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RE: Impedance matching batteries
Spent most of my life working with tubes. Try a 2.5 GW (that is 2500 MW) amplifier chain. One hundred Machlett ML8618s running 35 Kv plate pulsed to 800 A plate current each. Filament power 2.5 kW each.
Did the dump thing as a kid also.
Brings back memories!
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LScamper
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12/08/12 10:07am |
Tech Issues
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RE: Impedance matching batteries
Output cap ESR importance in constant current mode?
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LScamper
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12/06/12 10:26am |
Tech Issues
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RE: Solar Anxieties-
http://www.reuk.co.uk/Measuring-the-Power-of-A-Solar-Panel.htm
http://resources.solmetric.com/get/Guide%20to%20Interpreting%20I-V%20Curves.pdf
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LScamper
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07/30/12 11:29pm |
Tech Issues
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RE: Solar Anxieties-
BFL13 wrote:
“I am feeling especially stupid so need more elaboration, please.”
I feel your frustration. I admirer someone that ask questions when they don’t know the answers. I spent most of my life working with people like that. They asked the hard questions, they are called scientist and a lot of them tried to explain things to me that are far over my head.
I’ll try to shed some light on some of this. Some may laugh but I don’t care.
First Ipmax and Vpmax mean nothing unless you are using a MPPT controller, forget about them!
This may sound silly but I’ll try to use a car engine to explain some of it. Voc is like the engine running at max RPM out of gear, it makes noise (Voc) but does nothing. Isc is somewhat like the engine just as you stall it when pushing against a wall; max torque but you go nowhere. You use your engine in between these points. You use your panel in between Voc and Isc.
If you use a MPPT controller then this applies. Ipmax and Vpmax are the points that the panel puts out maximum power (maximum watts), the point that your engine puts out maximum horsepower. That is the point that torque time RPM is at maximum. Then there is this magic box – the MPPT controller. It converts the maximum power to the maximum current that your battery will accept. Sort of like your transmission converting engine power to maximum torque at the wheels.
Hope this helps.
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LScamper
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07/28/12 10:32pm |
Tech Issues
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RE: Voltage drop for AC whats acceptable?
wa8yxm wrote:
"It is possible, with a bit of electronics, to design a motor that would work and work well, at low voltages, say 90 or even less, and then with an electronic controller make them so they could survive an over voltage, say 150 or even 250 volts.
The cost of that controller is in the 10 dollar range."
I always like to learn. Could you elaborate on this some. $10 seems like they would do it if it was that easy.
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LScamper
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07/27/12 06:04pm |
Tech Issues
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RE: Voltage drop for AC whats acceptable?
There is no reason it would not work. The total rating would be the total rating of the transformer, that is a 500VA transformer will handle 500VA in each case. Each primary will carry the same current as the load. A 500VA transformer running at 120VAC with the primaries in series adding will provide 240VAC at 2.08A to the load. The source current will be two time the 2.08A or 4.16A. VA = 120*4.16 = 500VA and 240 * 2.08 = 500VA. There will be a little more current from the source due to magnetizing inductance of the primaries.
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LScamper
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07/26/12 04:10pm |
Tech Issues
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RE: Confession-How I Broke My Controller
Can’t believe that no one has taken a shot at this yet. So I’ll try to shoot myself.
First at this price the controller must be a pulse width modulated type. A fast review of what I think that means. When the controller is on the panel is essentially connected directly to the battery. When it is off it could be one of two different things depending on the type of controller. A series controller will disconnect the panel from the battery. The input voltage to the controller will be the open circuit voltage of the panel. A shunt controller will short the output of the panel to ground. The input voltage to the controller will be about 0 volts and the current from the panel will be the short circuit current. When the battery voltage drops below the lower set point the controller turns on. When it reaches the upper set point it turns off. This will continue as long as there is sun. I don’t like to call this a pulse width modulated type of controller. A more descriptive name would be a bang – bang controller. It is either on or off, no in between.
OK. The first and most important reason you cannot just connect them in parallel. The set points will not be exactly the same. This means that one controller will always be on when the other is off. In this condition it will have to handle the full power of the panel.
The second reason is a little more technical. Even if you could get them to switch at the same time the resistance of the two controllers will not match exactly. This mismatch will cause unequal currents. One controller would handle more power than the other.
Both of these problems can be over come if you designed a new controller but that is not what we are doing!
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LScamper
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07/22/12 07:15pm |
Tech Issues
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Of Interest To Electrical Engineers and Technicians
Just came across this. I spent many hours reading technical articles written by Jim Williams and Bob Pease. Learned a lot from them. This brought back many good memories from my working days. I suspect it will for some of you too.
http://www.edn.com/design/analog/4375963/Slideshow--Remembering-Jim-Williams-and-Bob-Pease?page=0
Moderator Note:
Made link clickable
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LScamper
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07/17/12 11:13pm |
Tech Issues
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RE: Water Pump is going backwards...
Put a cap on the water inlet, problem solved.
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LScamper
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07/16/12 03:53pm |
Tech Issues
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RE: Connecting Charger Diagonally on Multiple Batteries?
pianotuna wrote:
"I was not aware that 0.0015 of an ohm (or was it 0.015--working from memory in a low bandwidth location) was considered a "high resistance" connection."
Smartgauge uses 1.5 mOhms per link in their calculations. They use about 20cm or 8 inches of 35mm cable with a resistance of about .12 mOhms plus .2 mOhms for each of four termination interfaces on each link. I think that their termination resistance is reasonable. I add that up to .12 + .8 = .99 mOhms. 35mm cable is between AWG 2 and 3. Not sure how they get 1.5 mOhms to use in their calculations, the higher the resistance the worse it looks.
If you use eight inches of 2/0 cable with a resistance of about .078 mOhms and you use their termination resistance that would be about .88 mOhms per link or .88/1.5 = .59 of the resistance they use in their calculations. If you use these cables there will be much less difference with using an unbalanced connection.
High resistance is a relative term in this battery stuff. A 100AH battery has an internal resistance of around 20 mOhms. If the batteries are different by 1 or 2 mOhms that will have an equal or greater impact on balance than the interconnects.
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LScamper
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07/05/12 10:31pm |
Tech Issues
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RE: Connecting Charger Diagonally on Multiple Batteries?
The information on Smartgauge site about balancing is correct! However, there use of HIGH RESISTANCE interconnections make it appear much worse than it really is. If you use the correct interconnect wire size balance becomes a moot point. If you do a perfect job at balance there still is little chance that the system is really balanced. The reason is that the differences in battery internal resistance have more effect on balance than the low resistance interconnects.
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LScamper
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07/05/12 12:19pm |
Tech Issues
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RE: Converter to change modified sine wave to sine wave?
dclark1946 wrote:
"As you go up in frequency a signal seeing a high load impedance (compared to the source impedance) is reflected and this effect is quantified as the reflection coefficient. At low frequency such as 60 Hz you can think of the reflection as a load impedance that increases with frequency. This means the high frequency components don't see the same load as the fundamental which will cause these components to have a higher amplitude relative to the fundamental at the source because these frequency components are seeing essentially an open circuit load. This is similar to an RF impedance mismatch which results in reflected energy or a poor return loss ie. high vswr"
I somewhat agree with what you are saying. At resonance the filter acts as a lump sum transmission line and with an open output the voltage is reflected in phase thus doubling the output voltage. This voltage does come back to the source and is quantified as the refection coefficient (S11 anyone). A shorted output reflects the voltage 180 degrees out of phase and the output voltage is 0. I think we can agree on this.
The part I'm not so sure of is at higher frequencies. The filter is no longer in resonance and I don't think it will act like a transmission line just as an extension cord plugged into the inverter does not act like a transmission line. If it is an inductor input filter as the frequency goes up the the input impedance goes up as you say. But the source impedance is low and the filter no longer acts like a transmission line so I don't think this means the high frequency components will have a higher amplitude relative to the fundamental. A capacitor input filter will do the opposite. The impedance of the higher frequencies will be lower.
What I can see is that the inductor stored energy will have to go somewhere, back to the source I guess.
Either way I don't see this as a problem. The inverter output will handle this energy. Every time something is plugged into it there are reflections or energy coming back. A transformer with no load on it has a magnetizing field that stores and returns energy to the inverter without hurting it (unless it saturates from the square wave). A power supply generates high frequency reflections when the diodes turn on, this does not hurt the inverter.
The bottom line, I think as others have said, is that a filter is not a reasonable thing for this application. Buy a true sine wave inverter if you really need one.
I find this thread interesting but it has gotten way away from anything most people will be interested in. Hope I did not give too much bad information, have not thought about this stuff in many many years.
So go at what I have said or maybe we have answered the original question!
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LScamper
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07/03/12 10:41pm |
Tech Issues
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RE: Converter to change modified sine wave to sine wave?
Clay L wrote:
"In my case it was 40 years ago but I don't recall the rejected frequencies being reflected either. I think they just see a high impedance and don't get through the filter."
I think it depends on the type of filter, capacitor or inductor input. For the odd order harmonics the capacitor input case I think the inverter sees a low impedance and will supply current and with an inductor input it sees a high impedance and will supply voltage. So depending on the filter the reflections will act differently. In either case I don't think much energy is reflected. I maybe all wet about this.
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LScamper
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07/03/12 02:56pm |
Tech Issues
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RE: Converter to change modified sine wave to sine wave?
I'm sure I'll be flamed for this but I'll try to remember things from 30 years ago.
First a low pass filter is what is needed. There is no need for a high pass filter since you don't want the higher frequency components in the output. Remember that a bandpass filter is just a low pass and series high pass filter. No need for the high pass part.
From what I remember? When designing you normalize to some standard impedance and frequency, don't remember what they are. The shape of the filter response is dependent on the type of filter not the frequency, Bessel, Butterworth, or whatever. The design is DEPENDENT on the source impedance and the LOAD impedance. This is a problem since the load impedance in this case can be anything!! The design changes as the load changes! Have fun with that.
Don't really remember about the reflected voltage but I think that as long as the inverter can handle the reflected voltage all is OK. Power is not dissipated in the inverter. Could be very wrong about this. Like back in the days of tubes. On old radios if you disconnect the speaker some times you could draw an arc across the output transformer primary leads, maybe the same?
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LScamper
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07/03/12 10:46am |
Tech Issues
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RE: Converter to change modified sine wave to sine wave?
Depending on the Q of the filter which is dependent on the load you can get most any voltage that you want. For instance take a series resistor, inductor, and capacitor that is resonant at 60 Hz. Put it across the output and the voltage across the capacitor and the inductor will be a sine wave and try to go to infinite voltage as the resistance is reduced! DON'T TRY THIS IT MAY BLOW UP! Used this method many times to charge capacitor banks.
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LScamper
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06/30/12 12:06pm |
Tech Issues
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RE: Kill-a-Watt on modified sine wave inverter vs line power
For information only. Not trying to steal thread.
while not a Tek scope (I used Tek for over 35 years) the Rigol DS1102E 100MHz or the DS1052E 50MHz Digital Oscilloscopes are amazing for the price, about $350 > $400 new. I have the DS1102E.
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LScamper
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06/26/12 09:06am |
Tech Issues
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