Author Topic: pwm (Read 3320 times)

willib · « **on:** March 09, 2006, 04:15:35 AM »

I just simulated this pwm circuit .

it seems to work .

The pulse source , drives the gate of the FET, which turns on and off the current going through the battery.

V1 is simulating a three phase DC output voltage from a generator.

i gave the battery a internal resistance of .001 ohms.

and the generator a resistance of .053 ohms .

the diodes are models of real ones , so is the FET .

the FETs can be paralleled for more power if needed.

ideally you would monitor the rpm of the machine and vary the current to give you a constant rpm..

http://www.otherpower.com/images/scimages/2965/pwm_pulse.JPG

http://www.otherpower.com/images/scimages/2965/pwm_50mS_pulse.JPG

http://www.otherpower.com/images/scimages/2965/pwm_10mS_pulse.bmp

Nando · « **Reply #1 on:** March 08, 2006, 10:15:06 PM »

Willib:

Question 1) what type of Spice are you using ?. - I am interested, can you inform me.

The ckt presented is a simple ckt that basically produces a PWM charging current, if you ADD the diodes plus the generator plus the Mosfet resistances and compare with the battery resistance you may calculate the current plus what is lost in the added items.

What you have, basically, is like a Xantrex C45 or C60 or any of those equivalent charge regulators that when the battery is fully charged start to pulse the battery until the algorithm say stop and do diversion if such is the case.

Nando

willib · « **Reply #2 on:** March 08, 2006, 10:48:29 PM »

i'm using linear technologies , switcher cad III.

commanda · « **Reply #3 on:** March 09, 2006, 01:59:49 AM »

ideally you would monitor the rpm of the machine and vary the current to give you a constant rpm..

Silly question, but why? More rpm is generally more power.

Those of us working in this field with pwm are generally aiming to adjust the pwm to maximise the power out of the generator; MPPT.

Amanda

willib · « **Reply #4 on:** March 09, 2006, 10:12:39 AM »

You are right more RPM equates to more power,to a point though ,there should be some way to keep the stator from overheating and burning out..

I quess i was subconsciously thinking about a varible pitch control mechanism.

again...

How will your MPPT react to sustained high wind conditions?i suppose you would use a dump load .?

What do you think of my circuit?

I was thinking of using a PIC controller to monitor the RPM , in low wind conditions ,and decrease the average current when the blades start to slow down, and increase it when they speed up

Opera House · « **Reply #5 on:** March 09, 2006, 11:07:15 AM »

What circuit?

You have a FET, at least add an inductor so at least some interesting waveform is shown. Monitoring RPM has no advantage over monitoring voltage and current.

willib · « **Reply #6 on:** March 09, 2006, 11:50:17 AM »

lol !!!

after all, a FET is all you need to control the current .

its controlling the FET that is harder.

"Monitoring RPM has no advantage over monitoring voltage and current. "

Not entirely true , if you know your machine you will know the voltage and current at a particular RPM, although some battery monitoring would be a good idea.

SamoaPower · « **Reply #7 on:** March 09, 2006, 01:39:25 PM »

I can't figure out why you want constant RPM unless you want a synchronous machine for fixed frequency AC. By keeping the RPM constant your air rotor TSR will vary considerably with wind speed resulting in quite a rotor efficiency variation.

I applaud your efforts in applying electronics to try and improve the mismatch between alternator and load. This mismatch is the basic reason that brought down DanB's big machine.

We should probably take more lessons from the big boys. I believe the current trend in utility class machines is variable speed alternators with electronic load matching coupled, of course, with variable pitch to maintain TSR.

Keep designing and experimenting!

Opera House · « **Reply #8 on:** March 09, 2006, 03:56:59 PM »

"if you know your machine you will know the voltage and current at a particular RPM"

Big IF. Knowing the V & A means you don't have to know much about the machine.

Bruce S · « **Reply #9 on:** March 09, 2006, 04:28:12 PM »

Willib;

I took an extra long time to figure what you have in mind and I think I have it.

Here goes and forgive if I'm wrong :--) ok

I think what you're wanting to do is by monitoring the rpm which will relate to the amount of power your mill is set to capture. The faster the rpms the more output . BUT too much speed kills mills.

Using the PIC as a monitoring device you could make use of it along with the MPPT .

By default the MPPT is used to get the most possible energy by letting the mill see a lite load. Right so far?

With this in mind you (I think) want to match the best charging possibilities in the widest possible range.

The monitoring could also be used to control how much load the mill sees by use of a dump load?

The varible pitch my not be needed if you plan on making use of the monitoring to load the mill down in dangerose winds.

After looking at the circuit, I do like the simplicity of it, the fewer parts the easier to trouble shoot. Without the use of a FET or zeners it looks very much like a voltage doubler.

The MPPT and "pic" would be interesting to see working alongside each other.

Typically a MPPT is used for solar due to the flucuations in charging and those errant clouds and stuff, but I see no reason why they couldn't be used with mills.

The one item though would be the the "pic" will be drawing power and you'll need to see if it's worth the extra energy needs.

Thoughts?

Or did I just blow it?

Cheers!

Bruce S

willib · « **Reply #10 on:** March 09, 2006, 11:52:18 PM »

all i ment by a constant rpm was that if a machine is designed to charge a 12V battery , for example ,the average machine will become inefficient at charging the 12V battery at a higher rpm , because at double the cutin speed the current will also be double .If your stator can handle twice the current , you're ok..

AS far as the "circuit" is concerned , i was just happy that it worked well in simulation, and it did show the principle of pulse width modulation , which is, the average current can be controlled by varying the width of the pulse of the current going through the battery.

on a side note , i gave the power source the expected internal resistance of one phase of my #10 ga. coils (six coils = 0.05313ohms ) .(of 38 turns)more on that later

i havent finished testing it out yet , but i'm getting some interesting results.

let me give y'all peek at what i'm doing

below is what some might call a dead short , but in reality has resistance.

It is a 6.6" piece of #20 gauge wire soldered across the coil leads..

http://www.otherpower.com/images/scimages/2965/IM001612.JPG 67KB

it has a resistance of 0.0054996 ohms

the coil itself has a resistance of 0.005063 ohms, thats 24 turns of #10 wire a tad over five feet.

scope data.

Vp div RPM rpm / Vp average rpm / Vp

.2 7 71 rpm 355 333

.4 4 125 312

.6 2.5 200 333

the Rpm per Volts( peak ) is a little high but that can be corrected by using 38 turns instead of 24.

the interesting part is that i got a voltage at all !!

V/R = I and R = coil Resistance + load resistance = 0.01056 ohms

so at 200 RPM .6 / .01056

57Amps (peak value)

the rms value is 40 Amps

needless to say the load gets pretty hot , and the enamel is breaking down

http://www.otherpower.com/images/scimages/2965/IM001614.JPG 97KB

Opera House · « **Reply #11 on:** March 10, 2006, 06:59:21 AM »

Have you thought about the inductance of the generator? If you look at the curves of an alternator, you will see that the voltage peaks with increased RPM and then drops. If you are going to worry about 0.01 ohm, you better think about inductance too!

willib · « **Reply #12 on:** March 10, 2006, 01:43:24 PM »

I appreciate your reply.and all replys , sorry i didnt respond earlier.

"By default the MPPT is used to get the most possible energy by letting the mill see a lite load. Right so far?"

i'm not really sure , is that what MPPT does?

i am by no means an expert on MPPT.

"The MPPT and "pic" would be interesting to see working alongside each other.

Typically a MPPT is used for solar due to the flucuations in charging and those errant clouds and stuff, but I see no reason why they couldn't be used with mills."

it would be interesting , i have a PIC16F88 all set up for PWM , it is built into the hardware , as is, the analog to digital converter.

so when i turn a varible resistor the PWM varies from no pulses to full on.

i was using it to vary the current to an LED , but it can also be used for this purpose.

the outup from the PIC is 5V at 25mA source and sink.this means when the output is high it can source 25mA and when it is low it is also able to sink 25mA..

so it should be perfect when driving a IRF 3703 FET which has some very low "on" resistance ( i forget offhand)but it is in single digit milli Ohm range..

SamoaPower · « **Reply #13 on:** March 10, 2006, 04:50:54 PM »

willib,

I think you'll find that the PIC output isn't adequate to drive the gate of a low Rdson power FET directly. You'll need an additional driver stage. The problem is that typical power FETs with low Rdson have a rather large gate input capacitance (3-4 nf) which requires a large peak current to switch fast. That input C has to be charged quickly to keep the switching losses in the FET reasonably low. You probably need peak current capability in the order of 2-6 amps (FET dependant) to switch fast enough and I doubt the PIC could supply this. The driver average current capability doesn't tell you much - it's the peak current that's important. A bipolar complementary emmiter follower stage will usually do it.

willib · « **Reply #14 on:** March 10, 2006, 06:40:15 PM »

Hmmm bad news i suppose, well since i allready have them i guess if thats what it takes ,then thats what it takes..they list the gate charge in nano coulombs,insert confused smiley here.

this is part of the data sheet for the irf3703..

http://www.otherpower.com/images/scimages/2965/gate_capacitance.JPG

this is the whole data sheet..

http://www.irf.com/product-info/datasheets/data/irf3703.pdf

dinges · « **Reply #15 on:** March 10, 2006, 06:45:28 PM »

Willib,

I think you are getting 'charge' confused with 'capacity' here. One is in Coulomb (C), the other in Farad (F, usually though pF or nF)

Peter.

willib · « **Reply #16 on:** March 10, 2006, 06:52:19 PM »

they dont list gate to source capacitance , insert another confused smiley

dinges · « **Reply #17 on:** March 10, 2006, 09:35:57 PM »

I think Cgs is equal to Ciss (input capacitance), being 8250pF?

Actually, that looks like a lot to me, 8nF... Depending on the impedance of your driving circuit, it may definitely be something to take into account. Had never realized myself that these MOSFets had such big input capacitance. Despite that, they're even used in HAM amplifiers for shortwave (up to 20MHz, that I know of), so it should be possible to work around this.

Peter.

commanda · « **Reply #18 on:** March 11, 2006, 01:10:13 AM »

There's a lot more to the dynamic internal impedance of a generator than just the dc coil resistance. There's the inductance and a whole complex can of worms to do with magnetic paths & air gaps. Short story is it increases with both rpm & current, and is usually orders of magnitude greater than just the dc resistance. There are some on here who can do a fairly good mathematical approximation given enough information; me, I'd rather run it up with different rpm & different loads & measure it.

Amanda

willib · « **Reply #19 on:** March 11, 2006, 10:20:15 AM »

this is all frequency dependent .. , sure it takes more current to drive a capacative load ( the FET gate) at very high frequencies , but my pwm is only switching at 1KHZ , so i dont think that it is as much of a problem .

SamoaPower · « **Reply #20 on:** March 11, 2006, 03:12:43 PM »

Can't agree willib. What we're talking about here are the switching transitions themselves, not how often they're happening (although this does bear on the average loss). While the gate voltage is changing from low to high (or high to low) the FET drain current is going through a linear region where the drain to source voltage is not near zero. This results in apprecable peak power being dissipated in the FET at each transition. In order to minimize this loss we want to accomplish the transitions as quickly as possible. The less peak gate drive current available, the slower the transition. The higher the input capacitance, the greater the current required to switch in a given time. What this is saying is that the driver needs a low source impedance. This issue applies to turn off as well. It's necessary to dump the gate charge quickly which means a large current sink capability in the driver as well. It's pretty much a symetrical problem.

Like I mentioned earlier, a bipolar complementary emitter follower usually does the trick for a driver. It only takes two small transistors. I often use a 2N4401 NPN and a 2N4403 PNP. Many others will work as well.

SamoaPower · « **Reply #21 on:** March 11, 2006, 04:53:54 PM »

You might try this. Please use at your own risk.

willib · « **Reply #22 on:** March 12, 2006, 09:38:49 AM »

i got 5A into a .25 ohm load and the FET wasnt even hot, it was just warm ..the load on the other had was extreamly hot.

Please see

http://www.fieldlines.com/story/2006/3/12/5250/58585

as i will be posting my results there shortly.

photo's included

willib · « **Reply #23 on:** March 12, 2006, 11:56:10 AM »

Thanks for the circuit , i've posted some results in my diary .

when running the FET at 5Amps average current it did'nt even get hot.with no heatsink.

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Author Topic: pwm (Read 3320 times)