0
\$\begingroup\$

i am designing a circuit able to draw IV curves of LEDS and diodes through Arduino.

Basically i use the PWM function and an RC integrator to have an approximately constant voltage. Then i would simply connect the LED via a shunt resistor to measure current and voltage; however to not affect the integrator R_shunt must be greater than integrator resistor, but i can't use too big resistors otherwise current would be too small (max current = (5 Volt-V_breakdown,diode)/(R_shunt+R_integrator) right?). Then i put a transitor in emitter follower configuration to increase load impedance.

Problem is; i don't understand very well what i am doing (altough i thought i did). In emitter follower the voltage taken on a load after the emitter is V_base- 0.7 . I got this, but then i don't understand HOW the voltage at the collector influence the circuit, and why using now very big resistors in RC integrator (like 1 MOhm) current is low on the shunt resistor, say 0.7 mA instead of usual 14 mA. I would say: current in the shunt resistor is (excluding led for one moment) (V_base-0.7)/R_s but now seems that resistor at the base influences this current. How?

Schematic of the circuit

EDIT:The shunt resistor in the schematic is Rs, sorry for the confusing names. About collector voltage, i wasn't sure it would influence emitter current, that was the question, maybe that's the reason of misunderstanding. Actually i think i found the solution revising Horowitz's book. In emitter follower the output impedance (ie seeing emitter as the output with a Z_load) is Z_(load)*(beta+1), whereas the resistance on the base R becomes an input resistance of R/(beta+1). So for R in MOhm, considering the Thevenin equivalent i have a total resistance of R/(beta+1) + R_shunt in which a voltage V_base - 0.6 is flowing, that's how R of integrator influences the circuit. Am i right?

\$\endgroup\$
  • 1
    \$\begingroup\$ You talk about an R_shunt which is not in the schematic, that is confusing. How do you see that the voltage at the collector ( = 5 V) influences the circuit? How do you conclude that? When R becomes 1 M ohm the base current of the NPN becomes very small that means that the current coming from the emitter (through the LED) will also be small, at best it will be 500 times large if the NPN has such a large beta (not all do). It might be easier to use replace the NPN with a low Vt NMOS, for example an AO3400. \$\endgroup\$ – Bimpelrekkie Feb 19 at 14:09
  • \$\begingroup\$ Offhand, the only way I can see for the collector voltage to significantly affect the output is if your transistor is in saturation. If you measure the collector voltage and V1, you'll want them to be at least half a volt apart. \$\endgroup\$ – Cristobol Polychronopolis Feb 19 at 16:04
  • \$\begingroup\$ I edited the post, the comment was too long \$\endgroup\$ – Lenz Feb 19 at 17:19
  • \$\begingroup\$ How accurate do you want these curves to be? Is this just a "binning" process where you will sort them in some way? Also, why not consider a variable current source/sink for the LED and just measure the voltage? This can be simple but without a lot of initial accuracy to it and where ambient temperature may be another confounding problem; or you can add a way to calibrate it for better, more uniform results between multiple circuit-builds and where temperature is accounted for, better. Write more about what you need and why you need it? \$\endgroup\$ – jonk Feb 19 at 17:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.