More fun with a simple boost converter (3V--> 9V)

[Mr RB]

Carbonzit, no I didn't look at your LT Spice file in post #1, I don't use LT spice I use LT soldering iron, LT CRO and LT brain.

I put some parts values in the power mutivibrator, using the methodology and circuit I suggested all along the power multivibrator was tested first and some values chosen to give 9v and 44mA out with good efficiency. That is the "butter zone" I think for most 9v battery powered applications.

Then added the regulator that has worked so well in the past and did some voltage and current tests;

[b]Power multivibrator first test.[/b]
Vout    Iout      Vin     Iin     Eff%   Freq    Duty
9.1v    44mA      2.99v   160mA   82%    26kHz   68%

[b]After adding regulator Q3[/b]
Vout    Iout      Vin     Iin     Eff%   Freq    Duty
8.85v   0mA       3.30v   5mA     0      Chaotic
8.85v   0.8mA     3.27v   10mA    24%    Chaotic
8.86v   3.3mA     3.25v   17mA    53%    83kHz
8.85v   8.8mA     3.22v   35mA    69%    45kHz
8.85v   19mA      3.05v   65mA    83%    31kHz
8.80v   27mA      3.08v   86mA    88%!   25kHz   67%
8.75v   40mA      2.97v   140mA   53%    25kHz

This was all done very rough and ready but proved the methodology; 1 that the power stage was first optimised for good efficiency near the max required current, then, 2 the regulator once added worked flawlessly to control that power stage and maintain output voltage.

This was all done very rough by plugging some parts values in, and measurements too were rough so allow a couple percent on all measurements. However I think it should be capable of 90% efficiency through a tuned Iout band, which is definitely in SMPS IC territory.

Note the waveforms (Q1 collector) are very clean and efficiently switching and the frequency deliberately chosen by the multivibrator values to give good power transfer from that 420uH inductor. It ran at about the same frequency with no inductor at all (just a resistor) showing the multivibrator doing it's job (as I earlier suggested it an be tuned independent of the inductor characteristics, allowing tuning for good efficiency).

I don't think this simple circuit would good be for higher power output levels but it worked pretty good at the Iout I chose (for use to replace 9v battery).

(edit) Checking the photos the resistors on the Q3 regulator base are 100 ohms and 4k7, (not 2k7).

 

[carbonzit]

Carbonzit, no I didn't look at your LT Spice file in post #1, I don't use LT spice I use LT soldering iron, LT CRO and LT brain.

Far be it from me to tell someone else, especially someone like you who clearly knows much more about electronics than I do, how to work.

However, I'd like to suggest that maybe you shouldn't be so dismissive of simulators like LTspice. While ultimately there's no replacement for soldering iron, test instruments and brain, a simulator can help immensely with designing--and tweaking--a project such as this. Hell, I would never have gotten as far as I have here without it!

I put some parts values in the power mutivibrator, using the methodology and circuit I suggested all along the power multivibrator was tested first and some values chosen to give 9v and 44mA out with good efficiency. That is the "butter zone" I think for most 9v battery powered applications.

Then added the regulator that has worked so well in the past and did some voltage and current tests;

[b]Power multivibrator first test.[/b]
Vout    Iout      Vin     Iin     Eff%   Freq    Duty
9.1v    44mA      2.99v   160mA   82%    26kHz   68%

[b]After adding regulator Q3[/b]
Vout    Iout      Vin     Iin     Eff%   Freq    Duty
8.85v   0mA       3.30v   5mA     0      Chaotic
8.85v   0.8mA     3.27v   10mA    24%    Chaotic
8.86v   3.3mA     3.25v   17mA    53%    83kHz
8.85v   8.8mA     3.22v   35mA    69%    45kHz
8.85v   19mA      3.05v   65mA    83%    31kHz
8.80v   27mA      3.08v   86mA    88%!   25kHz   67%
8.75v   40mA      2.97v   140mA   53%    25kHz

Hate to say it, but my power multivibrator kicks your power multivibrator's butt!

I got a good solid 90mA output (into 100Ω) at a good solid 9 volts. Check my component values (our circuits are otherwise practically identical). efficiency was (I think, have to re-check it) somewhere around 70%, not bad for such a "primitive" circuit.

One thing I'm pretty sure of is that your inductor is waaaaaaay too big. Might I suggest that you retry this with a smaller L? (I'd start with 100µH, which seemed to be pretty much the "sweet spot" with my version.)

 

[Mr RB]

...
Hate to say it, but my power multivibrator kicks your power multivibrator's butt!

I got a good solid 90mA output (into 100Ω) at a good solid 9 volts. Check my component values (our circuits are otherwise practically identical). efficiency was (I think, have to re-check it) somewhere around 70%, not bad for such a "primitive" circuit.
...

Haha! Don't "hate to say it", be proud of the work you put in and what you have achieved.

Just bear in mind for comparison that I deliberately chose the 0-40mA output range and the very small amount of effort choosing parts values was to try and get about 80% efficiency near the top part of that range. Likewise the large inductor and slow 25kHz freq were all chosen for good efficiency. Remember everytime that power stage switches / or \ you get a fixed energy loss, so the less switching it does per second the higher the efficiency.

I have to ask why do you need such high output current? If this is for a general 9v battery replacer then 99% of people that use it will probably have currents about 5mA to 15mA range.

If you optimise to get 100mA output then it will work quite poorly for most of your users who only need 10mA and would prefer great battery life at 10mA. Just a thought to keep in mind...

 

[The Electrician]

I'm still not convinced that a MOSFET driver is not the way to go, but that's at least partly because I just want to finally work with one of those devices.

Here's a neat way to use a MOSFET driver.

Have a look at the first attached image. It shows a boost converter and some variations.

The first circuit shows a standard topology with a MOSFET lower switch and a diode upper switch. This circuit has some disadvantages; it isn't protected against short circuits on the output. If you short the output, the input source just flows through the diode and into the short. This circuit is also prone to runaway output voltage if there's no load on the output, and the ouput voltage depends on BOTH the load and the duty cycle of the switch.

Circuit 2 shows the MOSFET replaced with a switch to emphasize that the MOSFET is behaving as a switch. Circuit 3 shows a switch in both positions; of course, the switches must be operated alternately.

Circuit 4 is the same as circuit 1 except that the diode is replaced with another MOSFET, a P-channel type. This circuit has the property that power flow can be bi-directional (in fact if the source and load are interchanged, it becomes a buck converter), and the output voltage is almost completely determined by the duty cycle of the switches (there is a slight dependence on circuit parasitics; resistance of the inductor and the switches, etc.). If the switches are driven at a reasonable duty cycle, the output voltage cannot run away on no load. A problem with this circuit is how to drive the upper MOSFET.

Circuit 5 is the same as circuit 4, redrawn to emphasize that the two MOSFETs are connected in a totem pole configuration.

Now, have a look at this MOSFET driver:

**broken link removed**

In particular, look at the schematic of the internals in the Functional Diagram on the first page of the datasheet. Notice how the output is a pair of MOSFETs, an N-channel and a P-channel in a totem pole configuration. Those two output FETs are just what is needed to make a boost converter. They are rated for 3 amp peaks and there are two such drivers in the package. The internal electronics takes care of driving the upper P-channel FET. If you apply a logic level to the input (less than .8 V low to greater than 2.4 V high), the output FETs will be alternately switched on and off.

So, how do we use this for a boost converter? The junction between the output FETs apparently should be connected to an inductor, but where do we get the output? The output appears at the Vs connection of the device; connect the output filter capacitor there and take the load from there.

Circuit 6 in the image shows how to hook it up. Connect the two drivers in the IC package in parallel and drive the inputs with a variable duty cycle square wave. The input drive can be obtained in any manner desired; from a 555 oscillator, from a two transistor multivibrator, or as I did, from a function generator.

In order to get .5 amp at 9 volts from a 3 volt source, the 3 volt source will need to supply 1.5 amps; this is what conservation of energy requires. If you're drawing 1.5 amps from the input source, the switches can't have much on resistance. That's why a small power FET would be appropriate here. The output FETs in the MIC4424, which is what I used, are what I would call small power FETs; they can handle 3 amps peak. Paralleling two drivers would give 6 amp peak capability. But, even then the output resistance is rated 3.5/2 ohms, typical. This will cause some loss of output voltage when loaded at .5 amp. It's possible to solder several 8-pin DIP packages in parallel to get lower losses. I stacked 3 packages, for a total of 6 driver stages in parallel. If you want substantially less than .5 amps, one package would do.

I wired all this up and drove the inputs with a constant 33% duty cycle square wave, fed the input with 3 volts (from a bench supply rather than AA cells). Unloaded, the output voltage would be expected to be 9 volts, completely determined by the duty cycle. Without any feedback for regulation, the output voltage would be expected to drop a little, which it did. But by using an active switch (P-channel FET) for the upper switch rather than just a diode, the output voltage is almost completely determined by the duty cycle.

The second image is a scope capture showing some circuit variables with no load. The yellow trace is the voltage at the junction of the output FETs of the MIC4424; this is the voltage at the active end of the inductor. Notice how the voltage the the active end of the inductor goes from zero to 9 volts. The purple trace is the current in the inductor. Notice how the average value of the current is zero. The sawtooth of current is the current sloshing back and forth between the input source and the output capacitor. The green trace is the output voltage, at the Vs pin of the MIC4424.

The third image shows things with a nearly .5 amp load. The average current in the inductor (purple) has increased to nearly 1.4 amps; it never drops to zero. This is what is called continuous current mode. The output voltage has dropped to 7.76 volts and the peak voltage at the active end of the inductor doesn't get up to 9 volts anymore, but only 8 volts. All this is with no feedback for regulation. It would only take a little feedback to the circuit that controls the duty cycle to get good regulation.

 

[carbonzit]

Hey, thanks, "The". Even though that kinda violates one of my ground rules (no chips!), what you've shown here is fascinating. I'm still digesting it.

Couple questions: First of all, it seems very strange to take power output from the Vs pin of the driver chip. I guess this is really a "self-bootstrapping" circuit, where the power to drive the inverters, etc., inside the chip will gradually appear as the boost converter ramps up. Neat.

Regarding what happens with no load, even my primitive converter has enough regulation to guard against the voltage zooming up (I just tested mine with open output, and the voltage only rose to 10 volts), so that's not a problem.

Short-circuit protection? that's a whole 'nother thing altogether; frankly hadn't even thought about it (since this isn't even remotely a potential commercial device). But, of course, that would be nice.

Regarding efficiency: my understanding (which may not be correct) is that efficiency depends, at least in part, on having if not the maximum duty cycle possible, at least a sufficient Ton for the switch. Is this a really important parameter here? It seems to be.

It'd be interesting to get one of those drivers and play around with it, see what it can do with a couple of lowly BJTs driving it ...

 

[alec_t]

One of the uses of the boost converter is, of course, to squeeze more juice out of partially-depleted AA cells. In simulation I find that the converter output (pre-regulator) is very dependent on the internal resistance of the cells. My gut feeling is that the resistance will rise as the cells run down. Can anyone verify that? In practice, does the regulator cope well with such a change?

Alec

 

[The Electrician]

Hey, thanks, "The". Even though that kinda violates one of my ground rules (no chips!), what you've shown here is fascinating. I'm still digesting it.

Well, you did say "I just want to finally work with one of those devices."

Think of the driver as just being a couple of small power FETs with associated drive circuitry done for you. You've already run into one problem with FETs, which is their relatively large input capacitance. To get the rapid rise and fall times on the gate which are essential for good efficiency, you're going to have to use some additional circuitry anyway.

Couple questions: First of all, it seems very strange to take power output from the Vs pin of the driver chip. I guess this is really a "self-bootstrapping" circuit, where the power to drive the inverters, etc., inside the chip will gradually appear as the boost converter ramps up. Neat.

Yes. An important spec for this use, which is described in the datasheet, is the ability of the output pin to accept a reverse current without damage. The datasheet says a single driver can accept .5 amp.

Regarding what happens with no load, even my primitive converter has enough regulation to guard against the voltage zooming up (I just tested mine with open output, and the voltage only rose to 10 volts), so that's not a problem.

Did you do this with the regulation disabled as well as no load? Did you still have D3, D4 connected to the base of Q3 with the collector no longer connected to R6 for feedback? If D3 and D4 were still connected, that would hold down the output.

If you had D3 and D4 disconnected, the fact that the output only rose to 10 volts is an indicator of low efficiency.

I once designed a boost converter to convert the 12 volts from a vehicle battery to about 350 volts for a xenon flash warning light. The FET I used was rated to be safe if the internal body diode broke down due to applied voltage greater than the rated normal operating voltage of the FET (the datasheet shows the internal body diode as a zener). The converter charged a capacitor which provided the energy for the xenon flash tube. The capacitor was rated for 450 volts, and the breakdown of the FET body diode kept the voltage from rising above about 350 volts, which it would have since there was no feedback for regulation.

Short-circuit protection? that's a whole 'nother thing altogether; frankly hadn't even thought about it (since this isn't even remotely a potential commercial device). But, of course, that would be nice.

One way around this is to use a different topology, such as a Cuk, sepic or buck-boost converter.

Regarding efficiency: my understanding (which may not be correct) is that efficiency depends, at least in part, on having if not the maximum duty cycle possible, at least a sufficient Ton for the switch. Is this a really important parameter here? It seems to be.

This is quite true. When the pulse width becomes short in one direction, the peak current for the switch driven with the short pulse width will become large. This could be the diode in the case of a boost converter, even though strictly speaking, the diode isn't "driven".

When the current is a large, short pulse it becomes more important that the switch (possibly diode) carrying that pulse of current have a low "on" resistance. This is why it can be a good idea to use an active switch for the upper switch rather than just a diode. See:

https://en.wikipedia.org/wiki/Synchronous_rectifier

A small BJT such as a 2N2222 just can't deal with short, high current pulses.

It'd be interesting to get one of those drivers and play around with it, see what it can do with a couple of lowly BJTs driving it ...

The input to the MIC4424 is a schmitt trigger and is relatively high impedance. It's no problem at all to drive it with your multivibrator. Even with a slow rise time from the multivibrator, you will get fast rise and fall times on the output.

If you choose to use a FET for the bottom switch and keep a diode as the top switch, you will need to have much faster drive rise and fall times on the gate of your FET. You will probably need to use a complementary emitter follower which adds parts to your circuit. In this application, a medium power FET is suitable. It won't get hot because its on resistance is low (and assuming the input drive has short rise and fall times), so no heatsinking would be required.

If you use a FET for the top switch, you have the problem of providing "floating drive" with fast rise and fall times, which will require even more parts if you provide the circuitry yourself.

 

[carbonzit]

Regarding what happens with no load, even my primitive converter has enough regulation to guard against the voltage zooming up (I just tested mine with open output, and the voltage only rose to 10 volts), so that's not a problem.

Did you do this with the regulation disabled as well as no load? Did you still have D3, D4 connected to the base of Q3 with the collector no longer connected to R6 for feedback? If D3 and D4 were still connected, that would hold down the output.

If you had D3 and D4 disconnected, the fact that the output only rose to 10 volts is an indicator of low efficiency.

Yes; I was trying to get across that my regulator (D3, D4, Q3) actually worked. Of course, without it, the voltage could quickly rise to dangerous levels. But worrying about "fail-safe" protection is about the farthest thing from my mind at this point. I'm not trying to make a bulletproof power supply, only a simple, compact one. One has to trust one's components at some level ...

If you choose to use a FET for the bottom switch and keep a diode as the top switch, you will need to have much faster drive rise and fall times on the gate of your FET. You will probably need to use a complementary emitter follower which adds parts to your circuit. In this application, a medium power FET is suitable. It won't get hot because its on resistance is low (and assuming the input drive has short rise and fall times), so no heatsinking would be required.

If you use a FET for the top switch, you have the problem of providing "floating drive" with fast rise and fall times, which will require even more parts if you provide the circuitry yourself.

So I'm a little confused: if I use two switches (with a switch in the diode position), do I need to supply them with complementary pulses? (meaning opposite phases of my multivibrator, for instance.) If so, do I want the duty cycle for the 2nd (diode) switch to be small compared to the first switch?

 

[The Electrician]

So I'm a little confused: if I use two switches (with a switch in the diode position), do I need to supply them with complementary pulses? (meaning opposite phases of my multivibrator, for instance.) If so, do I want the duty cycle for the 2nd (diode) switch to be small compared to the first switch?

The switches, be they diodes, FETs, BJTs or actual switches conduct alternately; that's the key thing in converters like this. When the top switch is a diode, this property takes care of itself--the diode only conducts when the bottom switch isn't conducting. If the top switch is an active device, then the designer must provide drive to the top switch which is the complement of the drive to the bottom switch.

When using active devices for the switches it is typical for the on times of the two switches to be complements; when one is on, the other is off. This necessarily means that if one pulse is long (occupies most of the switching period), the other must be short. It's those very short pulses that are troublesome because the associated switch will be carrying a short pulse of relatively high current. The drives for the two switches don't have to be exact complements; it is possible to have some dead time when neither is on.

But, what must be avoided in most simple converters is to have both switches on at the same time. This causes "shoot through".

See: https://www.electro-tech-online.com/custompdfs/2011/07/AN-6003.pdf

and: https://www.electro-tech-online.com/custompdfs/2011/07/slyp089.pdf

 

[carbonzit]

Ah, yes, shoot-through; that's the same problem as with class D amplifiers, right?

So it seems I could easily run the switches from opposite sides of my multivibrator, correct?

 

[The Electrician]

Ah, yes, shoot-through; that's the same problem as with class D amplifiers, right?

So it seems I could easily run the switches from opposite sides of my multivibrator, correct?

In theory, yes. But, modern FETs are very fast and an overlap of only a few nanoseconds can be dangerous if the power levels are high. You should check with a fast scope if there is any overlap of the waveforms from the two sides of your multivibrator.

The design of the internals of the MIC4424 avoids shoot through in its output stage.

For your project if you were using FETs capable of handling tens of amps, such as the IRFZ44, a few nanoseconds of shoot through won't provide enough current to hurt the FETs.

One very undesirable effect of shoot through is the production of EMI. This can be a big problem if it happens in high power converters. Your converter is such low power that it probably won't be a problem.

 

[Mr RB]

Synchronous rectification adds a whole level of complexity that Carbonzit said he wanted to avoid as he wanted "minimum parts and no ICs".

Also the output diode is one of the most efficient parts of a 3v-9v boost converter seeing that the 1N5819 will have a forward voltage of only about 0.15v to 0.2v into the 9v load so diode losses will never be more than about 0.2 / 9v = 2.2%. Even with a FET synchronous rectifier the Vds might still be 0.05v to 0.1v so total efficiency savings from all that complexity and risk MAY approach a 1% overall improvement.

In a buck converter with a high ratio of Vin:Vout synchronous recitification can do wonders, but in this application it is just an unwarranted complexity and risk of potential problems.

 

[The Electrician]

Synchronous rectification adds a whole level of complexity

The little converter I built shows that no additional complexity is obligatory.

....hat Carbonzit said he wanted to avoid as he wanted "minimum parts and no ICs".

He also said with respect to MOSFET drivers that "I just want to finally work with one of those devices."

Also the output diode is one of the most efficient parts of a 3v-9v boost converter seeing that the 1N5819 will have a forward voltage of only about 0.15v to 0.2v into the 9v load so diode losses will never be more than about 0.2 / 9v = 2.2%. Even with a FET synchronous rectifier the Vds might still be 0.05v to 0.1v so total efficiency savings from all that complexity and risk MAY approach a 1% overall improvement.

Here are three 1n5819 datasheets, in decreasing order of forward voltages:

https://www.electro-tech-online.com/custompdfs/2011/07/ds23001.pdf
https://www.electro-tech-online.com/custompdfs/2011/07/1N5818.pdf
http://www.datasheetcatalog.org/datasheet/stmicroelectronics/6262.pdf

Even the most favorable one, the last one from STM shows that your numbers for forward drop are overly optimistic. You haven't said at what current you are giving the forward drop, but taking the highest output of 40 mA from post #21, the diode current when conducting will be equal to the input current of 140 mA. The STM datasheet shows a forward drop of .33V; the Diodes Inc. datasheet shows a forward drop of about .6V at 140 mA. These are all substantially greater than the .2V you cite.

Even more to the point, my converter had a current of 1.4 amp in the top FET with a drop of .24V. If the current in the top FET had been only .14 amp as in your example, the drop would be .024V instead of .33V as with the best 1N5819, more than 10 times less. You cite the FET drop as .05V to .1V, but in fact the actual drop in my hardware implementation would be .024V at .14 amp. And that's with the relatively low performance output FETs in the MIC4424. If a discrete power FET were used, the drop would be even less, only a few millivolts.

You understate the drop in a 1N5819 and overstate the drop in a FET.

You also mentioned twice some "risk" associated with a synchronous converter. What risk are you talking about.

In a buck converter with a high ratio of Vin:Vout synchronous recitification can do wonders, but in this application it is just an unwarranted complexity and risk of potential problems.

The way I did it involves no great complexity, uses a MOSFET driver, which the OP said he wanted to do, and involves no particular risk the I am aware of.

 

[carbonzit]

The way I did it involves no great complexity, uses a MOSFET driver, which the OP said he wanted to do, and involves no particular risk the I am aware of.

I should have clarified and said that I want to try using a MOSFET as a driver (meaning switch), not a separate MOSFET driver as you're proposing.

However, as I also said, your suggestion is sufficiently intriguing that it's on my to-do list.

 

[Mr RB]

...
https://www.electro-tech-online.com/custompdfs/2011/07/ds23001-1.pdf
https://www.electro-tech-online.com/custompdfs/2011/07/1N5818-1.pdf
http://www.datasheetcatalog.org/datasheet/stmicroelectronics/6262.pdf

Even the most favorable one, the last one from STM shows that your numbers for forward drop are overly optimistic. You haven't said at what current you are giving the forward drop,
... These are all substantially greater than the .2V you cite.

Yeah I was a fraction off there. I just checked my bag of 1n5819 it says; 1mA 0.153v, 50mA 0.258v and 200mA 0.303v. I was "guesstimating" based on my orig focus of a 9v battery replacer that for the majority of users will only be providing maybe 10mA average at about 33% duty so the 1N5819 drop would be maybe 0.24v. I haven't really been focussing on a high current 3v to 9v converter as nobody has yet explained what it would be used for.

...
You understate the drop in a 1N5819 and overstate the drop in a FET.

You're right, I was assuming a low current application and assuming the PFET would be driven from a low voltage battery maybe 3v or so but neither applies to your example. Sorry.

...
You also mentioned twice some "risk" associated with a synchronous converter. What risk are you talking about.

Risk of not driving the rectifier properly, not turning it off fast enough, shoot through etc etc all possible given the OP's stated lack of experience with FETs and desire for minimum parts solution that usually requires tradeoffs and imperfect driving waveforms. Again I think we had a communications issue you were talking about your suggestion using an IC and i was talking about doing something in line of the thread direction so far. Let's not sweat it any further.

In terms of ICs there are a few brands of dedicated boost converter ICs that just need an inductor and sometimes a diode, that will do the whole job, but if I read Carbonzit right he had a particular goal of something like a minty boost etc that hobby guys can build from a couple of transistors out of their junk box. Similar to my 2-tran and 3-tran designs that lots of hobby people have used over the years, hence my interest in the thread. I never released a 3v-9v high efficiency boost version and said to Carbonzit in the "Colin's thread" that it would be cool to work on an open source 3v to 9v converter to see what we could all cooperatively come up with.