# Voltage Regulators - Problem 4.a

#### PsySc0rpi0n

Joined Mar 4, 2014
1,706
This is my new problem to solve!

See attached picture!

The problem asks to calculate Vo and Ro for the given circuit. The zener has an internal dynamic impedance of 100 Ω. I'm considering that these 100 Ω is a resistor in series with a battery, simulating an approximation model for that zener. I'm only testing the voltage supply value of 15 V to check if the method is ok or not... Then I'll just plug in the 15 V - 1 V and the 15 V + 1 V to actually check the final results!

I already tried to solve this using the Thévenin equivalent method.

Vth would be the zener voltage plus the voltage drop across the 100 Ω resistor.
I'm calling this resistor as Rz.

So, I evaluated a symbolic equation for the net closed loop, without the Load resistor. Being I, the current flowing through this branch:

Vin = I*R1 + Vz + I*Rz

I = (Vin - Vz) / (R1 + Rz)

So Vth:

Vth = Vz + I*Rz
Vth = Vz + Rz*( (Vin - Vz) / (R1 + Rz) ) ---- V + Ω*(V/Ω) = V, checks
Vth = 5.9 V

And Rth:

Rth = R1 || Rz = (1 kΩ*100 Ω) / (1 kΩ + 100 Ω) = 91 Ω

But then I tried using net closed loop equations and nodal equations.

For net closed loop equations I called the current flowing through R1 as I1 and I did:

-Vin + I1*R1 + Vz + Iz*Rz = 0 V

Solving this for I1, Iz and I_Load, I'm getting Iz = 0A and that doesn't adds up. It can't be... Where am I going wrong?

Then I tried a Nodal equation for the node between the zener and R1 and I called it V1, which is the Vo I'm looking for.

I did:

or

as the current flowing through the zener is the same as the current flowing through Rz!

Then I wrote that:

I1 = (Vin - V1) / R1
I_Rz = (V1 - Vz) / Rz

(Vin - V1) / R1 = (V1 / R_Load) + ( (V1 - Vz) / Rz)
Solving this for V1 I'm getting V1 = 5.42 V

So, I'm not getting consistent values! I need some help here!

#### Jony130

Joined Feb 17, 2009
5,457
Well from this equation
(Vin - V1) / R1 = (V1 / R_Load) + ( (V1 - Vz) / Rz) I get V1 = 5.855V. So as usually you made a error in math.
And since Vth = 5.9V (without load). So with load connected V1 is :
V1 = 5.9V*(10kΩ)/(10kΩ+0.091kΩ) = 5.846V close enough (rounding error).

-Vin + I1*R1 + Vz + Iz*Rz = 0 V

Solving this for I1, Iz and I_Load, I'm getting Iz = 0A and that doesn't adds up. It can't be... Where am I going wrong?
For this circuit we can write only two independent equations for KVL and one for KCL.
And this is why try solve this:

-Vin + I1*R1 + Vz + Iz*Rz = 0,
I1 - ILoad - Iz = 0

Last edited:

#### PsySc0rpi0n

Joined Mar 4, 2014
1,706
Well from this equation
(Vin - V1) / R1 = (V1 / R_Load) + ( (V1 - Vz) / Rz) I get V1 = 5.855V. So as usually you made a error in math.
And since Vth = 5.9V (without load). So with load connected V1 is :
V1 = 5.9V*(10kΩ)/(10kΩ+0.091kΩ) = 5.846V close enough (rounding error).

For this circuit we can write only two independent equations for KVL and one for KCL.
And this is why try solve this:

-Vin + I1*R1 + Vz + Iz*Rz = 0,
I1 - ILoad - Iz = 0

Ok, I got what I did wrong with the single KCL equation! I also got V1 = 5.855V

But why can't I write the 2nd equation for the KVL method, -Vin + I1*R1 + I_Load*R_Load = 0 V?

#### Jony130

Joined Feb 17, 2009
5,457
Suppose a circuit has N nodes and B branches, so we can write N - 1 independent KCL equations. And B - (N - 1) independent KVL equations.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,706
Hum, ok I forgot about that but then how many nodes and branches does this circuit has? Does GND + Vin + Rz + RLoad common point counts as a node?

If so, we have 2 nodes. That one including the GND and the node I called '1'. So we would be able to write 2 - 1 = 1 KCL independent equation and then 3 - 2 = 1 KVL ???? independent equation?

Or if GND doesn't count as a node then we wouldn't be able to write any KCL independent equation because N - 1 = 1 - 1 = 0...

#### sailorjoe

Joined Jun 4, 2013
364
Current flows through branches and into and out of nodes. Does current flow into and out of GND?

#### PsySc0rpi0n

Joined Mar 4, 2014
1,706
Current flows through branches and into and out of nodes. Does current flow into and out of GND?
It doesn't? Yesterday someone told me so!...

#### PsySc0rpi0n

Joined Mar 4, 2014
1,706
Suppose a circuit has N nodes and B branches, so we can write N - 1 independent KCL equations. And B - (N - 1) independent KVL equations.
I got confused with that!

Can you read my post #5 and clear my mind on this, please?

#### anhnha

Joined Apr 19, 2012
902
Hum, ok I forgot about that but then how many nodes and branches does this circuit has? Does GND + Vin + Rz + RLoad common point counts as a node?
The circuit has three nodes but there are only two principal nodes.
A principal node or junction is a point where 3 or more branches join.
With two principal nodes you choose one node as a reference and assign it a voltage of zero.
So there remains only one node for KCL.
There are four branches so the number of independent equations is 2.
http://mathonweb.com/help/backgd5.htm

#### PsySc0rpi0n

Joined Mar 4, 2014
1,706
@anhnha I got it for the nodes. But why you say there are 4 branches? I see 1 branch for the zener, 1 branch for RLoad and one branch for R1 and the voltage source.

#### anhnha

Joined Apr 19, 2012
902
@anhnha I got it for the nodes. But why you say there are 4 branches? I see 1 branch for the zener, 1 branch for RLoad and one branch for R1 and the voltage source.
1. Branch with Zener
3. Branch with R1
4. Branch with voltage source

#### PsySc0rpi0n

Joined Mar 4, 2014
1,706
1. Branch with Zener
3. Branch with R1
4. Branch with voltage source

But isn't branch 3 considered to be the same as branch 4 since the current is the same because they're in series?

Ok, anyway... To find Ro, we are used to replace R_Load by a voltage source named 'vx' that will "generate" a current 'ix', like we do in amplifiers!

As we are considering a zener model as a voltage source in series with a resistor, and as we use to replace DC voltage sources by it's internal impedance (0 Ω) for the evaluation of Ro, how can I write vx/ix to find Ro????

#### PsySc0rpi0n

Joined Mar 4, 2014
1,706
I edited my previous reply! I'm struggling to know how to start evaluating Ro as we use to do it for amplifiers!

#### Jony130

Joined Feb 17, 2009
5,457
Start by drawing a diagram, do not forget to replace DC voltage sources by it's internal impedance (Vz is also a voltage source).
Also notice that you already find Ro. Ro = Rth.

#### hp1729

Joined Nov 23, 2015
2,304
This is my new problem to solve!

See attached picture!

The problem asks to calculate Vo and Ro for the given circuit. The zener has an internal dynamic impedance of 100 Ω. I'm considering that these 100 Ω is a resistor in series with a battery, simulating an approximation model for that zener. I'm only testing the voltage supply value of 15 V to check if the method is ok or not... Then I'll just plug in the 15 V - 1 V and the 15 V + 1 V to actually check the final results!

I already tried to solve this using the Thévenin equivalent method.

Vth would be the zener voltage plus the voltage drop across the 100 Ω resistor.
I'm calling this resistor as Rz.

So, I evaluated a symbolic equation for the net closed loop, without the Load resistor. Being I, the current flowing through this branch:

Vin = I*R1 + Vz + I*Rz

I = (Vin - Vz) / (R1 + Rz)

So Vth:

Vth = Vz + I*Rz
Vth = Vz + Rz*( (Vin - Vz) / (R1 + Rz) ) ---- V + Ω*(V/Ω) = V, checks
Vth = 5.9 V

And Rth:

Rth = R1 || Rz = (1 kΩ*100 Ω) / (1 kΩ + 100 Ω) = 91 Ω

But then I tried using net closed loop equations and nodal equations.

For net closed loop equations I called the current flowing through R1 as I1 and I did:

-Vin + I1*R1 + Vz + Iz*Rz = 0 V

Solving this for I1, Iz and I_Load, I'm getting Iz = 0A and that doesn't adds up. It can't be... Where am I going wrong?

Then I tried a Nodal equation for the node between the zener and R1 and I called it V1, which is the Vo I'm looking for.

I did:

or

as the current flowing through the zener is the same as the current flowing through Rz!

Then I wrote that:

I1 = (Vin - V1) / R1
I_Rz = (V1 - Vz) / Rz

(Vin - V1) / R1 = (V1 / R_Load) + ( (V1 - Vz) / Rz)
Solving this for V1 I'm getting V1 = 5.42 V

So, I'm not getting consistent values! I need some help here!
What am I missing? That circuit can't work??? 5 V Zener with a dynamic resistance of 100 ohms looks like it will have 50 mA of current through it. That makes sense. But just that 50mA through the 1K ohms drops 50 Volts, but only 15 V is applied.

15 V applied. 5 V across the Zener means 10 V across the1K ohm, or 10 mA. Reasonable. 5 V across the 10K means 0.5 mA leaving 9.5 mA through the Zener. Zener effective resistance is 526 ohms??? not 100. What do I not understand about the effective resistance of a Zener? Is that 100 ohms a minimum value?

#### PsySc0rpi0n

Joined Mar 4, 2014
1,706
@Jony130 Ok I got it! I've done it.

@hp1729 don't take my words as if I was an expert. Further, we are in an academic context so some terms maybe not correctly applied by me. In fact, that 100 Ω resistor is to use when replacing the zener by a voltage source so that the equivalent model we use will be a voltage source in series with that resistor. So, give no importance to that dynamic term my teacher used.

#### hp1729

Joined Nov 23, 2015
2,304
@Jony130 Ok I got it! I've done it.

@hp1729 don't take my words as if I was an expert. Further, we are in an academic context so some terms maybe not correctly applied by me. In fact, that 100 Ω resistor is to use when replacing the zener by a voltage source so that the equivalent model we use will be a voltage source in series with that resistor. So, give no importance to that dynamic term my teacher used.
rthabnks for the patience.
" that 100 Ω resistor is to use when replacing the zener by a voltage source"
That's where you lost me. I can understand the Zener having an effective resistance. Just not an absolute value. A 5 V Zener with 100 ohms means about 50 mA. That makes sense but it will not always be 100 ohms. I can see that as a minimum resistance. The 1K resistor to 15 V means 10 V across the 1K, or 10 mA.
The 10K load with 5 V across it means 0.5 mA through the 10K. Leaving 9.5 mA through the Zener and it now has an effective resistance of 5 V / 9.5 mA, or 526 ohms, not 100 ohms.
????
Real world parts ... that 5 V Zener will not always have exactly 5.0 Volts across it. At 10 mA it will have a lower voltage than at 50 mA. A reality not apparent in your analysis. How can you make precise calculations, down to mV, with such conditions. See attached pdf excel file on a zener diode exercise. 1N5228.

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#### PsySc0rpi0n

Joined Mar 4, 2014
1,706
@hp1729 I understand what you're trying to say! And I know you're right! I don't know where you from and how these subjects are taught in schools of your country! I have already noticed that in my school there is a lack of precision when it comes to problems to solve. We ignore a lot of technical details that cannot be neglected in real life and in fact, some shouldn't be neglected even in the most basic problems!

But I'm sure you'll also understand that we are in an academic context where everything is possible and what really matters is to understand the theoretical working mode of the components. I have already debated this several times here in the forum. But that's how things are, at least in the school I'm in, unfortunately.

So, for solving this problem, our teacher used that technical term "dynamic" but he didn't meant that we should treat that resistance as a "dynamic" parameter. I think he just wanted to make a note that a zener has a resistance that is not constant for different conditions. And that value would be for a specific set of conditions and that we should only consider that set of conditions to be able to replace the zener by the approximation we are used to, that is a voltage source in series with that resistance!

Anyway, at the end of yesterday's class I asked the teacher to help me to understand how to find the output impedance and then when I started to talk to him, I immediately realized what I was missing that was preventing me to solve it and get the same result as I got using Thévenin theorem!

Now it's time to move to the next circuit that is yet part of this same problem nº 4 and it is supposed that we notice that between this circuit and the new one (that I'm going to start to analyse soon) there is an huge difference and that the regulation of this already finished circuit sucks and that the new circuit is a lot better when it comes to regulate voltage!

Thanks!

#### Jony130

Joined Feb 17, 2009
5,457
What am I missing? That circuit can't work??? 5 V Zener with a dynamic resistance of 100 ohms looks like it will have 50 mA of current through it. That makes sense. But just that 50mA through the 1K ohms drops 50 Volts, but only 15 V is applied.

15 V applied. 5 V across the Zener means 10 V across the1K ohm, or 10 mA. Reasonable. 5 V across the 10K means 0.5 mA leaving 9.5 mA through the Zener. Zener effective resistance is 526 ohms??? not 100. What do I not understand about the effective resistance of a Zener? Is that 100 ohms a minimum value?
Well, you miss the point completely. Also do not mix a static resistance with dynamic resistance.
As everyone knows the Zener diode has a knee-voltage, and we sometimes " mode" this voltage (in hand calculations) as a voltage source or we treat it that way when we write KVL. And of course everyone should understand the limitation of this basic "model". As we know the Zener voltage is not constant but it's slightly changing with the current. And to include this "effect" into our model we add a resistor in series with the voltage source. To model this "effect".
http://www.ittc.ku.edu/~jstiles/312/handouts/Zener Diode Models.pdf
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