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About Ashtead

  • Birthday 09/13/1961

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  1. This looks like it would run a little 3-phase motor at variable speed, continually. Such motors are used in hard disks to turn the platters, it is basically a small AC motor meant to run continually, and not really a stepper motor. Like other AC motors, it may likely have a lower frequency limit below which it will not be able to start or remain turning, and when stopped, there is little or no holding torque. The commonly seen stepper motors that are used for finely positioning things, have 2 or 4 coils that are driven with quadrature signals (90 degrees out of phase). However, hooking three of the 4 coils in such a stepper-motor might make it run, though somewhat unhappily here... Also, there is the matter of the transistor threshold voltages that Audioguru mentions. Using a CMOS 4584 at 12V instead of the 74AC14 might work better. with the transistors shown. Ashtead
  2. The schematic shows a DAC-08, (Analog Devices) which National Semiconductor calls DAC-0800, and this one is similar to, but not quite identical with the DAC-0808, or MC1408/MC1508 that other manufacturerers may call it. The DAC-08 is somewhat more flexible, in that it may be used in some different configurations with non-grounded outputs, and with different, though now mostly obsolete, digital logic levels. It is somewhat unfortunate that the part numbers are so similar... The differences have to do with the electrical connections of pins 1 and 2, pin 1 being one of the "mystery pins". Pin 1 is "VLC" on the DAC-08 and simply not connected at all on the DAC-0808. The purpose of this pin, on the DAC-08 is to be able to set different logic thresholds for the digitat inputs, that is, the voltage, at which the circuit determines whether the inputs mean "0" or "1". The data-sheet that I have here indicates that the threshold voltage is 1,4V plus whatever is applied to the VLC pin. For low-voltage logic (3.3V-5V CMOS or 5V TTL) the 1.4V is just about right, so the VLC pin can be tied to 0V for this use. The datasheets shows examples of using 15V high-voltage logic, and then the threshold voltage should be raised to 7.6 V, by feeding 6.2 V generated with a zener diode into VLC. The DAC-0808 has a fixed theshold, at around 1.4 V, which makes it suitable for being driven by 5V TTL or CMOS, just like the DAC-08 when this pin is at 0V (or ground). As this pin is not connected on the DAC-0808, it doesn't really matter. Pin 2 is "/IOUT" on the DAC-08 and identified as GND on the DAC-0808. What this means is that the connection from this pin to the non-inverting input on the following op-amp should be at a hard 0V level for the DAC-0808, but does not have to be on the DAC-08. Your circuit diagram shows this connection at a different level; thus in this circuit, the device must be the DAC-08, and a DAC-0808 may not work very well. (it probably will still put out something that looks kinda right, but as it's GND pin isn't really grounded, all bets are off when it comes to accuracy or linearity or such things that are important with D-to-As). The other pin that needed explanation is pin 16, "COMP", which is there for the same reason on either the DAC-08 or the DAC-0800. This capacitor provides frequency compensation in the reference amplifier inside the converter, where the reference voltages and current between pins 14 and 15 is distributed to the 8 current sinks, one for each bit, that may be turned on or off, to make the resulting combination of current be drawn into IOUT output. It is likely that the transistors associated with this amplifier stage will be self-oscillating if this capacitor isn't present. Its value depends on the resistances connected to pin 14, the data-sheet indicates minimum values of 15pF for 1 kOhm or 75 pF for 5 kOhm. This may be of interest if the DAC is to be used as a volume control or multiplier for a fast-moving signal. Many of the examples show this capacitor with a value of 10 nF or 100 nF, which is useful when Vref is supposed to be a stable DC voltage. Here are links to some of the manufacturers' datasheets: http://www.national.com/pf/DA/DAC0808.html http://www.datasheetcatalog.com/datasheets_pdf/M/C/1/4/MC1408-8.shtml http://www.analog.com/en/prod/0,2877,DAC08,00.html http://www.national.com/pf/DA/DAC0800.html Hope this clarifies things. Ashtead
  3. The 74193 is a 4-bit counter so it can only count to 15. But you can cascade two of these by feeding the CARRY and BORROW outputs from the first unit into the COUNT UP and COUNT DOWN inputs of the second unit., and then make some kind of circuit that feeds clock-pulses to the first (least significant) unit's COUNT UP or COUNT DOWN inputs, depending on the position of your switch. The 74193 counts on L-H transitions on one of these inputs, while the other input should be at H. Then what is to happen when it has reached 59? Two 74193s will go on until 255 (all outputs H) and then continue from 0 again, unless some logic looks at the output lines and stop the count-up signal when it is at 59. (Obviously, it should still be able to count down from 59). Similarly, the count-down may have be blocked when the counter is at 0 (all outputs L). This all depends on what exactly the circuit is supposed to do besides counting up and down. Ashtead
  4. Rob, I have usually fed the output of my optocoupler circuits into some logic circuits, and used a simple resistor-transistor combination driven from some other part of the logic. I have typically used something like in the attached figure, where the transistor's base is left open, emitter is grounded (to 0V), and the collector output is returned to +5V via a 2.2 kOhm resistor, for an approximate 2 mA transistor current ICE on. (This current can be lowered, meaning this resistor can be increased, 10 kOhms for 0.5 mA for example). The following 74HC14 Schmitt-trigger gate will make a nice transition when the transistor goes into or out of conduction, and then it drives the 2N3904 that drives the relay itself. I have shown a possibly different voltage for the relay, again, this can be the same or the 5V for the optocoupler transistor can be obtained from a voltage regulator. Or the 74HC14 can be replaced by a 4584 which can work anywhere from 3 to 15 V, and then feed this off the same power supply as the relay for a 6V or 12V relay. Just increase the values of the 2.2 k and 1k resistors accordingly. These values are not terribly-critical, anything that provides between 0.5-2mA for the optocoupler transistor and at least 1 mA base drive for the 2N3904 (or another simliar general NPN type such as BC108 or BC237) transistor will do. Another part of the 74HC14 or 4584 can be used to drive a locally visible LED. And of course there should be some resistor or current-limiting for the LED in the optocoupler, as for every other LED. Hope this is useful. Ashtead
  5. LDR means Light-Dependent Resistor. These are made with some material that changes conductivity when illuminated -- the typical Cadmium-Sulfide LDR will have less than 1 kOhm when in the light, and something like 1 MOhm when in the dark. This variance in resistance can then be used to control a driver transistor and a relay. However, there is another option where there is an LED that can be removed, which is to use an optocoupler. This device contains an LED and a phototransistor inside a small package resembling an IC. When there is current passing through the LED its light will cause the phototransistoer to conduct, and as with the LDR, it can be used to drive a heftier transistor and a relay. There could also be another transistor driving a visible replacement LED if so desired... Another advantage of the optocoupler, similar to the solution using and LDR, is that it provides isolation, so it will have the possibility of working without problems no matter what kind of circuit is driving its LED -- as the optocoupler merely replaces it, and the currents and voltages associated with the relay can be on a completely different power circuit, so as to avoid disturbing whatever unit it is hooked up to. The 4N35 optocoupler or one of its equivalents would be useful here. Ashtead
  6. Could it be possible to record the pressure and shaft positions as they change over time using the sound card in the PC? There are two channels readily available and the sample rate available there would seem to be sufficient. It might be that the rate of change of the pressure signal is too slow for the low frequency cutoff of an ordinary sound card to record directly, however, I would think that some kind of voltage-to-frequency conversion (using something like an LM331) could be used to overcome this. There is also the matter of transmitting the shaft position in some useful form, though a shaft encoder generating a pulse train with some kind of index pulse (of say, twice the amplitude or of opposite polarity) at every TDC and BDC would be a possible way to make this easy to record. Then the resulting lists of pressure and shaft position values can be plotted and used for further calculations. Ashtead
  7. To me the 2k2 part looks like it should be 2200 ohms, (2.2 kilo-ohms) and the final m could suggest 20% . There evidently is a British standard that specifies this, as opposed to the usual color coding. I found something about that here: http://xtronics.com/kits/rcode.htm This way of specifying resistor values with an R, k or M in place of the decimal point is rather common in European magazines. 2k2 4M7 3R9 etc. would be 2.2kohm, 4.7Mohm and 3.9 ohm (they use R for the units of resistance). A similar usage is also seen with capacitors, batteries and zener diodes, where there are letters p, n, mu (or u) or V used for pico-farads, nano-farads and micro-farads and volts. 4n7 and 5V6 would be 4.7 nF or 5.6 V for example. Ashtead
  8. Hi gogo, The 2003 and 2803 that I mentioned are the NPN transistor arrays, grounded emitters and open collectors, and which have some resistors connected to the base so that they will work with input voltages as high as 5V. I sometimes prefer to think of them as sets of high-current open-collector inverters rather than a row of transistors, since the inputs are directly compatible with logic circuits. The 2803 has 8 drivers in an 18-pin package, the 2003 has 7 drivers in a 16-pin package, they are otherwise much the same. I see them both advertised as "high voltage high current Darlington drivers" which really means that they probably would be slightly slower than a real inverter such as a 4069 or 74HC04, but I don't think this is any problem here. You mention you drive your LEDs from a 74HC595. The main outputs of this can sink or source 6 mA from the outputs for the voltages to remain within the correct limits and 35 mA as the absolute maximum current. This certainly can drive the inputs of a 2002/2803 driecty, so that part is no problem. Also, and I have to guess, that these are multiplexed so both anodes and cathodes of the LEDs in your array are driven by 74HC595s? 220 Ohms for discrete LEDs sounds about right for LEDs running from a 5V supply continuously. When multiplexing, the LEDs can tolerate larger currents for the short time that they are on, so you may be able to reduce this value, down to 68 Ohms or so and get more light out of your LEDs, but this of course depends on the driving circuits being able to deliver the required current. Whichever their value, consider these series resistors as part of the LED array in the following discussion. The problem on the output side is that the 2803/2003 can only sink current and not source it. This is not a problem when driving the LEDs at the cathode (negative) end, but you will have to use some other driving transistors at the positive, anode end. I don't know of a PNP equivalent of the 2803/2003, so I have had to use individual PNP transistors with resistors in series with the base, and with open collectors that can source the necessary current to the display LEDs. Transistors such as 2N3906 should work fine here with a series resistor in the base of 4.7 kOhms or thereabouts. Connect the emitters to the positive supply; these become just like "upside-down" versions of the 2803/2003 innards. Adding all these drivers will result in a logical inversion, which means that when your HC595 formerly drove an anode-end positive to turn it on, it will now have to drive the base of a PNP transistor to LOW to turn these diodes on via the transistor. Likewise, at the cathode side of the LEDs, the formerly LOW output that turned on the LEDs directly will now have to be HIGH to turn on one of the NPN transistors inside the 2003/2803. There are at least three ways I can think of getting around this: 1. put extra 74HC04 or similar inverters between the shift-register outputs and the driver inputs/transistor bases. This is simple enough, but it is also rather inelegant and adds even more components to the design. 2. Insert the drivers and re-program the PIC so that the polarity of the outputs from your shift-registers changes. 3. Leave the PIC program and the shift-registers the way they are, change the polarity of all the diodes around, and put the drivers in at both sides so that the former anodes and now cathodes are driven by 2003/2803s or other NPN transistors, and the former cathodes, now the anodes are driven by the PNP transistors. If there are more rows than columns or vice versa, I'd recommend arranging the diode polarities such that you get the fewest PNP transistors and associated resistors. Just keeps the part count down. Hope this is useful. Ashtead
  9. Hi Autir, With logical circuits, there are a number of standard specs that apply to most of the circuits. For TTL as we are dealing with here, an input senses a LOW level when the voltage is from 0 to 0.8 V, and a HIGH level when the voltage is from about 2 to 5 V. When an input is driven LOW, there is a current flowing out of the circuit of about 1.6 mA, and when the input is driven HIGH, it only draws 0.04 mA, this time the current flows into the input. Notice that the current drawn is a lot larger when the input is LOW than when it is HIGH. A similar difference is seen with the output voltage specs: A HIGH output can source (deliver) about 0.4 mA, but sink (pull in from a load returned to Vcc, such as an input on another gate) as much as 16 mA. When sourcing a current larger than 0.4 mA, the voltage on the output may drop below the specified standard minimum HIGH output voltage, which is 2.4 V, and if we sink more than the 1.6 mA into an output, its voltage may exceed the standard maximum LOW output voltage which is 0.4 V. Notice that the outputs can sink a lot more current when they are LOW than they can source when they are high. A number of TTL chips have "open collector" outputs, that cannot source any current at all, but they are good at sinking current. There is an NPN transistor in there, which turns on, and connects the output to ground, but when it turns off the output floats as if disconnected. To make such outputs lines go to a HIGH level when the transistors are off, we put "pull-up resistors" between these lines and the Vcc bar, so that the resistor causes the voltage on the line to be pulled up to a level that other connected chips can understand as HIGH, above 2 V. Now to your questions, 1. The outputs of 7446 and 7447 are designed as "open-collector", which as mentioned above, means that behind each of a segment output, there is a transistor between it and ground, that turns on and conducts current, when the corresponding segment is to light up. These transistors are somewhat heftier than the standard ones, they can sink as much as 40 mA before the voltage rises above the spec limit of 0.4 V. This means that if we pull 50 mA out, we may get 0.5 or 0.6 V on the lines, as well as a hot chip. The recommended maximum load is 40 mA. That the outputs can sink 40 mA doesn't mean that we absolutely have to load them this hard; if we put on a load that sources 10 mA, the 7446/7447 will happily pull the output voltage down to somewhere below 0.4 V. In the 7448, there are the same kinds of open collector transistor outputs, but here, the "internal pull-ups" are connected between the outputs and Vcc to make the outputs go HIGH instead of open-circuit when the transistors turn off. The logic of the 7448 is also inverted, the outputs are HIGH when segments are on and LOW when they are off, just the opposite of the 7446 and 7447. Like audioguru said, the purpose of this is to be able to drive other, perhaps even stronger, circuits. 2. As mentioned the 7447 pulls its outputs down when segments are turned on, and that means that the LEDs of a 7-segment display should be connected between these and a positive voltage. This may be the 5V bar, or another power source at up to 15 V for the 7447. In addition, each of the segment LEDs must have a current-limiting resistor connected in series with it. In practice, this resistor will be between the cathode of the segment LED and the output pin of the 7447. The anodes of the segment LEDs are common connected to the high voltage, therefore the display will have to be of the common-anode configuration. The display segments should only be forced to conduct 10 mA, and with a typical display powered from a 5V supply and where there is 1.5 V drop across the LED, and we can assume the output voltage of the 7447 to be 0V when turned on, the remaining 3.5 V has to be dropped by this resistor, and that suggest a resistor value of about 350 Ohms, where the closest standard values available are 330 Ohms or 360 Ohms. There is no need here for any enormous precision, values as high as 560 Ohms will result in a lower current and a slightly dimmer display. You will need 7 of these resistors, one in series for each segment. 3 As we just saw, the common-anode display is used with the 7447. The CMOS 4511 is very similar in function (in fact, it has almost the same pin-out) but it drives its outputs HIGH when turning on segments, so this can be hooked up to a common-cathode display, using series resistors, much the same as the 7447. I have no idea which one is the most commonly used; the larger displays tend to be common anode, simply because it is easier to drive them with a higher voltage (the segments may be 2 or more LEDs in series) with open-collector drivers like the 7447 or even just buffers like the 2003 or 2803. Smaller displays are common as both kinds; my experience with multiplexed ones is that the common-cathode ones are slightly easier to deal with, with a hefty 2803 or similar driving each entire digit, and the segments driven in turn by something like the 4511 ...
  10. Oops! Sorry for the mis-information, I was wrong about the pin numbers: The correct ones would be Pin 8 Ground (or common emitter) and Pin 9 where the diodes go, for the 16-pin 7-channel parts. Pins 9 and 10 would have been right for the 2803 with 18 pins and 8 drivers in it... I don't think there are any variant pin-outs on these, my older 2203s and the more recent 2003s all are the same.
  11. An LED will only draw the current it needs to light up. Usually this is something around 5-10 mA, so if connected to a power supply able to deliver 1A, it will still leave most of that 1A for powering other loads. As for printer cables, the usual 25-pin D to 36-pin Centronics has 17 signal lines, plus ground, so the number of cores in the cable can be anything from 18 to 25, and there may or may not be a screen, depending on the quality of the particular cable under consideration. On the D25, pins 18-25 are all ground, while 1-17 are various data and control signals. The cables usually come with molded-on connectors, so if one cuts it open, it will be a matter of using a multimeter, ohm-meter, or continuity meter to figure out which pins in the D25 goes with which core. Ashtead
  12. Perhaps I can help out, I have used 2203s and relays in a similar way and they worked well. The 4511 has its outputs active High (voltage level is close to the positive Vdd) when segments are turned on. Each one of the 7 transistors inside the 2003 work as an open-collector inverter, which means that the output connects to Gnd when the input is driven High. So the 4511 can drive the 2003 directly, and the relays should be connected between each one of the 2003 output and some positive supply voltage. The 2003 pin 9 is the common emitter returns for the transistors, this should be connected to the ground, (same as pin 8 on the 4511) Pin 10 on the 2003 is the common for all the protective diodes, this should be connected to the same supply as the relays. This supply for the relays should preferrably be separate from the 5V supply for the 4511 and its friends, so that the noise from the relays turning on and off do not affect the logic circuits there. Besides, 5V is a little too low for turning 6V relays reliably on. If the 5V power supply for the rest of the circuit is taken from a regulator, maybe the relays can be powered from the supply voltage appearing before that, or a separate 6V to 7.5V supply could be obtained using another regulator circuit. To test this, hook up the 2003s to the relays and the power and ground but not to the 4511 (or pull the 4511 out of its socket if possible, that will make this easy), and put 5V (from pin 16 on the 4511) on each of the inputs, to cause the 2003 to turn the corresponding relay on. Unconnected 2003 inputs will behave as Low with the transistors Off, since there are pull-down resistors inside the 2003. Ashtead
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