Measuring Input impedance - Page 1
Forum section: electronics / beginners

author: boriskorbel
author_rank: Newbie
timestamp: August 09, 2019, 03:10:48 pm
content: Dear friends,In my project, I deal with the design of the EH circuit for thermoelectric generators. As a reference I use the circuit from this publication (https://ieeexplore.ieee.org/document/8119850) - see appendix. My goal is to adjust the circuit so that the internal resistance of the thermoelectric generator corresponds to the internal resistance of the load. If this equality is achieved, the best performance can be obtained from TEG.Could you help me with the procedure of how to measure the input impedance of the circuit and eventually identify elements that can alter the internal resistance of the circuit? (I tried to connect more NFET transistors in parallel but without any significant results).So far I tried to connect function generator (DC bias voltage (50-ohm output load)) on the input and measure the internal resistance as a voltage drop across the resistance divider (50 ohm and internal resistance) but I was getting weird values.Thank you in advance for any ideasBoris

author: Rerouter
author_rank: Super Contributor
timestamp: August 09, 2019, 03:32:25 pm
content: A TEC has a non linear impedance, unless you have very steady state conditions, I do not know if you can find a singular impedance,it would be closer to a maximum operating point converter for a solar cell, they operate very similar to a current source, with a voltage dependant on the temperature delta, so it would be more working towards the best combination of voltage and current.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 09, 2019, 03:36:32 pm
content: So, a boost converter driven with constant frequency and duty cycle?For Vo >> Vin / D, the inductor will fully discharge every cycle, i.e., operate in DCM (discontinuous conduction mode). That means every cycle, the inductor is charged to;Starting with the inductor equation:V = L dI/dtV = V_in, dt = t_on, and L are given. Rearrange:dI = V_in t_on / LInitial current is zero, so dI is the peak current at turn-off.The energy in the inductor is:E = 0.5 L I^2Substituting in I = dI,E = 0.5 L (V t_on / L)^2= 0.5 V_in^2 t_on^2 / LThis is delivered at F_sw pulses per second, or a power ofP = V_in^2 t_on^2 F_sw / LThe input current is I_in = P / V_in orI_in = V_in t_on^2 F_sw / LThe input DC resistance isR_in = V_in / I_in= L / (t_on^2 F_sw)Since D is given, I shouldn't have dragged along t_on; we can substitute t_on = D / F_sw, where D = 0.5 for this part.R_in = L / (F_sw / (2 F_sw)^2) = L / (1 / (2 F_sw))= 2 L F_swFor 1mH and 32kHz, we get 65.5 ohms.In general, we use a switching controller rather than a fixed oscillator, giving us a fully controllable R_in (typically, D is varied, or F_sw as well).Note that a supply bypass capacitor is required. If the AC input current is dissipated by the TEG, lower efficiency will result. Simply enough, place a capacitor across the TEG, with value C >> 1 / (2 pi F_sw R_in), or, say, 100uF. Choose a low-ESR type (ESR << R_in).Tim

author: boriskorbel
author_rank: Newbie
timestamp: August 10, 2019, 09:32:47 am
content: Thanks for the detailed explanation, so if I understand it well boost converter with fixed frequency is not a good solution, right?QuoteBy switching controller, you mean controlling the duty cycle by MCU according to connected load?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 10, 2019, 10:25:45 am
content: Yes. Or an MCU may be difficult to write a correct control for, in which case an integrated controller is easier to use. TI, Analog Devices, and others, make controllers and regulators of this sort.Tim