EEVblog #1046 - Mysterious Digital Voltage Doubling

[IanMacdonald]

Other small point, inverters don't cause a phase change. Only circuits with an inherent time delay can do that. I avoid saying things like that because it confuses beginners. 

It may look like a phase change on a scope, but that is only the case with a symmetrical waveform. A pulse put through an inverter doesn't come out shifted in time by half the pulse interval. It comes out at the same time, opposite polarity.

[Cerebus]

Other small point, inverters don't cause a phase change. Only circuits with an inherent time delay can do that. I avoid saying things like that because it confuses beginners. 

It may look like a phase change on a scope, but that is only the case with a symmetrical waveform. A pulse put through an inverter doesn't come out shifted in time by half the pulse interval. It comes out at the same time, opposite polarity.

It's that kind of pedantry that gets you killed on the next Zebra crossing because it isn't *properly* black and white.

Oh, and you're ignoring the inherent delay in any inverter, so its output is inverted and there's a phase change. If you're going to be unnecessarily pedantic, please do the job properly...

[rrinker]

Oh, and you're ignoring the inherent delay in any inverter, so its output is inverted and there's a phase change. If you're going to be unnecessarily pedantic, please do the job properly...

 And that, I suppose, is one good reason to run BOTH sides through a buffer, one inverting and one non-inverting, so you don't end up with one signal out of phase with the other.

[free_electron]

NOPE. you don't get 10Vpp ....
you are bouncing the reference around. What you see on the oscilloscope is a time artifact.

Try rectifying that signal .... see what you get ( bridge rectifier ). put a true rms voltmeter across it. see what you get.

If this is a true 10vpp signal and you rectify it ( assuming lossless diodes and no load) and buffer it in a small capacitor you should get 10V dc.
if you do this with this construction you will get .. 5V DC !

There is no magic 'voltage doubling' .... it would be too easy to take a 1 volt battery and a bunch of inverters and pump to whatever voltage you want. Doesn't happen, not without charge storage.

The scope projects an image of a 10 volt signal because it is unaware that the point of reference has shifted. ( when it shows -5 on the display it actually measures 0 volts but with its 'ground' tied to +5 volts... so it shows '-5'. When it shows +5 volts its 'ground' is tied to 0 volts. )

[Peabody]

The scope projects an image of a 10 volt signal because it is unaware that the point of reference has shifted.

But the LCD is equally unaware of that.  So it also sees a +5V to -5V swing.  It sees exactly the same thing as if its COM pin were grounded, and the SEG pins were driven from real +5V and -5V rails.  So it's kinda an "effective" voltage doubling.  Maybe we just need to call it something different.

[SparkyFX]

The scope projects an image of a 10 volt signal because it is unaware that the point of reference has shifted. ( when it shows -5 on the display it actually measures 0 volts but with its 'ground' tied to +5 volts... so it shows '-5'. When it shows +5 volts its 'ground' is tied to 0 volts. )

When being teached this, i misunderstood that part because the gates only switch through their 5V and 0V rails (for TTL, or Logic High and Logic Low), making the whole thing Logic ground referenced, but the catch here is that the LCD/load is itself not ground referenced, it is connected across input and output, making the crystals following a +5 -> -5V span in potential. Of course the 10V are not present at the same time, as in a 10VDC source, they are only present when transitioning/comparing two halfes. Like in Simons example for mains AC. The 620VPeak-to-Peak are not present concurrently, but between half waves, making it a possible scenario to come across that voltage e.g. between two charged capacitors in one of those voltage doubler configurations. Closest to the example at hand might be the Cross-coupled switched capacitors.

Depending on logic family you might get a different power output per edge as well, e.g. if you use a logic family with a high impedance high level the ability to drive a load changes.

I still wonder how much charge needs to be shifted in such a voltage doubler, and in case a part of the circuit becomes ground referenced. Probably only the parasitic capacity of this artificial/virtual ground between driver and load, which sees the same 10V potential? Now you attach an oscilloscope to see that waveform (assuming the circuit runs of an isolated supply)
a) direct, passive probe
b) with a differential probe

Does a) mean the gate needs to push the rest of the circuit ground to +5V and the included capacities to make that -5V reference shift, now that the ground reference shifts at the logic gate? Might be some fun to measure the current of the gate´s supply while it switches.

[free_electron]

the lcd itself is a capacitor ...

[cracked_machine]

Here's the thought that popped immediately into my head...Would you be able to use this technique to power a dual supply op amp from a DC supply?

Be nice. I ask with genuine curiosity

[HKJ]

Here's the thought that popped immediately into my head...Would you be able to use this technique to power a dual supply op amp from a DC supply?

Be nice. I ask with genuine curiosity

Yes and there are chips specially designed to do it, they are called charge pump. You can either double the voltage or generate a negative voltage. The advantage by using a chip is higher output current and lower component count.

[free_electron]

Here's the thought that popped immediately into my head...Would you be able to use this technique to power a dual supply op amp from a DC supply?

Be nice. I ask with genuine curiosity

Yes and there are chips specially designed to do it, they are called charge pump. You can either double the voltage or generate a negative voltage. The advantage by using a chip is higher output current and lower component count.

No you don't ! you can not rectify this signal. this is NOT how charge pumps work ! you need multiple capacitors for a charge pump and a switching system.

take that inverter put a bridge rectfier across it and see if you get 10 volts out of it ... you won't. this is not a 10vpp signal ! it looks like one bcause the reference is bouncing around and the 'moment in time' view of the scope let's you think ( it is visual persistence ) that this is 10 vpp. It is not. this is a case where the scope lies to you.

[Camellia30]

I think pouinting out that it is a full H bridge might help people to understand. you get to have negative voltages without a negative supply at the expense of floating the load. Widely used in controlling brushed motors and the basis of VFD's in the form of 3x 1/2 H bridges.

Can you explain this doubling effect in terms of H-Bridges and brushed DC motors? Or are you simply talking about reversing the current ?

[cracked_machine]

Here's the thought that popped immediately into my head...Would you be able to use this technique to power a dual supply op amp from a DC supply?

Be nice. I ask with genuine curiosity

Yes and there are chips specially designed to do it, they are called charge pump. You can either double the voltage or generate a negative voltage. The advantage by using a chip is higher output current and lower component count.

No you don't ! you can not rectify this signal. this is NOT how charge pumps work ! you need multiple capacitors for a charge pump and a switching system.

take that inverter put a bridge rectfier across it and see if you get 10 volts out of it ... you won't. this is not a 10vpp signal ! it looks like one bcause the reference is bouncing around and the 'moment in time' view of the scope let's you think ( it is visual persistence ) that this is 10 vpp. It is not. this is a case where the scope lies to you.

I wasn't expecting a charge pump nor a 10v signal either.  I was actually hoping this could be an alternative to using a charge pump (and therefore avoid the switching noise associated with a charge pump).

So I think maybe my question still remains unanswered...would you be able to use this configuration to convert a DC signal into an AC signal for powering a dual supply op amp? Doesn't this count as a bipolar signal? Wouldn't this allow the op amp to swing from positive to negative rails?

I fully expect there would be fairly severe current limits...i'm not hoping for more than 100mA anyhow.

[retrolefty]

So I think maybe my question still remains unanswered...

Yes, your question still unanswered.

would you be able to use this configuration to convert a DC signal into an AC signal for powering a dual supply op amp?

I say no... Someone show a schematic if disagreed.

Doesn't this count as a bipolar signal? Wouldn't this allow the op amp to swing from positive to negative rails?

I say no negative voltage available... unless adding a 'rail-splitter' ?.

[cracked_machine]

Yea fair enough. I should get the breadboard out and give it a go. Although Dave has made me paranoid about frying my scope now 

[The Soulman]

This is the way I've explained this to myself:

I'm in a two story house.  I'm on the first floor and I measure up to the next floor and get +10 ft.   Then I (change my reference and) go up stairs.  Now I measure the distance to the bottom floor and get -10 ft.  The difference (Vpp) between the two measurements is 20.

Think about that and watch the double NOT gate part again.

First one to nail it on the head. 

[cv007]

my 2cents-

wouldn't this be a good opportunity to demonstrate the TrueRMS feature of a multimeter? (or lack of it)
finally, a good use for TrueRMS

by the way, if you place the terminals of a 9v battery on your tongue, then quickly rotate it 180degrees, you will get the effect of 18 volts. Right? 

[@rt]

No, but if you rotate it quickly enough you’ll be able to measure it between the two points that are not the origin

[KeithBrown]

It is amazing how difficult it is to convince people of the facts. (Insert global warming here.) The popular "ping" ultrasonic sensors use a neat trick to get (almost) 40 V from a 5 V supply since they use outputs from a TTL to RS-232 converter, each at +/- 10 V. An inverse pair of these driven across the transmit piezo device is +/- 40 V in the ideal case (the charge pumps inside the MAX232 are not perfect, so its likely more like 35 V or so).
So there's this guy "Emil" who did a service to the world and reverse engineered these cheap modules on his site (see https://uglyduck.ath.cx/ep/archive/2014/01/Making_a_better_HC_SR04_Echo_Locator.html), but he would not believe me when I tried to explain this fact to him and he actually deleted my comments!  I have a screen grab before he deleted the exchange somewhere here...

[thm_w]

It is amazing how difficult it is to convince people of the facts. (Insert global warming here.) The popular "ping" ultrasonic sensors use a neat trick to get (almost) 40 V from a 5 V supply since they use outputs from a TTL to RS-232 converter, each at +/- 10 V. An inverse pair of these driven across the transmit piezo device is +/- 40 V in the ideal case (the charge pumps inside the MAX232 are not perfect, so its likely more like 35 V or so).
So there's this guy "Emil" who did a service to the world and reverse engineered these cheap modules on his site (see https://uglyduck.ath.cx/ep/archive/2014/01/Making_a_better_HC_SR04_Echo_Locator.html), but he would not believe me when I tried to explain this fact to him and he actually deleted my comments!  I have a screen grab before he deleted the exchange somewhere here...

Still a comment in there but I guess not yours:

One question regarding the max232 output voltage -
if T1OUT swings between -10V to +10V while T2OUT sends opposite phase pulses of +10V to -10V, then when any of these outputs is taken as reference, the other pin voltage is +/-20V.

That shouldn't be considered a 40Vpp output ?

Thanks

Emil Submitted at 21:59:41 on 21 September 2016
Vpp means "point to point" - when one voltage is 10V above ground the other is 10V bellow ground so the absolute difference between them is 20V not 40V.

You must have been calculating peak to peak, which is clearly wrong 

[KeithBrown]

I'm pretty sure it was in 2014 that I was wiped from his forum. Interesting that somebody else tried to correct him and that he still hasn't learned. Not sure where the "point to point" idea came from: is that another common misunderstanding? And I guess if that is what he thinks it means then he will never "see" 40 V since at any one time the magnitude of the voltages are no more than 20 V between two points.
Has anybody tried an alternative explanation: that in North America at least we have TWO 120 lines coming in to buildings but that clothes dryers definitely run off of 240! How is that possible if out of phase signals don't inherently double the voltage (rms, peak-to-peak, whichever).

[KeithBrown]

Found it! This was what was deleted:

I don't think I was rude, does anybody? Did I stray too far from the topic by mentioning the transistor drop? What DID I do wrong?
And is there any point trying to inform him again (with Dave's video as back-up, of course)?

[Bud]

It is amazing how difficult it is to convince people of the facts. (Insert global warming here.)

This one is easy to answer - because people are resistant to bullshit.