DC-DC CMOS (HEX INVERTER) REVERSE BIAS?

[tedsorvino]

Hi everyone.

I'm building a condenser mic and I supply it with about 80V dc from an Hex Inverter cmos circuit. The circuit I followed is attached. The cmos is supplied by +12V dc. I wonder if there is a circuit that can provide -80V dc from the same +12V dc source -since it's provided by phantom. power -(maybe by using the same circuit and changing something to give me negative bias).
I don't think that reversing the diodes' bias would work.

Thanks in advance

Ps.The voltage regulator and trimmer in the voltage input of the circuit are ommited.

 

[davezawadi]

This is the way that Rode derive a supply of about 60V from their mics when in battery mode, and why the 12-50V phantom range works. They use an oscillator and a multi-stage active voltage multiplier to give a square wave of about 60V p-p and at a high frequency. This is then smoothed etc to get a very good DC supply for mic bias.

 

[Khron]

Actually, reversing the diodes is exactly the way to get a negative voltage out of that circuit.

 

[rogs]

It seems as if many of the CMOS voltage multiplier circuits used for this task are based on THIS magazine article from the 1980s..
Figure 3(b) in that article shows a useful variation for creating a negative output from the circuit...

 

[tedsorvino]

Actually, reversing the diodes is exactly the way to get a negative voltage out of that circuit.

Great news Khron. Will this voltage be -80V dc ( I mean the exact equivalent of the positive biased)? I wonder if the diodes are blocking any voltage.

 

[abbey road d enfer]

Great news Khron. Will this voltage be -80V dc ( I mean the exact equivalent of the positive biased)? I wonder if the diodes are blocking any voltage.

The negative output needs one more stage, since the positive output adds the source voltage.
The example in the link uses zener diodes for adjusting voltages.

 

[tedsorvino]

The negative output needs one more stage, since the positive output adds the source voltage.
The example in the link uses zener diodes for adjusting voltages.

I think the example on fig 9 shows a circuit that needs a voltage drop from -17V to -10V, that's why there is the need for the zeners. In my case (and also on the fig. 3b example) I can't see any extra stage since the Hex inverter multiplies the 12V to 80V. And in case there is no blockage on the voltage amount from the reversed diodes I should take -80V on the output.
Am I correct?
In case there is the need for an extra stage what that could be and where should be placed on the circuit?

Apparently Table 1 shows a difference of about 10V between positive and negative, for seven stages of CMOS supply

 

[abbey road d enfer]

I think the example on fig 9 shows a circuit that needs a voltage drop from -17V to -10V, that's why there is the need for the zeners.

Not only. The positive output, for N stages is (N+1)Vin, while the negative output is N. Vin. They put the zeners to balance the voltages.

Am I correct?

No.

In case there is the need for an extra stage what that could be and where should be placed on the circuit?

Add one stage to the negative side.

Apparently Table 1 shows a difference of about 10V between positive and negative, for seven stages of CMOS supply

The difference is one Vin. If Vin is 5V, the difference is 5V, if Vin is 18V, the difference is 18V.

 

[tedsorvino]

Thanks Abbey road denfer. Really good explanation

Add one stage to the negative side.

How can I add one stage since all the stages of the Hex Inverter are used to provide around 80V (positively biased) and -68V (negatively biased) ? Will a 12V zener (like the figure 9 example) do the trick? There is no combination of positive and negative in my case that's why I assume I don't use 2 zeners.

 

[abbey road d enfer]

How can I add one stage since all the stages of the Hex Inverter are used to provide around 80V (positively biased) and -68V (negative biased) ? Will a 12V zener (like the figure 9 example) do the trick?

You need to add an inverter. Actually I would suggest that you make a more symmetrical circuit with the first inverter as an oscillator that drives 6 stages (yes you need another chip) and apply positive and negative rectification to both.
Or you do it like Neumann in the U89 or AKG in the 414, LC-based oscillator.

 

[tedsorvino]

Excellent schematic. Thank you Abbey road denfer.
But this extra stage makes things way more complicated since the IC is on a pcb (and it's an SMD). I just can't add an extra IC of that size (14pin) without having housing problems.
Is there an appropriate smaller IC or transistor inverter in a normal pinout package - just 1 or 2 stages- (in order to add it out of the pcb), to combine with the hex inverter?

What about a converter for -12V in the input and then all circuit the same way as the positive?
Or a voltage booster regulator for 14-15v so this way I will have 5 stages of 14v-15v which is equal to 6 stages of 12v?
Both small circuits can be found for around 3 euros in really small packages,


If there's no such solution as the three above mentioned, I will have to try without the extra stage and see how the sound is with less voltage on one side in the reverse phase (I mean it's useful only for the figure 8 mode). I guess it will be like a padded sound in a way. Maybe in practice the difference is not that huge.

 

[rogs]

Why not take the simpler option of using a 15v zener to regulate the supply to the Hex inverter.
That should allow you to use one inverter as the oscillator, and the remaining 5 inverters to generate a negative 80v rail.
Then simply use only 4 inverters to generate the positive 80v rail?...

 

[tedsorvino]

Why not take the simpler option of using a 15v zener to regulate the supply to the Hex inverter.
That should allow you to use one inverter as the oscillator, and the remaining 5 inverters to generate a negative 80v rail.
Then simply use only 4 inverters to generate the positive 80v rail?...

Thanks Rog for the excellent idea.
The problem I'm facing is that my hex inverter is a tiny SMD IC and I use a pcb, so I would like to avoid not using the whole IC, since it's already soldered in its place.
Is it an option adding the zener cathode on the junction of D6 with C7 to 0v, in order to have a voltage drop of about 9V (from 24V to 15V) and then have 4 stages of 15V by using the whole IC?

 

[rogs]

Just supply the device with 15v and simply don't fit the last stage charging capacitor on the positive output side....

 

[tedsorvino]

I just can't supply the positive side with 15V. The design I follow converts the 48V phantom power to 12V on the positive side. The negative is a custom modification I plan to implement. Too many mods may destroy the initial design and then make my life way too hard to troubleshoot.
Maybe a small voltage booster on the negative side is the easiest option.

 

[abbey road d enfer]

I just can't supply the positive side with 15V. The design I follow converts the 48V phantom power to 12V on the positive side.

Can't you tweak it for outputting more?

 

[tedsorvino]

It's too much hassle since the same 12v go to the other pcb that feed an OPA 1642. So...it's an absolute no. I do prefer a voltage booster or something, or even less voltage on my negative, than risking a complete mulfunction (of an otherwise good circuit for omni and cardioid) or total modification just for Figure 8 function.

 

[abbey road d enfer]

It's too much hassle since the same 12v go to the other pcb that feed an OPA 1642. So...it's an absolute no.

Why? There is nothing wrong with supplying an OPA1642 with 15V (it can withstand 36V).

 

[tedsorvino]

Thanks everyone for the input, a lot of things became clearer on how this circuit functions from this thread. II think I don't want to modify the positive 12v supply. Personal preference.
Since I have quite limited time to invest on building this simple schematic, I need to recap the options I prefer otherwise I will lose it.

So the options are:

1) 12v to -12v converter and then the whole circuit as it is on the schematic. Cost around 3.5 euros.

2)12v to 15v booster and then the whole circuit with the diodes reversed. Cost around 2 euros.

3)15v zener diode, cathode of the zener on the junction of D6 with C7 of my schematic. All diodes reversed, the rest of the schematic as it is. Cost around 20 cents.

4)Just reversing the diodes and see how it goes with around -60v. No extra cost.

Will any of these 4 options work better?

Thanks again

 

[abbey road d enfer]

Thanks everyone for the input, a lot of things became clearer on how this circuit functions from this thread. II think I don't want to modify the positive 12v supply. Personal preference.
Since I have quite limited time to invest on building this simple schematic, I need to recap the options I prefer otherwise I will lose it.

So the options are:

1) 12v to -12v converter and then the whole circuit as it is on the schematic. Cost around 3.5 euros.

This is not an option. You can't power the inverters with a negative supply. Unless you reference the whole circuit to the -12 rail.

2)12v to 15v booster and then the whole circuit with the diodes reversed. Cost around 2 euros.

Seems to me more complicated than changing to 15V. Just sayin'...

3)15v zener diode, cathode of the zener on the junction of D6 with C7 of my schematic. All diodes reversed, the rest of the schematic as it is. Cost around 20 cents.

Doesn't make sense.

4)Just reversing the diodes and see how it goes with around -60v. No extra cost.

You could do that, and delete one stage on the positive side or referencing the first diode to ground instead of +12, so then you would have equal voltages, which iscorrect for gigure-8 pattern. Remember the big advantage of Fig-8 is the very high rejection at 90°, which you would loose with unbalanced bias.