 can handle 10mA.   The actual DC current draw on the inputs is tiny.   Later in the datasheet you can see 0.1uA max and several orders or magnitude less than that for typical.

[Paul Marossy]

So I built a Valv-E-Tizer for fun because I was intrigued by the 60V from 9V via 40106 thing. I have one unexpected outcome. Does anyone know why the output voltage of a 40106 (around 62V) would drop down to about 45V when you add a 6V voltage regulator to the power supply? I'm a bit perplexed by why that would happen.

It's not clear where you are putting the regulator.

If the 40106 converter was powered by 9V and now it's powered by 6V you will get a voltage drop 62 * (6/9) = 42V.

If you are just placing a 6V across the 9V supply, and the 40106 is still powered from 9V, and the voltage drops it sounds like a problem.  Measure the 9V rail and see if it is dropping.   For a large drop I'd suspect a wiring fault on the 6V circuit.   (At a stretch the 6V regulator might oscillating an pulling a lot of current.)

40106 has its own 9V supply and is measuring 9.2V on Pin 14. 6V regulator branches off the 9V supply for the tube filaments, just like on the schematic. 9V goes into the 7806 and 5.8V comes out. That's why I am perplexed. Can't understand why the voltage drop on the output of the 40106 when that is added to the power supply.

[Rob Strand]

40106 has its own 9V supply and is measuring 9.2V on Pin 14. 6V regulator branches off the 9V supply for the tube filaments, just like on the schematic. 9V goes into the 7806 and 5.8V comes out. That's why I am perplexed. Can't understand why the voltage drop on the output of the 40106 when that is added to the power supply.

Try pulling the tubes out.

When the tube heaters are off they pull no plate current.  But now when the heaters are powered the tubes start pulling a plate current.  That plate current loads down the B+ rail and it sags down.   The output impedance of those converters is fairly high so the voltage has a tendency to sag under load.   If you can see ripple on B+ rail under load then beefing up the caps on the converter cap can prevent sag as the ripple causes the average DC voltage to drop [20V drop would need 40V ripple].   It can turn out the ripple is low and the voltage still sags in these cases the larger caps don't help much.   What's happening is the on resistance of the switches (in the 40106 gates) have unavoidable voltage drops and that "loses" voltage to the caps.    You can lower on-resistance by paralleling inverters.   Paralleling some of the inverters earlier in the converter might have more effect than paralleling later ones - I haven't analysed the details.