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Design a Zener regulator, regulated power supply using a 12V zener diode. When source voltage range from 15V to 20V and the load current ranges from 20mA to 100mA.

a. Determine proper choice of Ri

b. Determine maximum power dissipated in the Ri

c. Determine the output voltage Vout min and Vout max if Zener Diode has a resistance Rz = 3 ohms

Here is the circuit diagram :

schematic

simulate this circuit – Schematic created using CircuitLab

My question is how do I determine RL? As far as I can tell the information about range of input voltage and range of load current will help me calculate RL. But how to do it? Also, the min and max limit of current through load is given, for what value of Vi will load current maximum and for what value of Vi will the load current be minimum?

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Look over the following two schematic conditions:

schematic

simulate this circuit – Schematic created using CircuitLab

Your value of \$R_i\$ must be small enough to support a load current of \$100\:\textrm{mA}\$ (which implies a load of \$\tfrac{12\:\textrm{V}}{100\:\textrm{mA}}=120\:\Omega\$.) You also have to support the minimum needed current through \$D_1\$ at this time, as well. For a typical zener like this, that value can span over some range. But for good regulation, you probably want \$I_z\ge 10\:\textrm{mA}\$. And \$R_i\$ must do all of this while the voltage supply is at its minimum value of \$15\:\textrm{V}\$. So:

$$R_i\le\frac{15\:\textrm{V}-12\:\textrm{V}}{100\:\textrm{mA}+10\:\textrm{mA}}\approx 27.27\: \Omega$$

A standard value of \$27\:\Omega\$ would do.

Note that I had to place this circuit at the extremes in order to make sure that \$R_i\$ was having to cope with the worst possible supply circumstance (minimum) it might experience when trying to deal with the worst case load circumstance (maximum.)

Now that we know that \$R_i=27\:\Omega\$, the second schematic allows us to change things up in order to compute the worst case power in \$R_i\$ (and in \$D_1\$.) The power in \$R_i\$ will be the same, regardless of the load current. But I'm setting the load current to a minimum now, while setting the supply to its maximum, in order to see the worst case for \$D_1\$ (whose power does depend on the load current.)

In this case, the current through \$R_i\$ is:

$$I_{R_i}=\frac{20\:\textrm{V}-12\:\textrm{V}}{27\:\Omega}\approx 300\:\textrm{mA}\approx 2.5\:\textrm{W}$$

The power in \$R_i\$ is then about \$27\:\Omega\cdot\left(300\:\textrm{mA}\right)^2\$

Since only \$20\:\textrm{mA}\$ of that is going through the load, the remainder, or \$280\:\textrm{mA}\$, is left to go through \$D_1\$. So the power required by the zener diode is \$12\:\textrm{V}\cdot 280\:\textrm{mA}\approx 3.4\:\textrm{W}\$.

That covers (a) and (b) pretty well.

Now the problem is about (c). Before I engage that issue, I want to say that the datasheet for the 1N4742A says that when using \$21\:\textrm{mA}\$, \$R_z=9\:\Omega\$. This would tend to mean that if someone is telling me that \$R_z=3\:\Omega\$, the value I used for the minimum \$I_z=10\:\textrm{mA}\$ wasn't enough. It should be much higher. And this would tend to mean that I didn't compute \$R_i\$ correctly. However, your problem doesn't require this particular zener and so I can't debate or argue the facts, as given. The problem also didn't state a minimum zener current to use. So this allows me to keep my calculations and just move on.

Since the dynamic resistance of the zener is given as \$R_z=3\:\Omega\$, and since I can see that the dynamic change in \$I_z\$, from minimum to maximum, is \$300\:\textrm{mA}-10\:\textrm{mA}=290\:\textrm{mA}\$, I can multiply these two to get \$870\:\textrm{mV}\$ variation at the zener. Or, if I'm lucky with the zener, I may get \$12\:\textrm{V}\pm 440\:\textrm{mV}\$ at the output over all design circumstances (excepting thermal drift and aging and part variation, which also don't seem to be part of the problem here, but would be in the case of a real circuit.)

But let's do this a different way. We can insert \$R_z\$ into the circuit:

schematic

simulate this circuit

Now, that's not an exact reality. I just picked an arbitrary \$11.55\:\textrm{V}\$ for the ideal zener value in order to make things work out to center on \$12\:\textrm{V}\$ at the output. The reality will be different. Worse, the dynamic range of current through the zener is a factor of 30! There is no possible way that \$R_z=3\:\Omega\$ for such a dynamic range of current. It might be valid had I chosen a minimum \$I_z\ge 100\:\textrm{mA}\$, perhaps. Which would have resulted in a different value for \$R_i\$. But the problem doesn't care, so I don't care. Had it listed a specific part where I could examine a datasheet, a more accurate idea might have developed from here.

But the above works for homework, I suppose.

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  • \$\begingroup\$ There is an easier way \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 14 '16 at 21:20
  • \$\begingroup\$ @TonyStewart.EEsince'75 To teach? That's all I was trying to do. Or to calculate? (If "calculate," then of course there are easier ways. I could have written all I wrote in a couple of sentences, at most, then.) But briskness doesn't teach well. \$\endgroup\$ – jonk Nov 14 '16 at 21:21
  • \$\begingroup\$ No to teach, and this method works for high power LED's with an ESR. Rs= ΔVi/Imax -ESR \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 14 '16 at 21:24
  • \$\begingroup\$ @TonyStewart.EEsince'75 Cool. We each have our own ways of thinking and I guess its up to the OP to figure out what works for them. I gave my shot at teaching, already. \$\endgroup\$ – jonk Nov 14 '16 at 21:26
  • \$\begingroup\$ you explained very well, I"m still trying convince the masses how simple this works with LED, batteries and any shared application where there can be thermal runaway. In this case the differential voltage and both voltage source+sink add in series, same for driving LEDs compute with the Vzk@10%If and all the ESRs of driver, supply and LEd are subtracted from Ohms Law result \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 14 '16 at 21:28
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Since you know the regulated voltage is 12V you can calculate the minimum and maximum resistance of the load, which will change, from the currents given. If you design the regulator correctly, the minimum and maximum of Vi should not matter. That is the purpose of a regulator. If you have the minimum to the maximum load current anywhere from the minimum to maximum input voltage, the out put should always be 12V.

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  • \$\begingroup\$ Yes I agree, but to calculate Ri, we will have to have a fixed RL, isn't that correct? \$\endgroup\$ – Ankit Sahay Nov 14 '16 at 17:25
  • \$\begingroup\$ In this question, i have calculate the RLmin = 120 ohms and RLmax = 600 ohms But without having a fixed RL, how do I proceed for calculation of Ri? Should I calculate it for both values of RL to get a range of Ri as well? \$\endgroup\$ – Ankit Sahay Nov 14 '16 at 17:26
  • \$\begingroup\$ The zener diode will shunt any additional current around the load to keep the voltage at 12, so long Vi-I*Ri>12. You do not need to have a fixed load. You just need to make sure for all loads, the voltage at the cathode is trying to be higher than 12. If it is less then 12, it is the same as if the zener was not there. If it is more than 12, the zener will take enough current to keep its cathode at 12V. \$\endgroup\$ – owg60 Nov 14 '16 at 17:36
  • \$\begingroup\$ But neither do I know 'Ri', nor do I know 'I' how do I proceed with just range of Vi? \$\endgroup\$ – Ankit Sahay Nov 14 '16 at 17:42
  • \$\begingroup\$ If the circuit is regulating then the cathode is at 12V. If the load is taking 100mA then at least 100ma is flowing through Ri. Do you see that you can use Ohms law to calculate the value of Ri? Now if you calculate it at both 15 and 20, which do you have to use to make sure the circuit is 12V over all the conditions given? \$\endgroup\$ – owg60 Nov 14 '16 at 17:54
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My solution without a Load regulation voltage tolerance in the question.

  • RL(min) = 120 ohms= Vz/ILmax = 12/100m
  • Ri(min)+Rz = ΔVi/Imax= 30Ω = (15V-12V)/100mA
    • n.b. notice how I included the Rz with the Ri for voltage drop.
      • this works for LEDs where a single LED is ~1/Pd (package) using Vth 2.8V for most White LEDs, then Vf increases with ESR*If then x by number in string, divide by strings in parallel and few people appreciate this Rule of Thumb
  • thus Ri = 30-Rz= 27 Ω
  • P(Ri)max = (20-12)^2/27 Ω = 2.37 W

In the real world , 1st we define Vout, Iout and Vtolerance then Ri from ΔVi, ΔIL.

Then power dissipation will determine what type of Zener is needed. The load must not produce an R ratio below the \$V_{Zmin}\$ nor exceed Rmax such that the \$V_Z*I_Z*R_{jc} \$exceeds the allowed temperature rise of the case for a thermal resistance of \$R_{jc}\$ and heatsink , \$R_{ca}\$

But you have an existing circuit with fixed Vi,Ri and Vz one way is to determine\$ R_L(min) \$ is where \$\frac{R_L*Vi}{R_L+Ri}=V_Z @I_z \$ Such that If =(Vi-Vf)/Ri and \$I_i=I_Z+I_L\$.

Since Rz is never fixed. See below for a suitable Zener for this question.

enter image description here

Rz is defined as Zzt as Zener Threshold at some test current @10% to 25% of Imax

Ztk is the knee threshold at 1mA which has a much higher Ztk=125 Ω here vs Zzt = 2.5 Ω

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    \$\begingroup\$ stewart I calculated RLmin by RLmin = Vz/ILmax = 12/100m = 120 ohms RLmax = Vz/ILmin = 12/20m = 600 ohms Is this correct? If yes, what should be my next step? \$\endgroup\$ – Ankit Sahay Nov 14 '16 at 17:49
  • \$\begingroup\$ So what is Ri for ((120+Ri)/120=15/12V ? Ri=30 Ω next? subtract Rz=3 Ω and Ri=27 Ω next Pmax ok? its that easy. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 14 '16 at 19:56
  • \$\begingroup\$ -1 to the non contributors \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 14 '16 at 22:48

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