Riddle (Am I going crazy?) - Page 1
Forum section: electronics / general

author: staze
author_rank: Frequent Contributor
timestamp: December 28, 2014, 07:30:02 am
content: Okay, have a bit of a riddle that I can't seem to figure out. So, here goes.Device has 3 test points, we'll call them A (V+), B (Gnd), and C (?).Measuring between A, and B, I have +3.4V. Measuring between B and C, I have ~2V. Now, logic would say between A and C, I should have 1.4V, right? Nope. I have 2.3mV. Between A and C, there is an LED. 1.4V makes sense for the drop across an LED. But why would I see that drop by measuring each side of a diode relative to GND, but NOT when relative to itself and/or V+? And no, the diode doesn't light. And yes, checking it with a meter diode check makes it light nicely. Also checked for shorts, there aren't any I can find.What am I missing? It's got to be something obvious...* For those interested, this is a cheap Chinese toy top that has several LEDs on it that are supposed to light when turned on. My 16 month old son loves it, and I thought I'd try to fix it, to no avail (so far). And thinking maybe it was doing some PWM for LED brightness, I checked my scope, and nope, nothing (that I can see).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 28, 2014, 08:02:51 am
content: Is your meter 10Mohm?If it were unlimited impedance (truly zero current draw), C-B should read very near 3.4V, and A-C should read very near 0V. It may not be quite 0V (e.g., the 2.3mV) because of leakage current drawn through the LED: I presume there is a switch somewhere to cause the LED to light, which will draw some leakage current. If it's actually a mechanical switch, this will be near zero, so the voltage will be very small; if a semiconductor (transistor collector/drain, or the same inside an IC), perhaps on the higher side (mV to fractional V?).Real meters do draw some current. How much should be stated in the spec sheet, usually as a typical (if low precision) resistance (you find current by I = V/R), which may vary depending on range (often, the most sensitive range is "Hi-Z", meaning it has a very much higher equivalent resistance).Imagine what you should measure if the node is not an ideal voltage source, but both the meter and the node have series resistance...If Thevenin were correct in this case (unlikely, given the nonlinear nature of the LED), A-B is 3.400V, A-C is 0.0023V (therefore, B-C from the "high" measurement was 3.3977V), and B-C is 2.000V exactly, and your meter is 10M exactly, you would expect the "C" node has a Thevenin equivalent voltage (relative to B) of 3.944V and a series resistance of 6.98M.Tim

author: staze
author_rank: Frequent Contributor
timestamp: December 28, 2014, 08:58:39 am
content: Quote from: T3sl4co1l on December 28, 2014, 08:02:51 amYes, the meter I was using is 10Mohm. Wow, okay. So to check that, I just tried on my Keithley 196 in the 3V range (which is a ">1Gohm input impedance"), and indeed, I'm seeing more like 0.9V drop across the diode. To clarify, this is when the device is powered on (with a mechanical switch).I really thought most of this stuff only mattered when dealing with small voltages/currents. I'm still not sure I understand WHY this is happening with this measurement.And to clarify, just incase the A,B,C's are annoying.Anode of LED to Gnd: 3.4V. Cathode of LED to Gnd, 2V (2.5V at >1Gohm). Anode to Cathode, 2.3mV (3.8mV on Keithley 196 in 11Mohm, 0.9V in >1Gohm). So yes, obviously this is happening, I'm just still not clear on WHY.OH... wait. Is it the fact that a CC source/sink (what they're probably using to drive the LEDs (I can't tell, it's potted on the board)) is so weak (voltage wise) that by putting a meter across the LED, I've basically dragged its voltage UP to 3.4V (basically) because the meter does indeed let SOME of that voltage through (it's not infinite impedance) (and by dragging it up that much, there's "no" PD)? Then I would have to assume that I'm also dragging it down when I go Cathode to Gnd, but it must have a "bottom" of about 2V?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 28, 2014, 09:43:53 pm
content: The LED is unlit during this test, right?An LED is very nonlinear; its current varies as I = Is * exp(Vf*N/Vth), where Is is a constant determined by material properties and varies somewhat with temperature, Vf is the forward voltage, N is a tweak ('emission coefficient', usually 1-2) and Vth is the thermal voltage (= k_B*T / q_e ~= 26mV at room temperature). If you define the resistance as R = V/I, the resistance varies strongly with current and voltage. At low current, it's very large, so the voltage varies all over, while current remains low.There's also leakage current due to ambient light; an LED is a poor photocell (solar panel), but it does generate current nonetheless. 0.9V seems consistent with leakage current in the circuit or photocurrent.Tim