Hi Welcome to Fundamentals! Friday Couple of episodes back we looked at the microcontroller voltage doubler otherwise known as the Dixon doubler or the diode charge pump voltage doubler and quite a few people asked about the inverter configuration of that. So let's run through. It should be pretty quick. If you haven't seen the previous video, watch that.

it'll be linked down below. So this is what we had previously with the Dixon doubler. we had our VCC voltage. It might be say 5 vol or 3.3 Vols or whatever your system voltage is.

Um, and then we feed in a square wave from our microcontroller. It doesn't have to be from a micro, could be from something else, but let's just say it's from a microcontroller. And based on the Dixon dou here with the output filter, the output voltage is uh, V out equal 2 * VCC. So you double your VCC voltage minus your diode losses of course, but we won't get into that yet.

We're talking about the ideal case. Now, how do you get a voltage inverter? Well, it's very simple. Instead of doubling, we want to invert. So what do we do? H We erase these dodes here and we draw them the other way around like that.

but that's not all we have to do. We also have to get rid of VCC here that's no longer VCC that is ground like that. And Bingo! We keep our Um signal from our Square wave from our microcontroller. there goes from zero to VCC and in this case our output voltage V out is going to be not doubled.

It's going to be minus VCC Once again, ideal case assuming no diode losses Bingo that's our voltage inverter. That's all there is to it. Sorry I Don't think this one has a fancy name, but it can go under other names like uh, Charge Pump Voltage Inverter Diode Voltage Inverter Diode Charge Pump Inverter All sorts of combinations like that. but eh, it's a voltage inverter.

Great if you want to. Um, if you have just a single uh Supply it can be operating from batteries or just from a uh, single regulated uh main Supply or something like that and you want to generate a negative supply for an opamp, or maybe an offset voltage for uh, a voltage regulator to go down to Zer volts which we've talked about in the past or something like that. Very useful. but like all these uh, diode charge pumps, they're pretty low power.

You know you're only going to get, you know, a couple of milliamps out of this, sort of, you know, tens of milliamps absolute top. So really, it's not for anything high power. And once again, you can also follow this with a negative voltage regulator if you need regulation. There is, however, one more thing that we have to invert which I haven't shown here yet.

invert no pun intended before with our doubler. Our capacitors were like that. If you used polarized capacitors, that'd be positive there and positive there in this case. No, we have to reverse those.

So if you're using polarized capacitors, positive is here because our output voltage is negative. Got zero volts here. This is more negative than this, so it doesn't make sense unless you actually think about it connecting the positive of the capacitor to ground. But because that's negative, of course, the capacitor is still has a positive voltage or the correct polarity across it.

And likewise, this one here. So how does this work? Well, you guessed it. Very similar to how our previous greenre and Dixon doublers have worked. Instead of doubling and level shifting, we're actually inverting in this case.

So let's assume that our current is flowing through the diode like that. So our diode is conducting once again. Ideal diode. No losses at all.

Zero volts, voltage Drop on that. And let's assume that we've got uh, where our input waveform is at VCC and this capacitor has had time to charge up. Now, when this diode conducts okay, what is it? It's an ideal diode. It's got no losses.

This point here is going to equal this point here. and here it's going to be ground. So our reference point one which I've shown in green and this will be the green waveform here. By the way, that the Um output filter, we're just ignoring that at the moment, we're only looking at Uh, signal number one here.

the green waveform I should probably draw that in there. number one. There it is. Then out this point, number one is z volts there.

But remember I said our capacitor is charged up and if you remember, capacitors can't change their voltage instantaneously. So we've got what have we got? A charge capacitor here with 0 volts here? 5 vol here. What happens when this now switches down to zero like this? We've now got Zer volts here. What point does this become? Well, our diode is no long is now going to be reversed biased.

So our current, well it's going to be reverse via. So our current's trying to flow through like that. but that this point it can't flow through the diode because it's open. So if we've got um now Z vol here, but we've got plus 5 Vols Well, uh, VCC not 5 volts whatever your VCC voltage happens to be.

So what happens when this is now at Z volts and our capacitor is charged to positive here and negative here? Well, this point is a nice solid zero volts Now, because our driving circuit, our microcontroller or whatever it is, it's got a reasonably low output impedance, it's going to be a nice solid Zer volts here. So this point has no choice now because we've got Um plus VCC across this capacitor, this is zero. That VCC voltage just doesn't suddenly vanish. What happens is our current tries to flow in this direction, our Dio becomes reverse biased and this point I.E When it Di's reverse biased no current flows at all and this point has no choice.

But if we got zero volts here, but we still got VCC on our capacitor, look zero then this point becomes minus. There's that negative it becomes minus. It drops down like that to our minus VCC and that's all there is to it. And just for completeness, we should actually draw this input waveform on here as well.

So let's call that number three, shall we? And what happens here? It's not zero. Remember we our condition was starting out at VCC to charge that cap. So it's like that. So remember we said when this point drops down to zero here, it inverts.

There it is. It drops down to zero and bang. It inverts produces our inverted waveform output. and then of course we add on our output filter.

Yes, it's just a simple uh diode filter. you're familiar with those from your linear power supply. But yeah, the diodes backwards cuz we're dealing with negative Supply voltages. but it works exactly the same.

All it does is filter out this negative and produces our nice solid. If we got no load, of course, a nice solid negative output voltage at minus VCC Once again, assuming ideal diodes. Once you start putting a load on there, well, and real diodes as we're going to see, when we build up the circuit, it's going to drop. but that is the basic operation of our voltage inverter too easy once again, for the cost.

lousy cost of two dodes and two capacitors, you can generate a negative rail from any circuit that has a switching component like that. And of course, that switching uh component as we said, could be a microcontroller could be a triple 5 timer. um, or uh often. they will uh, do this as well.

If you got a DC a positive DC to DC converter, you can actually tap the switching signal off that and use this uh inverter circuit to generate a low current negative. Supply And of course, you can get a dedicated charge pump or capacitor charge pump chips to do this. You know, the classic 760 uh voltage inverter, which also you can configure that the other way. As we've said before, works as a voltage doubler as well.

but that's a classic inverter. I Think on the market, sort of the maximum output current like 100 milliamps or something. your usual jelly beam ones like 10 odd milliamps. uh Max output current Really low stuff because you really can't put much charge in these capacitors cuz that's what you're using.

You're doing. You're using the capacitor as an energy storage element and well, a little tiny wimpy cap. eh. And to the breadboard we go.

We've got exactly the same circuit we just saw on the Whiteboard buildup. Uh, we're going to use three channels of the scope to measure this thing. This will be channel one this point will be channel Two. This point, the output will be channel three I've got uh, 0.47 microfarad, uh caps here and here and I've just got uh, crappy, uh 1 in 4148 diodes in there.

So we're going to get a bit of loss on those dodes and we won't be able to drive much load. But we'll start out by viewing the waveforms with no load and you'll see We'll get exactly what we saw on the Whiteboard And this time, just for fun, we'll use our GW Inc Uh, GDs 2304a VP oscilloscope Now Uh, the yellow waveform here is channel one, the blue waveform is channel Uh two, and the purple waveform There is the output Channel three. So there's our input uh, waveform there 0 to 5 Vols uh, Square wave coming from my function generator and point number two. As you can see, it inverts just like we saw on the Whiteboard and then our output voltage.

Of course we got no load, it's just flat like that. So we're getting oh, by the way, they're all uh, sorry, that reference point there. they're all referenced to that point. there.

they're all DC coupled of course and that's our reference point. So 5 Vols Up here where all at at 2 Vols per division so 2, 4, 5 and then this one drops down to5 But you'll notice that if we Oh wrong control. when you change Scopes like this, you'll notice that Yeah, you can just see the diode clamping in there. You can see the diode loss in there.

It's not precisely zero. so the diode clamps it not to Z volts, but to you know. plus well, you know 0.6 6 Vols it's actually lower than that because we got bugger all current. but it does clamp it to that diode loss.

And likewise, you'll notice that the purple waveform there, even though they're all referenced precisely on the Zer volt line. Here, there's a diode loss in there from the blue waveform to the purple waveform. and once again, that will depend on the loss in your diode at a particular current. So, a particular output current.

So you would have to look up your diode characteristic curve to find out what that's going to be. Now let's have a look. What happens when we put on a lousy 1meg load? Lousy? I Mean it's really high. Okay, lousy amount of current.

We're only talking if it stays at five, if our output voltage stays at 5 volts, we're only talking five microamp. So we're drawing bugger or current. Here we go. This is with currently with no load and let me whack it on here.

Get the alligator clip Bang There it is. You could visibly see that, change that Chang quite significantly and we'll just add a few little uh measurements in here to make our life easier. So let's have a look and see what I've done here. I Really like the Uh measurement uh capability of this GW Inc Works quite well in both adding and removing measurements.

and this little Uh window down here shows all our measurements. Now what I'm able to do here is: I'm actually able to add the Uh peak of what the max value up here. So if we have a look, that Uh is our maximum value of, well, you can choose your channel in this case Channel One. So our yellow waveform there I've got the well sorry, not the Uh, not the max value I've got the high value there which doesn't include any overshoot or anything like that.

So there you go. we're going to get our high volt there. our high value there 4.96 volts. As um I said, this is coming uh, direct from my function generator, so it's going to be pretty close to 5 Vols it's low impedance um output from the fun function generator.

So and well, it is very, very close to 5 Vol as you'd expect and then what? I'm able to add, there is Um for Channel 2. Now we're on to the blue waveform here. I'm able to add the low Valu so the bottom of the blue waveform down there. there we go 4.48 Vols and then I'm also able to add the high value up here which then can show our Um diode loss in that direction.

and there it is 320 m Vol And as you can see, it's above, it's 320 Ms. Well, it's just jumped up to 400 above that uh, reference point there. it's not going to be hugely accurate. Of course we've only got an 8bit analog digital digital converter in here.

Um, it depends on how you've uh, input, um, scaled the waveforms and stuff like that anyway, so that can show our diod loss there. pretty neat. And then our, uh, then I've got the mean value selected here of Channel 3, which is our output waveform. and there's our output voltage of - 4.2 Vol.

Of course we expect that to be Uh 5 Vol. but it's not because of our accumulative diode losses there. We got two diode losses in there. You remember this one.

The low value. Uh, there it is. It's only Uh - 4.48 -4.5 volts. So we've already lost Uh5 volts in our D drop going negative like that.

and then we lose another. in this case, about uh, from -4.5 going to the output here. So this point here is that. uh, blue waveform -4.5 So we've lost our diode drop there.

and then we lose our diode drop again on the output with Uh -4 uh 28. So we've lost another. 3 volts across that diode there. Oh, by the way, this is for a Uh 1 Meg load still.

and if we open our load, let's do that. Boom. There we go Go! They did jump up a bit. our low voltage here jumped up to Uh - 6.64 so it dropped up14 volts there and our output voltage jumped up a little bit to 4.4 But let's put the 1 Meg load back, shall we? And bingo, you can see those vales change.

Let's go to say 100K load. Okay, there we go. We're now getting an output voltage of -4 Vol with 100K load and then we could let's drop there horrible and drop that down to 10K And whoa, now we start seeing some Ripple Effects By the way, if you're wondering how I'm doing that just using my decade uh resistance box here. very handy to have.

uh, build yourself a decade resistance box just for this, uh, purpose so we can get a good look at that. uh, get rid of the menu there. get a good look at that Ripple Now that output Ripple du you can see the capacitor charging and then discharging on that purple waveform. so there's that little charge there and then whoop discharge.

and once again, this is going to depend on the value of your capacitors and your switch in frequency. So we're switching frequency at the moment with this uh 10K load is uh 1 khz. so um, as you can see, it's got the hardware uh frequency counter in there showing the 1 khz. but we can, uh, change that of course.

let's give that a go here. sorry I've got to reach across my bench and let's change to 10 khz. Here we go. Boom! let's expand that out a bit and you can see that we're getting no more Rippling there.

The Ripple's gone at 10 khz. exactly what you'd expect. Get a smoother response like that with less Ripple by either increasing your Um frequency or increasing the value of your capacitor capacitors or both. and let's get really nasty.

and uh, take that down to W. I just shorted that out there. we go and take it down to 1K Ooh, that's pretty horrible. Where's our trigger level? The reason it was jittery? there is cuz our trigger level was right up the top here, right at the top of that waveform.

So we bring that down to the center of course. Oh, you can just hit the 50% uh button on your scope, it wax it in the middle and there we go. That's it. 1K load and we're still getting out.

minus 2.76 volts there. So yeah, your diode losses are starting to kill you. Now down at 5 Vols as you'd expect. I Mean you can get better than this by using um, shock key diodes.

and we're only using 47 microfarad uh, caps as well. You know, typical ones you might have in there. or you might have a microfarad or something like a ceramic cap. Otherwise, you know.

Oh, you can get 10 microfarad Ceramics Typical in you know, a basic SMD design these days, but you know, sort of above that you sort of going to go into the electrolytic uh territory. and if we short that load out boom, look at that. We're even killing our input waveform. So there you go.

There's a diode voltage inverter you can build for practically uh, zero cost cuz you've probably already got some diodes and some capacitors in your billing materials anyway, so it can be an absolute bargain if you just need to generate a Um a simple low current negative voltage as I said for an opamp, or for a negative regulat, or for to get a regulator down to zero, or for any other purpose that you need need that? Um split Supply And once again, you can add a linear regulator on the output here. If your input voltage is high enough, you could use a low Dropout regulator. So if you had a 5vt Uh supply for example, then you could easily use even with Um at at low currents. Even with crappy 1 in 4148 diodes and low values of capacitors in here, you could you get a fully regulated and clean 3.3 volt linear Supply with a low Dropout regulator.

Not a problem, but uh, using this particular load which is 10K at the moment, there's our output voltage uh, 3.64 volts. That's good enough Um to give us basically uh, you know, a couple hundred uh microamps, um output current if we use a low power uh, low Dropout voltage regulator. At 3.3 volts, we'd get a nice clean Supply with even these crappy parts. So there you go I Hope you enjoyed that.

If you want to discuss it, jump on over to the Eev blog. 4 The link direct link to the Uh individual video thread is down below and as always, if you like fundamentals Friday Catch you next time.