flippityflop Posted January 9, 2015 Report Share Posted January 9, 2015 ok, i need a simple DC comparator. so would the attached pic do the job??all resistors are equalEDIT 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. Quote Link to comment Share on other sites More sharing options...
flippityflop Posted January 9, 2015 Author Report Share Posted January 9, 2015 or do we have to have wheatstone bridge and more complex ones for the general cases?? Quote Link to comment Share on other sites More sharing options...
Hero999 Posted January 11, 2015 Report Share Posted January 11, 2015 That's not a comparator circuit.This is a fairly simple comparator circuit. Quote Link to comment Share on other sites More sharing options...
flippityflop Posted January 12, 2015 Author Report Share Posted January 12, 2015 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. Quote Link to comment Share on other sites More sharing options...
Hero999 Posted January 13, 2015 Report Share Posted January 13, 2015 How on earth is that a comparator? It's a potential divider.A comparator compares two inputs and turns on when one is greater than the other. It has a digital output and two analogue inputs. Quote Link to comment Share on other sites More sharing options...
flippityflop Posted January 14, 2015 Author Report Share Posted January 14, 2015 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: Quote Link to comment Share on other sites More sharing options...
pebe Posted January 14, 2015 Report Share Posted January 14, 2015 The current through resistor Q is zero. Your circuit is meaningless. Quote Link to comment Share on other sites More sharing options...
Hero999 Posted January 16, 2015 Report Share Posted January 16, 2015 No, I don't know what you're talking about. The circuit doesn't have a common 0V reference.Yes, the current through Q is zero because the two batteries are only connected together via one wire and there needs to be two wires to complete the circuit. Quote Link to comment Share on other sites More sharing options...
flippityflop Posted January 22, 2015 Author Report Share Posted January 22, 2015 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=39704.0basically, 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_gnd2or (V_ref_gnd2 - V_ref_gnd1) - (V_ref_pos2 - V_ref_gnd1) if V_ref_gnd1 < V_ref_gnd2notice 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. Quote Link to comment Share on other sites More sharing options...
pebe Posted January 22, 2015 Report Share Posted January 22, 2015 Where are you going with this discussion? Your posting is not meaningful unless it is aiming for a conclusive point.So far, I have not seen anything about the subject of the thread, ie. "comparator for DC voltage". Quote Link to comment Share on other sites More sharing options...
Hero999 Posted January 23, 2015 Report Share Posted January 23, 2015 It isn't a comparator.A common ground is needed because all voltages in a circuit are relative to one another. Measuring potential difference is like measuring distance: you need to start from somewhere. Quote Link to comment Share on other sites More sharing options...
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