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Everything posted by flippityflop

  1. so it's Ve = Vc - Vbe? ok... i missed that from just reading from wikipedia... i read the voltage on the load is *mostly* is constant for most of the operating current range... then again the examples from wikipedia has the load on the collector side... collector side, then...
  2. well, emitter in saturations is still slightly bit higher... (Ie_sat = Ib_sat * (1 + hfe))
  3. answering my own question again: it just struck me, but maybe the capacitor was not charging not because of properties of currents in collector and emitter for any NPN, but because i've been thinking in terms of conventional current... Let: Q be a sub-circuit of an avalanching transistor and capacitor in parallel. in my case, the avalanching transistor is 2N2222; the "generic NPN" is not the 2N2222 mentioned. whenever Q was in the emitter side of the generic NPN, the negative terminal of the Q's capacitor is already in the highest negative charge. the momentary opening of the generic NPN's collector applies a "more electropositive voltage" to the capacitor's positive terminal, but there are no charge carriers migrating to said terminal. hence when the generic NPN stops conducting, the "more electropositive charge" is not held. if we were to put Q on the collector side, the positive terminal of its capacitor is always what it is supplied to it. when the generic NPN starts conducting, electrons rush to the negative terminal of Q's capacitor and when NPN stops conducting, the electrons stay there. hence, we have negative charge there. is this correct? i'd appreciate it if somebody would confirm this.
  4. i'll solve the specific circuit problem on my own... although, i would like an answer on the first question... when switching, is it better to go with the emitter on NPNs and the collector on PNPs?? do their electrical characteristics differ more than slightly higher currents?
  5. i'm asking as i've been working on a design that to my understanding should've worked. i have a generic NPN restoring the supply voltage and feeding it's emitter output to a 2N2222 parallel to a charging capacitor for a delayed avalanche. somehow the avalanching only works when i swap the 2N2222 + charging capacitor sub-circuit to the collector side. it surprised me as when it was connected to emitter, it was also directly connected to ground. and when it was on the the collector side it was directly connected to the supply. so it was not isolated and would be biased to a common ground. although, as a side note, when it was on the emitter side, the charging capacitor might have caused a reverse current through the generic NPN and leaking to other parts of the circuit. so i probably should've put a diode in there. well i didn't get the chance to as i blew my 2N2222, so now i have to go to my local supplier to have more. so i'm not completely sure what to make of it... EDIT: the generic NPN is "on" strictly less than the time needed by the 2N2222-capacitor sub-circuit to avalanche.
  6. alright, continuing the series... another question i would like to as is on NPN BJTs, would it be better to put the load on the collector side or the emitter side?? in schematics, i almost always see it on the collector side (Ic = hfe * Ib), but the emitter side also works and when doing plain switching (in saturation) the emitter side works at least as well (Ie = Ib * (1 + hfe))... so would there be any difference? the same with PNPs, when it comes to switching, would the collector side (Ic = Ib * (1 + hfe)) work better than the emitter?
  7. i need to know what is the lowest current that is reliable for cases such as: a.) charging a capacitor b.) when the current is being read by an instrument such as an ammeter c.) logic signals what i mean is that even outside normal conditions -- exposed to hot summers and cold winters, being touched by bare hands when testing, or when it's humid, etc -- what are the minimum low currents that will cover a, b, c above?
  8. are these NPN BJTs: http://pdf1.alldatasheet.com/datasheet-pdf/view/17918/PHILIPS/MMBT2222A.html the same as those 2N2222 that has been around for since forever? can i trust that the reverse active and avalanche breakdown profiles are the same??
  9. ok i actually just realized the flaw after i last posted. i thought of mentioning it, but i thought it will have you guys think that i am trolling. (i actually just checked again for additional posts from you guys) yes, hero999 there needs to be a common ground here for this to be a universally applicable (aside from the aforementioned caveat that you have to balance the R1 and R2). to see why, we need to visit what was posted in this thread (hence my fear of this being seen as trolling): http://www.electronics-lab.com/forum/index.php?topic=39704.0 basically, V1 and V2 are just voltage drops, but the actual voltage potential of both grounds of V1 and V2 may be different (can be taken by using a third reference voltage or if you want to be absolute, the negative elementary charge). so in actuality: |V3| = (V_ref_pos1 - V_ref_gnd2) - (V_ref_gnd1 - V_ref_gnd2) if V_ref_gnd1 >= V_ref_gnd2 or (V_ref_gnd2 - V_ref_gnd1) - (V_ref_pos2 - V_ref_gnd1) if V_ref_gnd1 < V_ref_gnd2 notice that (V1 = V_ref_pos1 - V_ref_gnd1) need not be greater than (V2 = V_ref_pos2 - V_ref_gnd2) to have a positive V3 = V1 - V2, the opposite can actually happen. it really is dependent on the relationship of V_ref_gnd1 to V_ref_gnd2. anyways, this can be further compactified as: |V3| = | (V_ref_pos1 - V_ref_uni_gnd) - (V_ref_pos2 - V_ref_uni_gnd) | where V_ref_uni_ground = min(V_ref_gnd1, V_ref_gnd2), which is obviously the offset or translation along the x axis our values. but if we allow V_ref_gnd1 = V_ref_gnd2 = 0V, then we simply have |V3| = | V1 - V2 | as for my part on drawing 2 independent voltage sources, i was trying to be general. but, i probably should've explained or added after i realized it; that we still need a common ground. of course, we can't simply add a shorting wire on both grounds of V1 and V2, as that would bypass the V3 path. a high resistance short also would not work, basically you just have another non-functioning comparator. what we can do is ALSO have V1 and V2 be sourced from another common voltage source V0, so practically they are biased in a V_ref_uni_gnd which is V_ref_gnd0 from either a single V0 or if you have several individual sources, their grounds must be shorted.
  10. i was wondering why is it, for just half duplex systems, don't we have a dedicated transceiver and a dedicated transmitter antenna. ok, just hear me out... assuming we are using grounded monopoles, we can have a resonator connected to a diode whose anode is connected to the transmitting antenna, and have a receiving second antenna connected to the cathode of another diode then going to the resonator. the transmitting and receiving antennas would also be made of different materials that are nonreciprocal. ferrites that are biased to transmitting and receiving (higher gains). since they're half duplex, we don't even have to have 2 different resonator circuits. so basically, if this is feasible, we don't have to change anything in the current systems, just plug in these dedicated antennas... am i wrong here?? also, why do we not have a "Radio" section?? i think it warrants one.
  11. you know what i'm talkin' about... anyways, just in case somebody stumbles upon this simple thread in the future, i'll attach a more concise diagram:
  12. also, why i am convinced that sometimes it might be helpful to consider "pulls" is that i see a lot of voltage supplies apparently provides negative voltages.
  13. ok, i guess i lose here... in all cases "conventional current" and the idea of 0V grounding that comes with it abstracts all cases when it comes to passive components. in active components as long as it is only immediately in series with a passive component, then it *should be safe*. if 2 or more active components are in series, then i'd still raise questions if it could work. ok, look at the attached diagram and apply conventional current and see if we can make it all work. suppose the 2 identical avalanche diodes in series with a active component "Q". we also define an "ambient" voltage potential in the non affected/conducting components such as leads, and wires, any conductors. the voltage drop from each end, is of course, the whole voltage applied to the circuit. the voltage potential from the left of "D_av1", is say, 3V above ambient. the voltage potential from the right of "D_av2" is -4V below ambient. that's a voltage supply of 7V. say the avalanche diodes have a breakdown at 5.5V across. how this would *usually* work is when we have *single active component* with at least 2 leads and have their electric potential on their immediate 2 leads interact, either by voltage potentials and/or current created... so simply as connecting a single 5.5V avalanche diode to a 7V will be enough to have avalanche breakdown. this is the same for all passive components -- resistors, inductors and capacitors. but what will happen if we cannot continuously pass the voltage potential all the way from the other end of the circuit to make them interact in any components?? such is the case when we have Q as a hypothetical active component that DOES NOT maintain voltage potential across it the way passive components do. the voltage across D_av1 is only 3V. and, Q is not letting it's left lead have a voltage potential negative from ambient. the same would be true for D_av2, it only has 4 volts across it and "Q" will not let it's right lead have a positive potential from ambient. of course, this is hypothetical, as i don't know of a Q component that fits what i'm describing.
  14. no, i was reffering to a "passive" comparator, sadly also being a very limited one. there's very little voltage drop and current used in your "active" comparator... it's something very discrete that can be added with very little effect to the rest of the original circuitry. mine is something that comes up not very often, but a cheap one if it ever does. you'd hardly come across R1 and R2 that are equal. maybe you can readjust the rest of the original circuits, so that they are, but there's no guarantee it will work out. probably where we can see this arise often is when we tap this onto voltage references, instead of signal or power lines. add decoupling capacitors, if you do, though.
  15. uhhh... nobody's explaining this satisfactorily... c'mon... i need to know so i can ask further questions about semiconductors and VH electronics... EDIT: "VH speed electronics" forgot the "S"
  16. so, looking at the attached diagram, is there a "pull" on the cathode side of the capacitor or a push from the other side?? or both? if it's a little hard to explain, then tell me how it happens in actual physics, not in conventional currents... i need to know, because, in the future, i will be designing circuits that depend on there being a "pull". if there isn't for ground, then i guess i'll have to employ negative voltages...
  17. ok so i have questions with electronics that doesn't seem to be covered by tutorials that i've scanned throughout different websites, or i just don't want to read through a lot of material to find that small footnote that i needed to find. i'm basically going to be posting a series of threads named "the finer points of electronics" to have people explain things to me. this may sound like trolling (it's not), but consider this a way to get more activity in these forums. ok so question I: so the tutorials lead us to the idea that common ground is just the charge of the source equal to the "ambient" charge of environment, in conventional currents... but i've never been sure of it. maybe it actually exerts a "pull"... or maybe it doesn't and that's why some voltage sources provide a negative current, usually along a positive one. can anybody tell me which is it??
  18. ok this is my solution... though i'm not sure it'll work as i don't know if ground actually "pulls" into it, or is everything a result of positive voltage pushing everything in.... in terms of conventional current... "R_z" limits the current so the "Z" zener diode can produce a voltage reference. the reference is then picked up by the "C1" and "D_av" avalanche diode (whose breakdown is equal to Z) compares it with the actual voltage from source in C2. when the voltage goes below a certain limit, as referenced by "Z", the avalanche diode breaks down and switches the thyristor "Q", which then shorts the power supply. causing a momentary voltage cutoff to the load. we assume the load, once disconnected will remain open, as if it was turned off by another thyristor as well. ok. assuming this even works (i don't even know), being a complete newbie, in general case, my design is very limited... and, inelegant, even. we'd have to assume that the load will remain off, once turned off (as the case when it is turned on by a thyristor). can the supply handle momentary shorts?? we'd also have to find a way to turn off Q once it shorts. we'd have to assume that the voltage is falling to reach that certain voltage limit and not rising to it, etc, etc.... so i'd be happy if somebody can help me simplify this design. EDIT: another limit is that the zener diode has to very, very narrowly provide voltage reference only when the voltage supply is very close to reference.
  19. so i got this small circuit that plugs to the mains as power source. as for its DC converter, it's one of those transformer-less ones that drops the voltage with a resistors and filters before rectifying and smoothing it. the ones that can only deliver small currents. well, after that, it uses two IN4737 zeners in series to create a reference of 15V. i needed to find how much current is being drawn between the 1) DC converter (as described above) and 2) the zeners and the rest of the main larger circuit. so i disconnected the positive wire between these two and put in series an ammeter. the measurements were odd. using my multimeter, it's here as follows (as i recall it properly): 20A range - 0.06 A 200mA range - 0.006 mA 20mA range - 0.060 mA 2mA range - i can't remember 200uA range - i can't remember it was a weird set of readings (and yes the number i gave above for each range are exactly as they showed up in my display, even the units). i restored the original connection and the whole thing won't work anymore. i then found out that the IN4737's were blown and shorts at a few milliohms in both direction each. so what went wrong?? the only difference was the ammeter in series for the positive wire between the converter and main circuit. aren't ammeters specifically designed to be passive and attempt to be a perfect conductor in all cases??
  20. or do we have to have wheatstone bridge and more complex ones for the general cases??
  21. ok, i need a simple DC comparator. so would the attached pic do the job?? all resistors are equal EDIT from above ^: my mistake, R1 and R2 (not labelled below) are the only ones that have to be equal. R3 (not labelled below) ideally has high impedance.
  22. i thought i had a solution, but it wasn't doing the right voltage comparison... good thing there's a "remove post" option here in this forum.
  23. i need a component or circuit that cuts off current when voltage drops from a certain level... or basically an SCR (which are usually designed to have a low holding voltage) that instead has a high holding current/voltage (cuts off earlier).
  24. a simple question... so i have to create 8A DC for a very simple section of a circuit for 2-3 seconds. it draws it's power from the 120 VAC mains. i will rectify the AC and condition it witha capacitor, but i cannot use a voltage regulator and/or a transformer before or after the rectification (don't ask). so, i'll simply use a resistor to drop the voltage by 72V. the remaining, to the load, so the whole current is 8A. also, the load needs no regulation. the question is, can this resistor: (http://www.digikey.ca/product-detail/en/PF2472-8R2F1/PF2472-8R2F1-ND/2448327) handle nearly 600W, when this is only specified for 100W, considering it will only be conducting for only 2-3 seconds? duty cycle is 1-2 every other hours, just in case if the resistor will deteriorate anyways, i have an estimate how long it will last.
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