1
\$\begingroup\$

enter image description here

I am an electronics student, and I was set some homework online. I am not having trouble with the question itself but the theory. It says that the current through the resistor is the same as the maximun current that the diodes can safely handle, but as you can see, there are (\$V_{out}\$) volts being dissipating to the right, and a minimum required current going through the diode to maintain consistent voltage.

The thing is, how could the current through the resistor be the same as the current through the diode, as \$Voltage=J/C\$, or, joules per coulomb, so that means there must be some energy carried by the coulombs, and ultimately, some electrons. But then this is contradictory, wouldn't the current going through the resistor be the total current, the current dissipated to the right and current through the zenor? Or... is it that there isn't, then, any current passing at \$V_{out}\$? and it doesn't represent voltage being dissipated? I cant wrap my ahead around it...

\$\endgroup\$
2
\$\begingroup\$

When designing a zener circuit I usually try to meet the following requirements:

  1. The zener circuit works without load. This means that the zener diode must be able to dissipate the power that is normally dissipated in the load. Check the datasheet for the zener for maximum dissipated power. A regular small zener diode is often rated for 400mW. At this point you know the zener voltage and its maximum power, therefore you can calculate the maximum current through the diode: \$I_{Z,max}=\dfrac{P_{max}}{V_Z}\$.
  2. The zener is in its regulating mode when maximum load is attached When a load is attached, the current through the zener is lower as a part of the current flows through the load. However a zener needs a minimum current to 'actively' regulate the voltage across it. For an average zener diode this is around 1-5mA, but again you should check the datasheet for this.
  3. The known maximum load current Of course if the maximum load current is lower than the maximum zener current (from #1), there is no need to stress the zener to its max. As long as the minimum zener current is guaranteed, because otherwise it stops regulating the voltage across it.
  4. Maximum load current From the above it can be determined that \$I_{L,max} = I_{Z,max} - I_{Z,min}\$. Or in words: the sum of load current and zener current must be lower or equal to the maximum zener current.

Now if VS is given, you can calculate the series resistor \$R = \dfrac{V_S - V_Z}{I_L+I_{Z,min}}\$

\$\endgroup\$
2
\$\begingroup\$

If there is no current drawn from \$V_{out}\$, then by Kirchoff's Current Law, the same current must flow in both the resistor and the Zener diode, and you want that current to be not more than the Zener's maximum rated current, to prevent damage to the Zener diode.

If some current is taken from \$V_{out}\$, then the current through the resistor will indeed be greater than the current through the Zener diode. If the question didn't say anything about current from \$V_{out}\$, then you have to assume that the current is zero.

In Real Life, if you were designing a circuit using a Zener diode, you would probably be aware how much current would be drawn from \$V_{out}\$, and would allow for that in your resistor calculation, typically aiming for a Zener current somewhat less than its rated maximum current.

\$\endgroup\$
  • \$\begingroup\$ Not to mention that the knee of a Zener diode is a curve rather than a sharp point, so the breakdown voltage is dependent on the current that passes through it. \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 15 '13 at 0:27
1
\$\begingroup\$

It says that the current through the resistor is the same as the maximun current that the diodes can safely handle

The text isn't saying the current through the resistor is equal to the current in the diode. It's saying the current in the resistor is equally to what the diode could theoretically (not necessarily at this time) handle.

And yes if the current through the resistor would indeed equal that in the diode, then current wouldn't flow through V out (Kirchhoff's current law).

\$\endgroup\$
0
\$\begingroup\$

Volts don't get dissipated. Basically, that Vout is hooked across an infinite resistor so all of the current that is going across the resistor must also go through the zener diode because they are in parallel series.

\$\endgroup\$

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.