1
\$\begingroup\$

i have searched in over voltage protection (like mosfet ) , i wanted to use zener diode parallel so it wont dissipate power unless in over voltage protection , but its power is low so i read that i needs resistor to control the current passing through it .

i found people put the resistor in this shape : enter image description here

but this resistor will dissipate power , so i thought if we can put resistor like this :

enter image description here

so in this situation resistor (R3) will only dissipate in case of over voltage but in desired voltage current wont pass through zener so wont pass through resistor .

so my questions are :

is the voltage going to led equals to source or to zener ?

is the new placement for resistor right? if wrong why?

is there another placement for resistor if this doesnt work ?

|improve this question
\$\endgroup\$
3
\$\begingroup\$

The first circuit you show, with the resistor in-line with the load current, is what you require. That way, the Zener diode will limit the voltage, and the excess voltage will be dropped across the resistor.

In your suggested circuit, the excess voltage will also be dropped across the resistor, but since the resistor is in series with the diode, and the combination is in parallel with the load, the load will still see the full voltage - the load voltage will not be limited by the Zener.

|improve this answer
\$\endgroup\$
  • \$\begingroup\$ thanks for your answer , i have more questions in the first picture ,does load current affects on resistance ? when choose resistance value i will take only the max current of load in my mind ? , and does resistor voltage drop affected by current drawn by load? \$\endgroup\$ – ahmed shafie Sep 23 at 0:39
  • \$\begingroup\$ In the first circuit, the load current flows through the resistor, so affects the voltage drop across the resistor - all determined by Ohm's Law. \$\endgroup\$ – Peter Bennett Sep 23 at 2:13
  • \$\begingroup\$ thanks sir @Peter Bennett \$\endgroup\$ – ahmed shafie Sep 27 at 12:24

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.