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## Re: [TowerTalk] SteppIR problem (K7NJ)(K4SAV)

 To: towertalk@contesting.com Re: [TowerTalk] SteppIR problem (K7NJ)(K4SAV) K4SAV Tue, 12 Jun 2007 08:55:30 -0500
 ```R. Kline wrote: ...."It follows from the above that if the amplitude of the current pulse increases, the average current will remain unchanged if the length of time of the current pulse is decreased. Conversely, if the amplitude of the current pulse decreases, the average current will not change if the length of time of the current pulse increases...." You have almost got it, but you didn't list the other corollaries. One is: if the amplitude of the current pulse is held constant and length of time for the current pulse increases, the average current increases. That is exactly what is happening. This operates a lot like any other switch-mode system. Many of them trade current pulse width for average current. Even though the switch may be in series with the output, the peak and average currents in different parts of a series network may be completely different (although you do need some reactance in the network to make that happen). So why does the pulse width need to change? The regulator is going to hold the switch on until the current reaches a preset value. For the system it is driving, the time constant is = LR. If you increase either the inductance or resistance the time required for the peak current to rise to the preset value increases. Now see the previous paragraph. This causes the average current to increase, however the peak current in the system is still the same. The peak current is the same because the regulator IC always turns off when that peak is reached. And yes, this does bring up another possibility. If the motors have more inductance than normal, it can produce the same problem. This should all be part of the engineering for the system. Analyze all the contributors: motor inductance and resistance, cable resistance, capacitance, inductance and transmission line effects, power supply regulation and current capability both static and dynamic, driver IC peak currents and ability to sink transients. Then do some testing to verify that you haven't made any mistakes or overlooked anything in the analysis. Then you will know the capability of the system and you won't discover it when the end user exceeds the limits of the system. You will know what those limits are and it should be easy to identify which limit has been exceeded. I don't know if this was done or not, but if not it should have been. This is also why it is difficult to engineer anything by committee, especially re-engineer anything after the fact. If you download the spec sheet for the driver IC (L6219DS made by Allegro) you can see a little more about how the driver works and regulates the current. Briefly, after the current reaches the preset value the switch turns off, and there is a predetermined wait period before it is allowed to turn back on. This operation continues to cycle at a fast rate to maintain the current to the motor for a time period determined by the logic driving it. After the logic turns the control pulse off, the motor discharges its stored energy back down the cable into the clamp diodes inside the IC. Jerry, K4SAV R. Kline wrote: >R. Kline wrote: > > > >>There has been mention that the source is a current source, and depending >>upon the length of the control cable, the current would increase, etc. >> >> >This > > >>couldn't be correct, since the load (motor), cable resistance, and source >>are all in series - so the current would be the same everywhere. What may >>be happening is that the voltage increases in an attempt to keep the >> >> >current > > >>constant. Perhaps the supply "senses" the voltage output and shuts down if >>it exceeds a predetermined maximum voltage. >>73, >>riki, K7NJ >> >> > > > >Apparently you didn't read my explanation, in answer to WC1M's question >about how this works. Here it is again. > >An interesting observation. One can easily make the mistake of saying >that the current to the motor is regulated so the supply current doesn't >change due to the extra resistance, because the motor current doesn't >change, but this is not exactly true, as you are alluding to. > >Think of the load as an RL network. The pulses delivered to this load >are both width and frequency modulated, in other words they change to >provide a fixed amount of average current to the motor. So for an >increased resistance the pulse width has to be wider than that for a >lower resistance. This is because the rise time to this network will be >slower. Looking back at the power supply this requires more average >current from the supply because of these wider pulses. (The rep rate of >these pulses also changes but that's just another detail.) However the >peak current required from the supply is the same in both cases, because >the regulator IC shuts the current off when it reaches a preset value. > >It is this max peak value that is independent of the amount of >resistance, not the average current from the supply. > >That does raise the question of what the power supply's capability is, >in terms of peak and average current capability. It is possible that >the power supply filter cap can deliver the high peaks but that the >supply itself cannot sustain the higher average current required. > >Jerry, K4SAV > > >-------------------------------------------------------------------------- > >I've tried reading your message over and over, but just can't follow it. I >get the feeling that you know what you're talking about, and I would like to >understand what you're saying. > >Let's return to basics. Without getting involved in a lot of mathematics, >the equivalent graphical definition of average current for one period of the >waveform would be: >(1) Graph the current vs time >(2) Determine the area bounded by the current graph (step 1) >(3) Divide the area (step 2) by the elapsed time from the beginning to the >end of the waveform. > >It follows from the above that if the amplitude of the current pulse >increases, the average current will remain unchanged if the length of time >of the current pulse is decreased. Conversely, if the amplitude of the >current pulse decreases, the average current will not change if the length >of time of the current pulse increases. > >Please, then, explain the following points to clarify your previous >explanation: >(a) Why is the rise time "...of the network..." slower? > >(b) Why does shutting off the current result result in the SAME REQUIRED >peak current ? > >(c) Assuming the same average current, why are longer current pulses >required? > >I get the impression that you've put in place many of the pieces to the >puzzle, but the logical connections between them are somewhat fuzzy. > >73, > >riki, K7NJ > > > > > > > >_______________________________________________ > > > >_______________________________________________ >TowerTalk mailing list >TowerTalk@contesting.com >http://lists.contesting.com/mailman/listinfo/towertalk > > > _______________________________________________ _______________________________________________ TowerTalk mailing list TowerTalk@contesting.com http://lists.contesting.com/mailman/listinfo/towertalk ```
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