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awright
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Have you looked at thedata sheet for the tlp741g?
http://www.toshiba.com/taec/components2/Datasheet_Sync//209/4322.pdf
It is a low-current (150ma), moderately high voltage (400v) SCR with a photo sensitive gate that can be triggered by light from the embedded LED with LED forward current up to 10 ma.
A resistor of 27K to 33K should be installed from gate to cathode of the SCR to suppress self-triggering due to leakage from anode to gate.
It can be applied like any SCR except that you trigger it with forward current though the LED rather than by a voltage applied to the gate. The LED terminals are insulated from the SCR terminals for up to 4000 volts. If you don't require isolation from LED to SCR then there's not much point in using an opto-isolated SCR.
The Photo SCR will block current in both directions until triggered by light from the embedded LED. Once triggered it will conduct (conventional) current from anode to cathode until current drops below the holding current of 0.2 ma. If forward current drops below 0.2 ma (and there is no current through the LED) the SCR will "commutate," that is, turn off. Reverse voltage (i.e., SCR cathode more positive than SCR anode) will be blocked with or without LED current.
The photo triac operates like a conventional triac but can be triggered by current through the embedded LED. It differs from the photo SCR in that it can conduct in the "reverse" direction if triggered by LED current.
Applications for either device are similar as those for conventional thyristors except that the photo-thyristors can be triggered by current through the LED. I'm sure youcan find tons of applications for conventional SCRs and triacs on the 'net.
awright -
PWM is not the only way to soft start a DC motor but it is certainly the most efficient way in most cases.
If you just want to soft start the motor and have no servo control requirements, I would think any basic PWM generating IC driven by a ramp and driving a single IGBT or MOSFET of sufficient current/voltage ratings would do the job. I haven't actually done this myself, so do not have design details to offer but I don't think a great deal of sophistication is required.
If you want something that's been designed for you, look at this CANA-KIT <http://www.sparkfun.com/products/9668>, a 24 volt 50 amp basic PWM controller with built-in soft start function. A brief instruction manual (not including dircuit schematic) is offered at that site. $59.
awright -
Sorry to make such an obvious suggestion, but have you asked the manufacturer of the power supply how his unit will handle your load and what precautions to take?
Looks OK to me but then, it's not my $12K.
awright -
While it is true that a neon sign transformer is current limited by design, do NOT assume that you can be casual about dealing with the high voltage. It only takes 10-20 ma through the heart to throw it into fibrulation with possibly lethal results and I believe that neon signs operate on higher currents than this. I presume you do not have a spec sheet on your transformer telling you what the current limit is. In any case, the nominal current for fibrulation is an average. You or your guests may be more sensitive.
Be sure there is no possibility of human contact with the hot secondary lead and that the other end of the secondary is solidly grounded.
I urge you to read this Wikipedia entry on NSTs before proceeding.
http://en.wikipedia.org/wiki/Neon_sign_transformer
awright -
Why is this question coming up again? Did you post the same question on a different forum?
The answer is the same as before. There is nothing at all critical about the SCRs used in this circuit. Any voltage rating will do because you can't find an SCR with too low a voltage rating for a full wave rectified 17 volt supply. Look up the current ratings of the two specified SCRs and find replacements with at least those current ratings. Specifications for both SCRs are available on line. You can find satisfactory replacements for less than a dollar at digiKey, Mouser, and many surplus outlets like All Electronics and Alltronics.
awright -
Using a commercial HV probe like the Fluke suggested by Ken is definitely the best way to go. However, such an accessory must be used with a DVM having the proper input resistance. This is automatically taken care of if you use the probe with most Fluke DVMs. It will definitely NOT work with any arbitrary 30 volt analog panel meter.
The reason is that the probe contains a high impedance voltage divider consisting of an approximately 1000 Megohm series HIGH VOLTAGE resistor (one that resists flashover across the resistor) and a nominally 1 Megohm shunt resistor to ground. The actual values of the resistors form a 1000:1 voltage divider when the shunt resistor is loaded with the 10 Megohm input resistance of the DVM. Loading the probe with an analog d'Arsonval (moving coil) panel meter will seriously degrade the voltage division ratio to the point that you will have no idea of what you are measuring (in the absence of careful experimentation or calculations to determine the actual voltage division ratio).
Look at the spec sheet for the probe here:
<http://assets.fluke.com/manuals/80k40___iseng0900.pdf>
Note that the probe meets specs only if used with a DVM having an input resistance of 10 Megohms +/- 1%.
You can fabricate you own HV probe using an analog meter but you will be messing with potentially lethal voltages. In fact, you can buy relatively low cost HV probes with built-in analog meters. To make your own voltage divider for use with a meter you have on hand you must determine the exact resistance of the meter so that the required resistor values can be calculated. Do not try this with a VOM Ohmeter as you will probably burn out the meter under test.
Rather than me trying to explain all the details and the cautions of working with high voltages, check out the very useful and detailed discussion of high voltage probe construction here: <http://www.repairfaq.org/sam/hvprobe.htm> Incidentally, Sam Goldwasser has collected and created an enormous amount of useful information on the web site linked to above.
Note especially the safety considerations.
awright -
I wouldn't try to switch the secondary current. Switch the primary on and off, as is done in a soldering gun. You will pay dearly for a relay that will work in your secondary circuit. Also, you will be adding contact resistance to the secondary loop where it will affect performance. Remember that you have 12 volts supply in the car, but only a few tenths of a volt in your rig. The voltage drop across any contacts in the secondary loop would seriously affect performance.
You are not going to see anything approaching 1 volt across the tip of a soldering gun. After writing a rambling speculative comment here I decided to just measure the voltage across the tip of my 140 watt soldering gun. It was 0.2 VAC across the tip conductors at the point of connection to the gun (after a few seconds warmup - voltage increases with temperature of the tip) and about 0.3 VAC across the conductor bars emerging from the gun body (indicating a low but finite resistance at the tip clamping nuts which I recently cleaned and tightened).
Remember that the conductors inside the gun are a single loop of heavy copper alloy bar and the tip is a short, heavy copper conductor. You are not going to develop much voltage there. I wouldn't be surprised to find that the tip itself contains a section of some alloy with a higher resistance in order to concentrate the heat where it is useful, but I don'know if that's true.
The required conductor size has a lot to do with your duty cycle. My gun is rated at 120 volts, 1.2 amps primary, or 144 watts and I measured 0.3 volts across the gun conductor bars. That implies about 500 amps in the secondary loop and this is only a medium size gun. It would be quite impractical to design your secondary conductors for that continuous current. Look at the size recommendations for 500 amp welding cable that carries current for much longer periods than your rig will. Welding cable conductors for 500 amps is about the size of your thumb.
I'd take a clue from the soldering gun manufacturers, although part of their conductor selection may have to do with physical strength. Time for some experimentation on your part. Consider using copper tubing. Since you are going to be dirculating water anyway for cooling your mold, you could circulate water through your secondary conductors, allowing much higher current to be carried for the amount of copper in the conductors.
Use a switch on the transformer primary with the rating of at least that of the switch on the microwave. That's all the transformer can handle.
This looks like a good application for a SSR (Solid State Relay) with a current rating of, say, double your calculated primary current since you will be controlling it with a timer. You won't have as large an inrush current to deal with as you would with a motor or incandescent lamp load but you might have a brief surge and an inductive kick to deal with. Use a suppressor. The SSR manufacturers have app notes to guide your selection for various types of loads.
awright -
Google "Watchdog Timers." TI, Maxim, Analog Devices, and many other chip makers make devices with various levels of complexity and sophistication to restart a system if voltage levels drop below some threshold or if the watchdog input senses an absence of normal system activity or if it receives a reset pulse. They come with preset or adjustable time delays of milliseconds to seconds. They put out a delayed logic "0" or "1" that can be used for whatever you desire.
If you already have and are happy with a pulse from your system to command a reset you can use a TDR (Time Delay Relay) They also come in all flavors of adjustability and functionality including delayed on, delayed off, one-shot, and many other programs. Sounds like you may be seeking a one-shot function that will switch the contacts for a selectable time period from a fraction of a second to many hours upon receipt of a pulse or contact closure.
As I recall, Grainger's catalog offers a variety of TDRs and has a writeup explaining all the variations of programs available.
awright -
Your machine will run fine on the original fuses but you will lose protection in the event of a failure that would have blown the recommended fuses. I strongly recommend installing the recommended fuses but if it will take some time to obtain them from the states, I would go ahead and use the original fuses temporarily.
Utility line voltages are rarely exactly the nominal voltages stated by the utilities. The long-term average there is probably a few volts different from 230 volts and the instantaneous voltage will fluctuate with instantaneous loading on the line. If you can obtain a meter, measure the actual line voltage at your receptacles several times over the day and night, determine the average, and use the setting closest to your average. It is normally not critical.
For a reason that I cannot remember right now I measured the line voltage in my home recently and found 127 volts. Line voltage here is typically described as 120 volts.
If you don'e want to measure your voltage, I concur with Hero999 that you should start on 240 volt setting and see how your equipment functions.
awright -
That's exactly what the new "Smart Meters" that the utilities are at this moment investing multi-millions installing do. They are queried periodically and respond with the electrical usage information. I do not know what technology they use for the link to the utility. I also do not know how often the meter "reports in," but I believe that it will be several times per hour.
One of the guys installing "Smart Meters" here in the San Francisco Bay Area for PG&E told me he did not know anything about the technology, but that he thought a satellite link was involved. He did not know if the satellite communicated with each "Smart Meter" individually or if data from many meters was collected at a central point and the accumulated data was linked to the utility via satellite. Conceivably, the meters could use cell phone links locally.
The "Smart Meters" provide the utilities much more detail in the quasi-instantaneous time-of-usage of energy by each customer which some (including me) regard as yet another intrusion into the privacy of the individual. The flip side is that it facilitates time-of-use fee structures, i.e., cost per KWH based upon when you use the energy which can be advantageous to the thoughtful energy consumer.
awright -
What makes you think that you must smooth out the ripple on the full-wave rectified waveform coming out of the rectifier bridge for welding? Why would a smooth DC voltage/current work better than the raw rectified wave? The heating value of the arc will not be significantly affected.
Now, there IS good reason to provide an inductor on the output, but it is generally referred to as an arc stabilizer because it helps maintain the arc across the nulls of the rectified waveform. It also affects the handling characteristics of the arc, but not, I believe, because the DC is smoother (except that the nulls are rounded out).
Since you are not necessarily trying to "smooth" the ripple out, conventional L-C filter design concepts are not particularly applicable.
Wire size should be selected for the highest average (actually, true RMS of the AC+DC ) current you expect to use over any, say, ten minute period, taking into consideration the fact that you are dealing with a coil, not a straight wire which would have better heat dissipation.
You may want to provide an air gap in the magnetic core to avoid magnetic saturation of the core but I'm not sure that is really necessary. An "air gap" is normally provided by inserting a fiberboard shim between the "E" stack and the "I" stack.
I think your best bet is to look at the inductors in commercial welders of the same current capability that you want and try to roughly copy the inductor core size, wire size, and coil size that you see. I don't believe the inductor value is particularly critical to the quality of the arc or the resulting weld.
awright -
Are you still there and still interested, 89Panadol?
Most likely your relay has a coil with too low a resistance. What type of relay is it and what is its coil resistance? How much current does the coil draw if connected directly from your 12 volt supply to ground? How does that compare with the maximum current rating of your darlington output stage transistors?
What is the current capability of your 12 volt power supply? Is it capable of driving the relay coil directly? Measure the 12 volt supply while the device is trying to close the relay to be sure it stays at 12 volts.
Your darlington output stage would have a current gain of at least 10,000, which is plenty for the drive current available from the 4093 gate via the 100K resistor. The BC547/8 has a collector current rating of only 100ma which is OK for many small relays but not necessarily for a large relay with a low resistance coil. But I would expect the transistor to fry if you tried to pull too much current. Are both transistors still OK? If TR2 burned out TR1 alone would probably lack sufficient gain to drive a low resistance relay coil.
Try a more sensitive relay, perhaps a solid-state-relay, depending upon what your load is. And check your output transistors.
awright -
It would be more useful for you to tell us what you are trying to heat. It is straightforward to calculate the POWER dissipated in a given length of a given type of wire at a given voltage. Except for very standardized conditions and media, calculating the TEMPERATURE attained (In the wire? In the fluid medium? In the surrounding insulator?) is a very complex calculation involving surface areas and shapes, fluid flow, Reynold's numbers, thermal conductivities, etc.
Consider a small space heater. The thermal POWER put into the room is fixed by the electrical characteristics of the heating element and the voltage but the TEMPERATURE of the element or of the air could vary over an extremely wide range depending upon the airflow. Or consider an immersion type tea water heater. The temperature will stay below 212 degrees F until the water boils away, at which time the temparature will shoot up by several hundred degrees and the heater will melt.
So, what are you trying to do?
Oh, yeah. Wire. Almost any wire material CAN be used as a heater but only a few are practical. In general, copper makes a lousy heater because it is too conductive. 10, or 100 amps through (approximately) zero ohms produces (approximately) zero power. Nichrome alloys are most often used for electrical heating elements because they have relatively low electrical conductivity so you can generate a useful amount of power using a practical length and size of wire. It is also resistant to oxidation.
For special, short-term applications you can use iron wire but it will rust and has much higher conductivity than Nichrome. I have used a couple of inches of iron wire on a small very low voltage transformer as a quick and dirty nylon rope cutter for years. -
stube40, what ever happened with your high power test load?
awright -
I concur with Alex's recommendation to avoid oil as a coolant on the basis of mess and cost. Disposal of a tank of water is SO much cheaper and easier than a tank of oil. Is this a long-term or a single-shot project? Oil is used in transformers for reasons other than thermal efficiency, like corrosion, fouling, freezing, and boiling.
Heat capacity of oil (BTU/lb) is a little less than half that of water and it gets even worse in BTU/unit volume because oil is less dense than water. I haven't investigated the topic, but I'd guess that heat transfer from an element to water is faster than to oil also due to viscosity. That is, the convective flow of coolant past the element would be slower in oil than in water.
Consider using electric water heater elements in a water tank as loads to completely eliminate the problem of electrical insulation of the resistance wire from the coolant. Their resistance varies from about 10 ohms for a 240 volt 6 KW element to about 20 ohms for a 3KW element. Find out where all those rusted out water heaters go to die. The elements are probably still usable and are probably just a waste disposal problem for plumbers.
6 KW for 1 minute is 100 watt-hours of energy. 1 KWH equals 3412 BTUs so your 1-minute test will generate 341.2 BTUs. 1 BTU is the amount of energy that will raise 1 lb. of water 1 degree F. If we assume that a reasonable temperature rise for your test is, say, 50 degrees F (from 55 to 105 degrees F) you would need about 7 pounds or less than 1 gallon of water. You will have to have a 10 or 20 gallon tank just to have enough physical space for your water heater elements, so temperature rise is not a problem.
Sorry to use non-metric units.
awright -
You don't say anytiing about the size or ratings of the transformer but if you are hoping to use it as part of a welder it must be fairly large. That implies a very low DC resistance because the DC resistance is simply that of the (heavy) wire in the winding. In the absence of very sophisticated and precise low resistance measuring equipment (not a DVM) you won't learn much from a DC resistance measurement except for continuity or open circuit. For a transformer of a few KVA rating I would expect a primary resistance in the neighborhood of an ohm. Remember that if the winding had a significant resistance the primary current of several tens of ohms would dissipate a lot of power uselessly inside the windings.
Are you sure you have identified the windings properly (should be fairly easy)?
I'm not familiar with 3-phase equipment but I would think that a single primary of the unloaded 3-phase transformer could be connected across the line without problems. Shouldn't matter whether or not the neutrals are connected.
Your description of the incoming line hardware exploding is alarming because the main breaker and/or the branch circuit breaker or fuses should have opened long before the overhead line could be damaged even with a direct short across the branch circuit. That's what they are there for. Better check out your main panel.
Why are you planning on using a 3-phase transformer if you actually only have single-phase power available? I am aware of the advantages of 3-phase power in industrial settings (and, in fact, in my garage shop that has several 3-phase machines) but if you must create it with a rotary converter it would seem that you are unnecessarily complicating your life. A welder of any significant power rating is a pretty heavy load on a phase converter that was designed to run home-type machine tools - but I am making assumptions here.
High current draw in an unloaded transformer would seem to indicate that one of the windings is shorted. There may have been a good reason the transformer was on surplus.
Good luck.
awright -
As Hero999 says, cleaning with water is acceptable. I first remove or protect any components that could be damaged by water, spray with a detergent solution, scrub with a toothbrush, rinse very well, then finally rinse with distilled water (available by the gallon at your local supermarket - it's used in clothes irons), then dry gently but thoroughly.
Pay special attention to cleaning inside any pots or other components with cavities that might have gotten filled with dirty water or mud, both in cleaning and in rinsing. Use a low viscosity cleaner-degreaser detergent that will rinse easily, rather than a viscous dish soap that might resist rinsing.
Potential problem areas are transformers that can be very difficult to rinse and dry adequately because the solution will penetrate the coils and transformer laminations (but I imagine that you don't have transformers in an automotive device). Rather than trying to dry quickly with a hair dryer, I recommend slow but thorough drying over several days in a warm (not hot) place like in a sunny place indoors, near a lamp, or near a heater outlet that's not too hot. The object should never get too hot to grasp comfortably for an extended period (as a temperature test).
Still, no guarantees if the object had been submerged for an extended period that led to corrosion.
Good luck.
awright -
Hero999, the joule is a unit of energy and the watt is a rate of energy flow, i.e., power. One watt is equal to a rate of energy flow of one joule per second, as jt stated. A watt is a watt is a watt in any consistent system of units, be they thermal, mechanical, or electrical. Mathmatically, they are directly convertable.
John's problem is not a mathmatical conversion of units, but rather figuring out how effectively the thermoelectric device will convert the heat power in into electrical power out. I believe this can be quite a complicated function of the source and sink temperatures, the area and thermal conductivity of the device, the Seeback Coefficient of the device, and the electrical resistances of the device and the load. Seeback devices (not Peltier devices that convert electrical energy into thermal energy, although they are essentially identical except for the application and how they are optimized for the application) are very inefficient at converting heat energy into electrical energy. I believe one would require either a manufacturer's rating of the performance of the device at John's specific operating conditions or data derived from experimantation on the device to predict the output of the device. Alternatively, just assume an efficiency of about 10% and apply that factor to the input thermal power.
awright -
I agree with Alex. Not much to go on from your sketch. Detailed photos AND a description of the device you are trying to repair would be very helpful.
awright -
Well, I certainly agree with the undesirability of burying a relay in the garden, but why would you have to do that? Wouldn't the relay be located at the controller in your house, rather than in the garden? Would you care to explain more about the details your application?
awright -
Why don't you want to use a bistable relay? Seems like a natural for your application.
If you had the skill to design a 24 volt control system, why can't you do the same for a lighting system? Both require steady on controller power. Can't you use the same sensing of the 6V controller signals, only drive a 24 volt relay instead of a water valve? Depending upon the power required for your lights, you can cascade relays to drive a contactor of adequate current rating.
Can you tell us more about your controller and your limitations?
awright -
The answer used to be a simple yes, but no longer. High efficiency clothes washers like the new front-loaders often use switched reluctance motors instead of induction motors because switched reluctance motors driven by sophisticated electronics can provide easy reversability and a wide range of operating speeds, thus allowing elimination of transmissions otherwise required for the speeds and direction changes of the various washer functions.
Almost all refrigerators use induction motors sealed with oil in a hermetically sealed case. These refrigeration compressor systems are highly optimized for the function and should only be operated at voltages and frequencies mentioned on the embossed label welded to the case.
I don't know about dryers. Since older models tended to operate at a single speed, with only the heat or drying time varying, most of them had incuction motors driving a long belt wrapped around the drum. Newer, high efficiency models may also have multiple speeds, which would imply a switched reluctance motor.
Since a switched reluctance motor is electronically driven, it MAY be tolerant of power frequency variation. I do not know. Speed of a switched reluctance motor is not power frequency dependent. Best to check frequency tolerance with the manufacturer if at all possible.
You should be able to observe whether your appliance has a switched reluctance motor by looking at the wires going into the motor. Most appliance induction motors will have two conductors entering the motor (plus a ground conductor connected to the frame), although this is not universal. If the motor has many wires entering it and if there is a somewhat complex circuit board with some large power semoconductors on it, you probably have a switched reluctance motor. That's very good for efficiency and simplicity of design.
awright -
I haven't gone back over the entire thread to remind myself of how we got to this point, but I think that Gazza"s analysis is a little simplistic. I also do not claim to be a transformer engineer.
Pin = Pout ignores the losses in the transformer which can become significant for a small transformer near or beyond its rated load. Notice that a small transformer loaded to around its rated load gets warm. Some get downright hot. That power going into heat comes from the input power. While properly designed transformers are remarkably efficient compared to most electronic devices, they are not perfect and they become very imperfect when marginally designed and heavily loaded.
Part of the losses are in the hysteresis of the steel core and part of the losses are the I^2® resistive heating in the wire of the primary and secondary windings. My vague recollection from school decades ago are that the hysteresis losses in the steel are constant, independent of current, while the I^2® losses are obviously totally dependent upon loading.
The result of these losses is REGULATION, the deviation of true output power from theoretical output power, which is seen as the output voltage being lower than the turns ratio would predict. Prediction of the actual deviation from perfection involves careful evaluation of transformer parameters.
By the way, what is "peak power?" Are you selling shop vacs or home air compressors?
awright -
The answer is complex and requires some knowlege of the nature of the specific equipment you want to change power frequency on.
In general, induction motor driven equipment will be OK with higher than design frequency power, but not lower, but you have to check what the increased speed of a 50 Hz. motor powered with 60 Hz will be. If the higher motor speed can be adjusted out by, say, a pulley change, then the increased frequency will definitely be no problem. Some loads will overload the motor due to the increased power consumption at higher speed. The 50 Hz induction motor powered by 60 Hz can actually produce more than its rated power but it may not be enough to compensate for the increased power demanded by the oversped load. The increased power output of the motor is due to the larger amount of iron and steel that a 50 Hz motor of a given rated power is built with.
In general, a 60 Hz induction motor cannot be operated at 50 Hz unless it was designed as a 50/60 Hz, dual frequency motor, as many are. Look at the nameplate. The basic problem is that an induction motor designed for 60 Hz only will magnetically saturate on 50 Hz power because there is not enough copper and steel in the motor. Saturation leads to greatly increased current and burnout.
Most devices that rectify the incoming power to provide DC to the load will be OK going to a higher frequency because most power transformers will be fine with increased frequency and the ripple on the storage capacitors will actually be lower than the equipment was designed for. Going to a lower frequency will generally be OK if there is no transformer or if the transformer is rated 50/60 Hz. You have to check if the device can tolerate the larger ripple content of the rectified 50 Hz power. You should also check whether there is a motor in the device, like a fan. Here, again, if the fan is designed for 60 Hz only, it may saturate on 50 Hz.
Equipment powered by brush motors, like most hand-held power tools, will generally be tolerant of frequency change in either direction because, unlike induction motors, their speed is independent of power frequency and because they are not subject to magnetic saturation due to reduced power frequency.
Anything with a critical speed or timing function that uses an induction motor driven clock or timer will be off speed/time by the ratio of the frequency change.
A device with a large power transformer like a transformer welding machine cannot tolerate reduced power frequency unless specifically designed for 50/60 Hz. In general, going to a higher than design frequency would be OK, but be careful about overspeeding fans.
So the answer to your question is "sometimes," but always check with the factory, if possible, or make sure you know what is happening inside the box.
Good luck!
awright
Open hardware solar power generator for developing countries
in Inventive/New Ideas
Posted
One resource for information on systems installed in isolated areas with basic system data, but not detailed schematics, is Home Power Magazine. Lots of discussion and examples of system design and capacity calculation.
They offer CDs of multiple years of back issues on each. I certainly do not speak for the publisher, but perhaps you can appeal for a free set of CDs.
awright