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View Full Version : Idea I'm designing a 2-stage ultra-BEC :)


KreAture
Apr 17, 2004, 07:22 AM
Hi ppl! I finally found why my plane kept ending up with nose splintred. Turned out my manual was wrong on my ESC's BEC capability... I can barely support 2 HS-55's with it on a 3-cell LiPo :( (It's a TMM 1812-3s)
I was using a HS-81 and a W-084. Together I would assume they compare to 3-4 HS-55's.

I can't use another battery for RX as it would add same weight as regular flight-pack and last for shorter than main batt...

So I decided I needed a really good BEC-circuit.

What I found was the linear regulators can't deal with such high input-voltages at the same time as they can deliver full current-capacity. They will simply overheat like the BEC in my ESC did and cut the voltage to the RX and servos...

What I came up with is a 2-stage BEC that first uses a 555-timer to generate a sawtooth, and uses this to PWM-control a FET via a comparator. This would make a crude regulator to approx 7v. From here, the linear regulator takes over and can deliver 1 amp continous at 5v.

I will start with a 1 amp version, and improve it to 2-3 amps later, but for now, 1A would be more than enough for my needs...

Currently I am simulating the circuit to see how well it would work. I'll post updates as I progress.

KreAture
Apr 17, 2004, 10:21 AM
Well. Simulation shows it to be working :) I actually have so good line regulation at first stage, the second stage might not be needed :)

Basically I've made a switched mode inductorless drop-down regulator :)

KreAture
Apr 17, 2004, 01:08 PM
Dho! I couldn't stabelize it :(

Oh well, might just as well try something simpler then...

If you rig up a opamp, a FET and a voltage-reference so that opamp controls FET. Can you get it to oscillate abowe the voltage required for a linear regulator so that the reg can work from the dropped voltage without having to use linear drop from full battery-voltage?

I mean, take 16v input, use a 6v reference and let the fet and opamp osillate while trying to create a 6v supply. Now regulate the bad 6v to a nice 5v with a linear reg...

KreAture
Apr 17, 2004, 02:26 PM
Here's a picture showing the last test-circuit in the simulator.
It oscillates very nicely but I'm afraid it would converge in real-life...

http://www.unixcore.com/~kreature/images/kbec/kbec_01.gif

Any thoughts ?

Comatose
Apr 17, 2004, 04:57 PM
If you want to ensure oscillation, it'd help to add a cap between the input pin of the op-amp and ground. Personally, I'd use a little PIC running a software feedback loop controlling the duty cycle of a PWM.

KreAture
Apr 17, 2004, 05:32 PM
Yeh. I was just fiddling with this trying to make it without a microcontroller. That way more could build it. Any design with a micro would limit the building to those capable of programming one. As long as it's without anyone should be able to slap one together...

mmormota
Apr 17, 2004, 06:14 PM
Here's a picture showing the last test-circuit in the simulator.
It oscillates very nicely but I'm afraid it would converge in real-life...

http://www.unixcore.com/~kreature/images/kbec/kbec_01.gif

Any thoughts ?

Sorry, but this device is wrong. You just connect a battery to a capacitor through a FET. If the voltage on the capacitor is lower then the battery voltage (this is what you want), a very high current is flowing - it depends on the serial resistance of the FET, the battery and the capacitor. The product of this current and the voltage difference is power loss, heat. It means, that this device is not a switched mode high efficiency power supply, but a switched mode dissipative device with very high current spikes. But the dissipation is exactly the same as on a normal analog stab. So the device missed the point.

For high efficiency you need an inductor, or or switched capacitors - charge them in series, discharge them in parallel.

There are a lot of good step-down switched mode stabs from Texas, Maxim, Linear Technology etc. They need only a FET and some capacitors.

KreAture
Apr 17, 2004, 08:44 PM
I see your points mmormota but if there wasn't anything to gain from it, it would have current and voltage peaking at same time. Remember, the oscillation causes the capacitor to be charged just enough to keep it within good working-levels of the linear regulator. This is the voltage translation.

The dissapation is lower than using a simgle 16-5v linear drop regulator. When I let the FET open to recharge the capacitor with 1.5 volts more, it does so in 50us. It has the equivialent of 4.4 ohm resistance and dissapates 10 watts while doing it. But the on-time under full load (1 amp, 5v) is only 1/5 or so of the actual cycle, thus it only dissapates 2 actual watts to drop 16 volts to approx 6.5v. Equivialent linear drop would be 9.5 watts at 1 amp. The FET's ability to quickly change resistance to adjust voltage on capacitor makes it possible...

Exactly how effichient it will be is still to be determined, but it will beat a linear regulator.

KreAture
Apr 17, 2004, 08:46 PM
Here's the layout for the prototype btw...
http://www.unixcore.com/~kreature/images/kbec/kbec_03.gif

I'll post some results when I test it. I just hope it will keep oscillating hehe...

clipclop
Apr 17, 2004, 11:51 PM
Kreature,
Nice work , but I tend to agree with mmormota as useing a stepdown regulator chip requires a minimum of external components and they are cheap .
You are doing the same thing as a chip just with discretes .
Stewart

KreAture
Apr 18, 2004, 06:22 AM
The stepdowns cost bundles here in Norway and all use a coil. This coil is even more expensive than the chips. This is a work in progress though, to see how it would work. Maby it can be usefull.

Besides, theese were components I had here in my parts-bin. Not having to buy anything right now is a big pluss.

vintage1
Apr 18, 2004, 06:29 AM
You do need a coil to get efficiency, thats just in there in the nature of the beast. You could wind a few turns round a ferrite toroid they are not expensive.

If the switching speed is high enough, its pretty small and cheap. Or you could use a resistor if you didn't want efficiency and were trading e.g. heat in a chip for heat in a resistor.

I would be interested in URLS to any of the chips that are designed for this purpose. This is an ideal DIY project - simple to get working and predictable, and the parts shold not be THAT expensive.

KreAture
Apr 18, 2004, 06:37 AM
I should be getting 50-60% eff easily with my design as long as it oscillates. The coil-versions are ofcource a lot better. My aim was only to stay below the need for advanced heatsinks and coils that break and send off massive EM's.

mmormota
Apr 18, 2004, 06:46 AM
The dissapation is lower than using a simgle 16-5v linear drop regulator.

No, it is not lower.


Consider 1Amps continous load on the 5V side, and 100A pulse current.

First a simple analog stab. P_average=(16V-5V)*1A=11W

You need to keep open the FET in 1% of the time. In that 1% you dissipate (16V-5V)*100A=1100W but only in 1% of the time, P_average=11W

To save energy, to bridge the voltage gap without dissipation, you need a device which _stores_ energy for a short time.

KreAture
Apr 18, 2004, 06:55 AM
Your logic is flawed as it works backwards. You start by assuming a pulse-current and duty cycle equal to what you would get with 11w loss, then you proceeed to calculate the 11w loss.

KreAture
Apr 18, 2004, 11:50 AM
Well, as I feared... It stabelizes and acts as a linear regulator, even without the required capacitors on the feedback-connection.

It worked when I added a external 555 timer to excite the balance, and it then did as expected :) I used the 555 to create an inaccuracy in the feedback at approx 10khz. The FET did not get hot even when passing an average of 0.5 amp so the circuit actually works...

But, the 555 addition is too much. I guess I'll just have to make a inductor based one :( Oh well.

clipclop
Apr 18, 2004, 12:42 PM
Kreature,
Have a look at nationals LM2576S-5 its a 3amp switching reg comes in a TO-263 package , min input voltage is 4V , Max input voltage is 40V . Free samples available too
Stewart

KreAture
Apr 18, 2004, 01:19 PM
Thanks clipclip! I will!
But a sw reg with coil can be made with a LMxx circuit too :)

mmormota
Apr 18, 2004, 04:10 PM
Your logic is flawed as it works backwards. You start by assuming a pulse-current and duty cycle equal to what you would get with 11w loss, then you proceeed to calculate the 11w loss.

Kirchoff's law: current into a point and current from the point must be equal. That's why the average of the pulsed current must be equal to the load current. Without inductor, the pulsed current into the 7805 input and capacitor is equal of the battery current. If the fet is open only say 10 percent of the time, the battery current must be 10 times higher then the load current.

If it is a normal stepdown regulator with an inductor, the the last part is different: the average current of the inductor is equal to the load current, but: the pulsed battery current is equal to the _average_ current, and it is pulsed, when the FET is open, the current flows from the battery, when closed, the current flows from the _gnd_ (a Schottky diode or another FET opens).
Believe me, I am EE, and designed switched mode power supplies for years...

mmormota
Apr 18, 2004, 04:23 PM
A step-down regulator with inductor is similar to a square wave generator and a lowpass filter, the L and C part is the filter.
In case of a brushed controller, there is no inductor _in_the_controller_. The output is the square wave, and the motor itself is the "filter" as it is inductive.

KreAture
Apr 18, 2004, 04:27 PM
Agreed mmormota, but the law does not apply when the resistance can change dramatically like a FET can.

If we assume constant drain of 1 amp, 16v input, 5v output. That is 5w over load, and 11w on a drop-down converter using linear regulation. So far we agree right ?

If we pass this 1 amp through a FET 100 times the current, but only 1/100 of the time and the RdsON resistance is 6 mOhm we have: 100a^2*0.006 = 60 watts. 60/100 = 0.6 watts. The same power is now available over load but the power needed to be dissapated in the FET is much lower than the linear reg. To convert current to voltage the capacitor can be used. And yes, if the cap is not made for this it will heat up significantly... The voltage becomes irrelevant as Rds*I^2 dictates the losses in the FET.

This is after all how PWM based regulators work, and don't come and tell me they don't exist...

mmormota
Apr 18, 2004, 08:11 PM
Agreed mmormota, but the law does not apply when the resistance can change dramatically like a FET can.

Disagree, Kirchoff-law is universal. Consider, that electrostatic charge remains in the point (electrons if you like) if the rule fails.


If we assume constant drain of 1 amp, 16v input, 5v output. That is 5w over load, and 11w on a drop-down converter using linear regulation. So far we agree right ?

So far, so good. ;)


If we pass this 1 amp through a FET 100 times the current, but only 1/100 of the time and the RdsON resistance is 6 mOhm we have: 100a^2*0.006 = 60 watts. 60/100 = 0.6 watts. ...

First error. The voltge, current and resistance is not independent, Ohms law rules them. If you choose 6 mOhm (as a free parameter) for the FET, the current is not free parameter any more, because the voltage is fixed input parameter too: 11V
I = 11V/0,006Ohm = 1833,3 A
With this current, you need 1 : 1833 open/close time. P = 1833^2*0,006 / 1833 = 11W

If you fixed the current to be 100A, then the FET resistance oi not free parameter any more,
R_fet = 11V/100A=0,11Ohm
With this current, you need 1:100 open/close time, P=100^2 / 100 * 0,11 =11W


The same power is now available over load but the power needed to be dissapated in the FET is much lower than the linear reg. To convert current to voltage the capacitor can be used. And yes, if the cap is not made for this it will heat up significantly... The voltage becomes irrelevant as Rds*I^2 dictates the losses in the FET.


As the calculation was errenous, there is no gain in dissipation.

Capacitors really can be used to convert voltage, this is the switched capacitor method. But it is based on charg in series, discharge in parallel to step down, or reverse order to step up. In the circuit diagram (we are talking about that .gif) there are no such capacitors.


This is after all how PWM based regulators work, and don't come and tell me they don't exist...

A PWM regulator has a PWM signal inside, a square wave, toggled between Gnd and U_batt. To convert this square wave to DC voltage, you need a low pass filter, ie. a series inductor and a parallel capacitor. No way to avoid the inductor without power loss.
In the brushless motor regulators there is no inductor, but the motor itself acts as low pass filter. There is no lower smooth DC _voltage_ in the system. There is square voltage on the motor. It is only possible because the coils are inductive in the motor. The _current_ in the coils can be smooth, but not the _voltage_ on the coils, if you use that type on controller.
If you try to filter out the square wavw just by increasing the small noise filter capacitor, the regulator just went wrong - it is against the principle it works.

KreAture
Apr 18, 2004, 09:38 PM
I see what you are saying now mmormota. Must have been blind before.

Well, I'll just have to conseed (spelling?) and go for the coil then. Design is a lot simpler too, I guess I was just so caught up in not using inductor...

Here's something you will probably recognize:
http://www.unixcore.com/~kreature/images/kbec/kbec_2_01.gif

And the layout:
http://www.unixcore.com/~kreature/images/kbec/kbec_2_02.gif

It's just a first run to see how it would look. It might (read probably will) need some more filtering though.

mmormota
Apr 18, 2004, 10:19 PM
I see what you are saying now mmormota. Must have been blind before.

Well, I'll just have to conseed (spelling?) and go for the coil then. Design is a lot simpler too, I guess I was just so caught up in not using inductor...

Here's something you will probably recognize:


Is it the sample circuit from the LM340 datasheet?

The device is probably working.

As the driver IC is not so fast, the switching frequency has to be relative low (compared to the new dedicated switched mode circuits). Because of it, the coil has to store more energy (higher frequency: smaller parts of the same energy transferred more times in a second).
The coil has to be big (the iron saturates in a small coil). The footprint on the pcb is way too small. For this circuit you need a rather big ferrite toroid,
maybe bigger then the entire pcb.
Not easy to calculate the proper size, better to follow the factory recommendations - they usually offer coils for their schematics.

If you like to minimize the coil size, search for a modern switched mode driver, they are working above 200kHz. Maxim, Linear Technology, Texas and a lot of others are manufacturing good ICs. The pcb can be tricky too, because of the high current and the high frequency (pcb inductance comes to the game), better to start with the sample pcb pattern from the datasheet.

When I was about to design a new power supply, first allways built the sample circuit without modifications. Then tested it with a scope at different loads. It was a good reference, when I modified the circuit for the particular requirements, I allway used the sample as a reference.

Capacitors: first choice should be the new high capacity ceramics, they are available in the 10...100uF range. Alternative solution: a smaller ceramic, and a biger electrolyt, but it must be a low resistance one, designed for switched mode supplies. The higher frequency allows smaller capacitors too.

KreAture
Apr 18, 2004, 10:26 PM
A switched mode driver cost way too much here, and I really want this to be a slap-together for a buck thing...

The FET's are actually small signal FET's but the first version will only be for 4-600 mA or so. The small FET's are very quick in switching (talking RF here) and they can each handle over 400 mA continous current. The coil can't be large as it would be heavy. I guess high frequency is unavoidable... How high will actually depend on my tiny ferrite cores as they will be the basis of my experiments to come. They fit in the outlined footprint.

I'm rather satisfied with the actual layout though. Got it 1 sided and fairly coarse so anyone should be able to etch it.

mmormota
Apr 18, 2004, 10:34 PM
It is a very simple circuit, check it. If it is working properly - you did it.
If thre are problems - increase the coil first.
P channel fets: Fairchild has some really good types in small footpints, like the FDS6679.
The FDV304 has very low V_gate-source voltage, it seems to be a problem.

KreAture
Apr 19, 2004, 04:17 AM
True mmormota, only reason for using the FDV304 was I had a bag of them here.
Any other P-fet and I'd have to go buy one.

After I test and find what needs to be adjusted, I can get some really good FET's.

KreAture
Apr 19, 2004, 06:13 AM
Well. I tested the circuit now and it works just as it's supposed to :)
I stuffed on a small toroid from a broked ATX PSU (approx 1/3" diameter 24ga wound) and fired it up with 10v input.

It delivers 240 mA at 5v now and the circuit is oscillating nicely, although at 280 kHz!
The frequency needs to be reduced. I think I might do that by adding some resistors to provide hysterisis to the LM circuit. That would increase ripple though, would it not ?

KreAture
Apr 19, 2004, 06:14 AM
Oh, and the ripple at 240 mA is 30mv.

mmormota
Apr 19, 2004, 06:37 AM
Well. I tested the circuit now and it works just as it's supposed to :)
I stuffed on a small toroid from a broked ATX PSU (approx 1/3" diameter 24ga wound) and fired it up with 10v input.

It delivers 240 mA at 5v now and the circuit is oscillating nicely, although at 280 kHz!
The frequency needs to be reduced. I think I might do that by adding some resistors to provide hysterisis to the LM circuit. That would increase ripple though, would it not ?

Why needs to be reduced the freq? The drive signal on the gate seems to be pour, and the fet heats up, that's why? (Far from square shaped?) If it is not the case, the signal looks good, the fet is cold, then the high frequency is an advantage. For a given output power, the higher the frequency the smaller coil iron do the job. (I never built this very circuit - have no experience with it. I always built supplies with dedicated switched mode ICs.)

The hysteresis of course increases the ripple. Without hysteresis resistor, the frequency determined by the speed of the analog stab and the series resistance of the ouput filter capacitor.

30 mV is pretty good without additional LC filter.

Test it at the required maximum load current too - to be sure that the iron not saturating at higher current.

KreAture
Apr 19, 2004, 01:37 PM
Since the test-version worked so well I have prepared the PCB for the final beta version.

http://www.unixcore.com/~kreature/images/kbec/kbec_2.1_01.gif

The FET is a DPAK sized IRFR5505 and the coil is mounted on back of PCB. (Then the entire thing can be shrinkwrapped for protection.

I will see how the drive signal looks with the correct FET and see if it needs a lower frequency. Your point is ofcource correct and high frequency is also easier for the servos to filter out with their internal capacitor.

I have tested it a bit with transients and high loads but so far mostly in simulator as I don't have the correct FET yet. I expect 1 amp transients to be fine. For now I won't need too high currents.

Later, one might add a second LM stage and use a 6 or 7v LM for first stage... That would allow the switching noise (that needs to be there for the circuit to function) to be removed from the output.

KreAture
Apr 19, 2004, 04:41 PM
Here's the PCB all ready for it's FET and accompanying "linear-point resistor":
http://www.unixcore.com/~kreature/images/kbec/kbec_2.1_02.jpg

And the back:
http://www.unixcore.com/~kreature/images/kbec/kbec_2.1_03.jpg

With the current input cap of 33uF 16V it is limited to less than 16v, but the FET has an upper limit of 55V. I am so looking forward to getting my FET's tomorrow :)

clipclop
Apr 19, 2004, 05:04 PM
Kreature ,
its looking good .
Dont forget to test for RFI with your RX's . you may need to put a sheild over the inductor .
Stewart

KreAture
Apr 19, 2004, 05:27 PM
Yeah. I'll have to check carefully for that. I notice the Ultra-Bec sold at many stores do not have any shielding. Maby the frequencys are too low ?

vintage1
Apr 19, 2004, 05:37 PM
You will get more problems from the sharp edges inside the thing than anything else. Its advisable to put suppressor caps across the leads in and out exactly where the wires come in, with very short leads. Screening should not be necessary. Watch out for osciallting switching FETS too - in general 10-100ohms in series with the gate stops them hooting during switchover.

If screening IS necessary I'd use double sided board and make one side ground - where the coil is.

Then take all suppressor capss to that directly.

It looks very small and neat.

Nice work.

KreAture
Apr 19, 2004, 06:05 PM
Thanks vintage1.
As you can see I use rounded tracks not angled. This should help on some noise from the switching. Also, I will be trimming the edges of the PCB to not cut into the shrinkwrap.

The whole thing weighs less than 1 gram. (1/28 oz)

mmormota
Apr 19, 2004, 07:55 PM
Nice work.
The toroid has inherently small magnetic radiation, I agree with Vintage, the sharp edges are the main noise sources. Unfortunately it is practically impossible to suppress this radiation using shieldings (heavy, complicated etc). But usually the noise is very low at the high frequency range - most probably it will work without rfi problems. The motor controllers are very similar devices, working on higher power level without shieldings - and no rfi problems.

mmormota
Apr 19, 2004, 08:03 PM
Usually better to avoid Ta elco-s in switched mode power supplies, the ceramic capacitors are more reliable. But in a DIY device it is not so important as in high series production.

KreAture
Apr 19, 2004, 08:12 PM
true mmormota, I just slapped something on there. It was quicker than having to make a symbol in my editing software for a side-mounted electrolytic. It should have both a electrolytic and a cheramic for best noise-suppression though.

KreAture
Apr 20, 2004, 11:34 AM
My FET's are here!

And things are not as good as I hoped :(

Somhow I am having trouble getting it to oscillate as it should. (Again?)
This time it seams to be the operating-point of the FET that is the problem combined with the input-voltage. At only 10v input, it appears to need a 1.2 kOhm resistor as R1 to keep it oscillating. It keeps a 240 mA load going without heating the FET more than you can touch it, but it all seams to loose it if I increase the load.

KreAture
Apr 20, 2004, 11:52 AM
Yey! It is the input-voltage that is the culprit... Ofcource, this makes sence! The turn-on voltage of the FET dictates the minimum voltage!

My FET has a threashold of about -2.4 to -4v and my regulator needs 1 volt to be happy. This means I need 5+4+1v minimum if I want to use the FET at full saturation.

What I need is a FET with a lower threshold...

The good news is, the circuit handled 240 mA easily at 12v input (s3 LiPo) and the FET didn't even change temperature... At 500 mA it runns equally cold. No problem.
I had to try it, so I stuffed on a load of 750 mA... Running cool and happy at 750 mA with a switching frewuency of 6kHz with a 1000 uF cap on output. The ripple was 60 mV!

vintage1
Apr 20, 2004, 05:28 PM
You might be able to bootsrap the FET gate somehow. That's a neat trick to increase voltage on the gate wehnuseing N channel to regulate positivce volts.... I can't see cqt. diag. on this page to suggest how..

KreAture
Apr 20, 2004, 07:13 PM
Vintage1, maby you're right.
It would be easier to use a FET with a -0.7v gate instead of one with -2 to -4v but it would be a SOT8 instead of a DPAK. It must be this threashold that is causing this, as the gate would need to be pulled to atleast -2v and the regulator is doing th e pulling with it's input pin. This in turn is in need of atleast 1 volt higher than the output voltage. 5 + 1 + (2 to 4) is 8 to 10v and I have had problems getting it to oscillate at 10v. Also, at 12v it needs a considerable resistor to do so.

The schematic is very simple and looks like this:
http://www.unixcore.com/~kreature/images/kbec/kbec_2.1_04.gif

The incredible simplicity is the reason why I really like this design. A small p-fet that pulls down a N-fet would possibly allow better N-fet's to be used, but if a FET with low enough threshold is used, this should not be necessary.

The circuit in the schematic would be limited by the voltage-rating on it's input capacitor, FET and transient diode. Selecting theese within the max voltage one needs is simple, but for low input-voltages it becomes tricky...

To make such a circuit for 6v use, one merly uses a 6v equivialent type of linear regulator to do the error amplification.

KreAture
Apr 20, 2004, 07:52 PM
Update!

I replaced the FET (2-4v threshold) with a power transistor (working on current, not voltage) and now the circuit likes 8-cell packs :)

I can run it at 250 mA, 500 mA and 750 mA continous loads without any noticable difference on both 8 cell and 3s LiPo.

I'll just have to let it run down the pack now to see how low it will go :)

mmormota
Apr 20, 2004, 07:55 PM
A bipolar transistor instead of the Fet solves the problem.

mmormota
Apr 20, 2004, 07:56 PM
Wow, I late a few seconds...

KreAture
Apr 20, 2004, 08:34 PM
Yep. The circuit was originally designed for a bipolar and that is what I am now using. I was hoping to use a FET instead to lower switching-losses, but the circuit is more than powerfull enough for my needs with a regular transistor.

I am now running tests to see how well it holds up while Vin drops...

Input is now 9.78v, switching frequency at 30 kHz and output nice and stable at 5v +/- 20 mV. Load is 500 mA continous.

mmormota
Apr 20, 2004, 09:17 PM
The device will be very useful in powerful aeroplanes, when the battery voltage is high, and the controller is wihout BEC (too much power for a normal analog stab).

KreAture
Apr 20, 2004, 09:35 PM
Done testing voltage limit!
It can keep working down to approx 7.25v after that oscillation collapses. It still continues to try and deliver and can still deliver 5v at lower currents. (100-200 mA)

I'm very happy. over 750 mA and down to 7.25 was way better than I'd dare hope. I was guessing approx 7.5-7.6v but never in my wildest would I have guessed 7.25v!

KreAture
Apr 21, 2004, 07:31 AM
Some test-results:
8 NiMh: 9.96V 0.52A 5.1792W IN, 5v 0.76A 3.8W OUT = 73% eff
3s LiPo: 11.55V 0.45A 5.1975W IN, 5V 0.76A 3.8W OUT = 73% eff
10 NiMh: 13.24V 0.4A 5.296W IN, 5V 0.76A 3.8W OUT = 71.8% eff

Cool ! :D

clipclop
Apr 21, 2004, 08:05 AM
Kreature,
Nice work !! pretty good efficiency to for a small package .
Stewart

KreAture
Apr 21, 2004, 11:05 AM
If I could get it to run a FET instead of the transistor I am sure eff can be raised 10% or so...

Comatose
Apr 21, 2004, 11:54 AM
Maybe a bipolar transistor driving a fet?

KreAture
Apr 21, 2004, 12:25 PM
More like a FET-driver :) The faster (harder) switching, the more eff circuit.
However, 1 amp cont is more than enough for my needs at the moment.

vintage1
Apr 21, 2004, 01:12 PM
Er..R1 seems to be WRONG. You need about 100 ohms in series with the FET gate, and about a couple of K between gate and source.

Otherwise you are asking the chip to drive an awful lot of current to pull the gate down.

I did not realise you were using P channel FET. Ah. And an analogue regulator in self oscillating mode :)

What a glorious bodge! :D

Seems to me a zener and an opamp might do the same trick as well.

KreAture
Apr 21, 2004, 03:03 PM
The R1 there was chosen for a 0.7v threashold FET.
Obviously it got wrong when I used a 4v one.

The current required if it is a PNP power-transistor would be right on the spot. Well actually, 56 Ohm is what I use now to trigger at approx 10-15 mA.

The system is actually quite brilliant and works very well. It should be just as good with a FET if I could get the right one. Also, the FET's can take a lot of current in spikes. Thus using no series resistor should be fine even for a FET.

Hooters Driver
Apr 21, 2004, 11:49 PM
KreAture,

Can you give us a parts list (preferrably with part numbers for components used) for your final design.

KreAture
Apr 22, 2004, 09:13 AM
Here's the new version using a SO8 packaged -0.7v threashold P-FET.

http://www.unixcore.com/~kreature/images/kbec/kbec_2.2_01.gif

It should accommodate a aluminum heatsink over the REG and FET but it might need a cutout for the diode. This should further improve max current :)

Will test this layout and the new transistor tomorrow. Have to go visit gramps today :)

Partslist for those interested:

Part Value
C1 47u (30v capable)
C2 470u (minimum 5v capable)
D1 SM6T30A (30 V capable transient diode)
FET1 IRF7404 (0.7v threshold 20v capable. This limits max V-in to 20v.)
L1 Coil. Preferably toroid, atleast 5-600 uH, mounted on back of board.
R1 30 to 100 Ohm, 0805 size
REG LM340MP-5,0 (Or any pin-compatible 5v linear reg.)

FET1 can be replaced by any POWER-MOSFET transistor and will work well. Using a FET is simply an attempt to further improve eff and increase max current.

KreAture
Apr 22, 2004, 12:03 PM
I had some time before leaving and have etched the new layout, populated and tested it :) Works great!

It oscillates as it's supposed to all the way down to 8.5v at 750 mA load with no apparant heating beyond body-temp. A regular transistor is still better for low input-voltage, but this circuit is for those occations when a regular BEC is useless, not for wimply 8v input.

KreAture
Apr 22, 2004, 12:19 PM
Wow. It turns out the transistor version is more effichient ? Wierd.
The FET version has lower resistance in the switching and just a tad higher threashold?

Maby the FET-version will be more eff at higher currents. I did notice a lower heating of both input cap and FET when I increased load from 250 to 750 mA. I may not have reached max eff yet? I'll try 1-5 and higher hehe :)

Hooters Driver
Apr 22, 2004, 02:02 PM
Thanks for the parts information. I have a few of those FETs laying around so the new version would be even easier for me to build.

I'd be interested in seeing the results of your higher amperage testing on the FET version. How much less efficient did the FET version appear to be with the lower current draw?

KreAture
Apr 23, 2004, 05:19 AM
I measured an eff of 63 to 65% at 750 mA load. At 250 mA it was approx 61%.

I'm sure the eff would increase as I move over into a higher current-range. The important part here is to realize that low eff on low amps drain is safe, as the power is low anyways... We'll see when I get home. I'll increase amps untill it gets hot, stops oscillating, or ripple becomes unbearable. Wichever comes first hehe :D

KreAture
Apr 27, 2004, 06:56 PM
I made a new thread for my project due to the change in nature of it...

http://www.rcgroups.com/forums/showpost.php?p=2088169&postcount=1

It's gotten quite nice now :) 74% eff :)
With a high output opamp, FET and some sort of reference I'm sure it could be much more effichient as switching could be sharper with lower losses in transistor/FET.