2
\$\begingroup\$

I have to simulate the reverse bias curve in Multi Sim with a 1N4681 I understood the equation to work out the series resistor to be:

$$R_s = \frac{V_S-V_Z}{I_Z}$$

Which in my case would create a 101 Ohm resistor.

However, this doesn't seem to be creating the characteristic curve that I was expecting.

Datasheet I am using for reference

schematic

simulate this circuit – Schematic created using CircuitLab

Therefor, am I selecting the correct value for the series resistor? And when you're making the above calculation with the view of performing a DC Sweep what Voltage do you use for the source voltage calculation? Mean voltage? Max Voltage? Arbitrary value?

\$\endgroup\$
1
\$\begingroup\$

First, you have to consider that every (Zener) diode has a power rating. From this power rating, you can calculate a maximum current for the diode. E.g., if a Zener diode has a maximum power rating of \$ 1~\text{W} \$, and a voltage rating of \$ 2~\text{V} \$, the maximum current through the diode should be \$ 0.5~\text{A} \$. If you want to connect this diode to a voltage supply of \$ 5~\text{V} \$, then you need something (a resistor) that will create a voltage drop of \$ 3~\text{V} \$:

$$ R = \frac{5~\text{V} - 2~\text{V}}{0.5~\text{A}} = 6~\Omega $$

Power rating for this resistor should be at least \$ P_R = I^2 R = 1.5~\text{W} \$.

Now, you want to place some load in parallel to the Zener diode. This will reduce the current through the diode, which will result in a lower voltage. To calculate exact operating point, i.e., voltage drop at the Zener diode, we need more information on this diode, like its \$U\$-\$I\$ static characteristics.

As for the 1N4681 diode datasheet, you can see that its maximum voltage is \$ V_{\max} = 2.52~\text{V} \$, and its minimum voltage is \$ V_{\min} = 2.28~\text{V} \$. The actual operating voltage will depend on the current through the diode, which depends on the series resistor, as well as on the load that is in parallel to the diode. You can also see that the maximum current is \$ I_{\max} = 0.095~\text{A} \$. The maximum voltage would occur for maximum current through the diode. Also take into account that the power rating for \$ 100~\Omega \$ resistor should be at least \$ 1~\text{W} \$.

I've looked through your model - everything is fine except for \$ R_S \$ parameter in the diode. Change this parameter to \$ 0~\Omega \$, and everything will be fine. For this particular diode, you may want to set the series resistor to \$ 80~\Omega \$.


Voltage drop on the Zener diode at the breakdown voltage region can be approximated using a linear function, as follows:

$$ V_D(I_D) = k I_D + c $$

where \$k\$ and \$c\$ are the diode parameters. The voltage equation for the system is as follows:

$$V_S = I R + V_D(-I), \quad I_D = -I ,$$

where \$V_S\$ is the voltage source, \$ I\$ is the system current, and \$ R\$ is the series resistor. Combining these two equations, we get the \$U\$-\$I\$ static characteristics of the system:

$$V_S = I(R-k) + c$$

\$\endgroup\$
  • \$\begingroup\$ I think what is confusing me here is that in the example handout I achieve this graph gyazo.com/b918afb5d69ed8c48e2da99d1e3318ca whereas in the run the circuit I get this gyazo.com/f55b1c3585ffc770d37510c9a9ac8cbe - I understand in the first graph the diode has reached its breakdown point. \$\endgroup\$ – Dhatsah Oct 8 '16 at 12:00
  • \$\begingroup\$ I don't really understand these graphs, you should explain the axis. I extended my answer to explain how to calculate the U-I static characteristics of the system. \$\endgroup\$ – Marko Gulin Oct 8 '16 at 13:13
  • \$\begingroup\$ The Graphs are taken from a DC sweep 0v to 50v on the X axis and the Y Axis being the voltage at PR1 or PR2 \$\endgroup\$ – Dhatsah Oct 8 '16 at 14:44
  • \$\begingroup\$ Are you sure that both of these 2 circuits are the same? First, I can see that series resistor values are different (\$80~\Omega\$ and \$100~\Omega\$). Second, are you sure that you've removed \$ R_S \$ parameter in both circuits? \$\endgroup\$ – Marko Gulin Oct 8 '16 at 14:47
  • \$\begingroup\$ No - the circuits are not the same, they should both demonstrate the characteristic curve of the diode though. \$\endgroup\$ – Dhatsah Oct 8 '16 at 14:51
0
\$\begingroup\$

For optimal design , you need to know;

  • Iout Min/Max
  • Vout Min/Max
  • Vin Min/Max

    • also Zener Pd vs temperature rise limit and ESR or Rs of Zener is helpful for low drop R if low drop V
    • But in general, Zener current must dissipate only the difference between Iout max-min. plus the minimum current to stay regulated for Vmin.

Thus at Imax out Zener uses the minimum current for Vmax and when at Imin the Zener must dissipate VI max heat for Vout max.

  • e.g. Imax 0.1A , Imin=0.09A Iz=20mA @4.7V @19Ω therefore 4.5V@Iz=10mA
  • If Vmin=4.5 then Iz= 10mA +(100mA-90mA) =20mA

    • But if Iout min=0, Iz= 10mA + 100mA. n.g. since Iz max=75mA @400mW
\$\endgroup\$
0
\$\begingroup\$

To choose the right ballast resistor you need to start with four things:

  1. The supply voltage, \$V_S\$
  2. The load current, \$I_L\$
  3. The Zener voltage, \$V_Z\$
  4. The Zener test current, \$I_Z{}_T\$

The basic circuit is shown below, and shows the currents in the various parts of the circuit.

\$I_Z{}_T\$ is the Zener test current, and is the current through the Zener which will guarantee the specified voltage drop across the Zener.

\$I_L\$ is the load current, and R1 is chosen so that

$$ R1 = \frac{V_S-V_Z}{I_Z{}_T+{I_L}} $$

D1 is a shunt regulator, and R1 provides a quasi constant source of current, so if the load current decreases, the Zener shunts that amount of "lost" current to ground so the voltage across the load will remain constant.

For example, if the load current falls to 19mA, \$I_Z\$ will increase to 21mA.

The converse is also true, in that if the load current increases to 21mA, \$I_Z\$ will fall to 19mA.

The opposite is true if \$V_S\$ changes, with \$I_Z\$ increasing when \$V_S\$ increases and falling when \$V_S\$ falls.

At the end of it all is what happens if your load fails open and \$V_S\$ is worst case high.

In that case, the Zener would be dissipating $$ P = IE= 40mA\times 4.7V = 188 \text { milliwatts} $$ so, since it's capable of dissipating half a watt, it should be safe.

enter image description here

\$\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.