EEVblog #1046 - Mysterious Digital Voltage Doubling

[Psi]

You can demonstrate this with sound using a speaker or piezo
If you drive it with GPIO to GND vs between two GPIOs.

You get more physical movement and more sound driving between two GPIOs out of phase than one GPIO to ground

You're basically 'reference jacking'
Picking the right moment by timing or by design to change how you apply voltage and its reference in order to get a larger overall effect at the output.
Like measuring a battery around the other way and seeing a negative voltage.

[IanB]

@Dave: The coupon code for the differential probe mentioned in the video doesn't seem to work on the EEVBlog shop. It says "This coupon has expired." ? Is that correct?

[bitseeker]

Ian, the code worked OK for me — no spaces between words.

[IanB]

Yes, it works now. There may have been a time delay before it was active.

[bitseeker]

Ah! Just a little Gear Acquisition Syndrome making for an itchy mouse finger.

The probe works great. I got mine earlier this year and the 1/10 mode makes it as convenient as a passive probe.

Happy probing!

[EEVblog]

@Dave: The coupon code for the differential probe mentioned in the video doesn't seem to work on the EEVBlog shop. It says "This coupon has expired." ? Is that correct?

Fixed now.

[jstuart]

It's not a digital phenomenon.



What is the peak voltage across the cap? And what's the peak voltage across the diode?

If you think they're the same then welcome to the smoking diode club. Been there, done that, o yeah. Maybe you should build it and see. Be sure to use floating probes. Or buy one of Dave's diff probes.

[jstuart]

[edit]Nevermind.[/edit]

[firewalker]

I think the tern "voltage doubling" confuses many of the people.

Just use a battery and couple of relays to  build an h bridge.

Alexander.

[chriswebb]

If I am understanding it correctly mathematically you are doing f(x) = x - NOT x  which can be simplified as f(x) = 2x where x is channel 1 and NOT x is channel 2. And because your differential probe is floating you dont see the dc bias with respect to ground. Does that make sense?

edit: Frequency wise there is a 360 degree phase shift from the double inversion and the two signals are added to get 10 Vpp

[Simon]

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.

[OzOnE]

I tend to agree - this video was quite confusing, and I've been working on / designing / repairing electronics for 25 years now.

The video isn't so much "wrong" in what it says or demonstrates, but it completely threw me off, because of the term "voltage doubling".
It's not really a "voltage doubler" circuit in any real sense, and there will only ever 5 Volts across the LCD segment(s) at any given time.


This and me second-guessing myself for about half an hour. lol
I thought I'd missed something over the years, and had to test it for myself...

https://imgur.com/a/Wm0tH

Sure enough, yes, the o'scope will show 10 Volts P-P across the NOT gate, but that's only PEAK-TO-PEAK, which means that the only reason the o'scope appears to show the apparent doubling of the voltage is due to where it's reference point is set at "0V".

Plus, obviously the 'scope is designed to show negative voltages that swing below that 0V reference level, and that's what you would see when the signal across the NOT gate switches polarity.


I don't think the video explained clearly that there is not a "real" doubling of the voltage, as the signal only goes positive and negative with respect to the reference point on the 'scope, and doesn't actually create twice the voltage across the LCD (or any other load) in the circuit itself.


Granted, it was partly my own brain fart as well, but it appears that lots of people on the video comments are very confused by this too.

I even had to grab a cup of coffee and bacon sandwich before I finally realised that my original assumption was correct. lol


Quote from the vid at 1:21 (not gigawatts, sadly)...

"We do actually get 10 Volts peak-to-peak ACROSS this LCD for a 5-Volt supply. So it is actually really voltage doubling..."


I think statements like the one above are what's causing most of the confusion (and the video title).

Yes, the P-P voltage is effectively doubled, but the "voltage doubling" phrase makes it sound like there would be a true 10 Volts across the LCD segment.


OzOnE.

[OzOnE]

I think a good way to really show what's going on is to just hook up a 9 Volt battery to the scope probe, then show what happens when the polarity is reversed.

That would explain why the trace on the o'scope goes below the 0V reference point, and why it's called a Peak-to-Peak measurement.


I was even trying to think of how the apparent "doubling" of the voltage could be down to stored capacitance, so the reference point would change if the signal was toggling fast enough. But yeah, that doesn't really come into it at all in this case.


You fried my brain, Dave. lol

OzOnE.

[Peabody]

Well I'm going to defend the "actual voltage doubling".  From the point of view of the LCD display, you go from SEG1 being +5V with respect to COM, to SEG1 being -5V with respect to COM.  That's a 10V swing - a real 10V swing.  You get the same kind of thing at every capacitor in a Dickson charge pump stage - the references are floating.
 

[retrolefty]

Well I'm going to defend the "actual voltage doubling".  From the point of view of the LCD display, you go from SEG1 being +5V with respect to COM, to SEG1 being -5V with respect to COM.  That's a 10V swing - a real 10V swing.  You get the same kind of thing at every capacitor in a Dickson charge pump stage - the references are floating.
 

 There is no "10V swing" at any given point in time. This is just like and same effect as passing a 5vdc square wave through a series capacitor.

[bw2341]

I'm not sure if all my terminology is correct, but here's my take on it:

Let's compare the three different cases with a 5V digital drive:
1. DC drive (5V)
2. Unipolar square-wave AC drive (0V and 5V)
3. Bipolar square-wave AC drive (-5V and 5V)

We need to figure out the RMS voltage in each case. This way, we can directly compare the drive methods.

For DC, "Vrms" equals 5V by definition.

For unipolar, Vrms=5V*sqrt(0.5)=3.54V. Just think of it as connecting and disconnecting a battery at 50% duty cycle. You can only deliver half the power compared to a constant DC supply. Since V=sqrt(P*R), halving the power requires sqrt(0.5) of the voltage.

For bipolar, Vrms=5V. If you're swapping the polarity of a battery, it still connected (almost) all the time. You're delivering the same power as a constant DC supply.

There's no magical doubling of anything as Vrms will never exceed the original 5V. You're just maximizing the effectiveness of the drive method.

[Simon]

Well I'm going to defend the "actual voltage doubling".  From the point of view of the LCD display, you go from SEG1 being +5V with respect to COM, to SEG1 being -5V with respect to COM.  That's a 10V swing - a real 10V swing.  You get the same kind of thing at every capacitor in a Dickson charge pump stage - the references are floating.
 

 There is no "10V swing" at any given point in time. This is just like and same effect as passing a 5vdc square wave through a series capacitor.

Completely and utterly wrong, you completely miss the point, if you put 5V DC pulses through a series capacitor you will gety +/-2.5V and 5V peak to peak. A full H bridge does indeed produce negative pulses as much as it does positive ones, that is the whole point of it.

[IanB]

The way to settle this would be to rectify the pulses and use them to charge up a reservoir capacitor. If the capacitor settles out at 10 V DC it would be more convincing than doing math on an oscilloscope.

[Simon]

oh for Christs sake, I can't believe something so simple is causing so much confusion. Hasn't anyone use H bridges ?

[IanB]

LOL 

It's not confusion, it's healthy skepticism. If you want to convince people of something strange you have to present strong evidence.

The issue here is that there are no points in the circuit that have a potential difference of 10 V at the same time. If you put a voltmeter on the circuit there is nowhere you can measure 10 V. So it is reasonable to say that 10 V doesn't exist.

[Cerebus]

... then welcome to the smoking diode club.

Sounds like a dive bar.

Sounds like my kind of dive bar.

[Simon]

No 10V dose not exist, 10Vpp on an AC supply exists, it's the same as saying that the mains in the UK is around 620Vpp (2x240x1.414). The swing is 10V, now if you rectified it with a voltage doubler you will get 10V the same as you can get 620V from the mains that is 310V peak.

[ogden]

IMHO there would be not so much frustration around if Dave explained what LCD will see in case of "single ended" drive. Additional reading for those still unsure: check about "bridge tied load" audio power amplifier configuration. Start here: https://en.wikipedia.org/wiki/Bridged_and_paralleled_amplifiers

[jstuart]

My first comment on this subject was to say this isn't a digital phenomenon, it is absolutely an analog thing. My first thought was to compare Dave's inverter circuit to an differential output amp. But then if you have trouble with Dave's claims, you'll have the same trouble with a linear differential amp.

Or maybe not. Maybe a sine wave would be easier to understand.

If you've never used a differential output amp, the LME49724 is an example. It has two outputs. When one goes up, the other goes down. It produces a differential output proportional (by some gain) to the input. It can be powered from a single supply, say 5V.

If the differential signal is a sine wave (as opposed to a square wave), it has a zero crossing, where each amp output is at 2.5V (half Vcc). At this time the load has zero volts across it. As time goes on one output (call it "A") goes up, one (call it "B") goes down, symmetrically. The rate of change of the voltage across the load is 2x the r.o.c of each individual output. That should be easy to see.

So if you follow the outputs thru a quarter cycle, "A" rises to 5V and "B" drops to 0. At that time the load has 5 volts across it, yes?

Go another quarter cycle "A" drops and "B" rises till they are both back at 2.5V. The load is again at 0V.

Now "B" rises and "A" drops. After a quarter cycle "B" is at 5V and "A" is 0. If you're the load, this isn't 5V, it's the opposite. Current is flows the other way. The opposite of 5V "-5V".

During the last quarter cycle "B" drops and "A" rises till they both reach 2.5V. The cycle ends and the load is back at 0V.

The load experienced a voltage that went from 0 to +5 to 0 to -5 and back to 0. At the same time the outputs went from 2.5/2.5 to 5/0 to 2.5/2.5 to 0/5 to 2.5/2.5.

Using sine waves you can measure the load voltage directly using a simple handheld DMM. (Don't use a mains grounded instrument.) I have a few LME49724's on hand, maybe I'll do that. I don't do videos, so you'll have to take my word on the outcome.

Sine waves aren't special to the operation of differential amps. They're just easy to measure with handheld DMM's. The LME49724 can produce 100Hz square waves just as good as Dave's. Square waves behave just like sine waves, far as the load is concerned.

[JustMeHere]

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.

[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.