# Volatge Regulators - Problem 1

#### PsySc0rpi0n

Joined Mar 4, 2014
1,446
Hello!

I'm working now on voltage regulators. I'm trying to practice for my exam.

The attached circuit is my 1st problem.

Problem asks to evaluate the voltage at R_Load, and the Zener current and the Power at the Zener. I'm not sure about the Power on the zener because the Pmax is given and it's 0.25 W.

The voltage supply is supposed to be unregulated and the voltage can vary from 18 V up to 22 V.

We learn that we should work for the worst case scenarios and this said we should always find the minimum Zener current which should be 10% of it's maximum current.

Vz = 12 V
P = 0.25 W
and the transistor hfe is 50.

So, for the minimum current I did this:

0.25 = 12*Iz_max
Iz_max = 20.83 mA
Iz_min = 10% of Iz_max = 2.083 mA

Then:
I_R1 = (V1 - V2) / R1
I_R1 = (18-12) / 470 = 12.77 mA

Then I'm not sure what to do next... I already tried an approach but it is not correct but I'm not sure why...

I have written a nodal equation for the currents at node 2 as:
I_R1 = Iz_min + IB
IB = I_R1 - Iz_min

but the IB current I found is too high... 10.69 mA which, assuming transistor working on active zone, will turn into an IE current of 545 mA, ending up in a voltage drop across R_Load of 54.5 V.

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#### WBahn

Joined Mar 31, 2012
24,970
UNITS!!!!!

Where does the 470 Ω resistor come from? Did you select it, or is it a provided value that you have to use?

What will the power dissipation be in the zener at the max voltage input?

The problem with your nodal equation is that you are assuming that the zener current will be Iz_min. What is the basis for that claim?

#### PsySc0rpi0n

Joined Mar 4, 2014
1,446
R1 = 470 Ω is a value given by the problem itself!

I already did for the max Iz current... 20.83 mA...

We were taught that for the zener to work as a zener it should have flowing through it at least 10% of it's maximum current...

Anyway, I already did another approach.

I am considering only the 18 V as the worst case scenario. Mybe I'm also wrong here.

Anyway, for 18 V I already found Ie and I_R1 matching LTSpice values.

I_R1 = (18 V - 12 V) / 470 Ω = 12.77 mA

Then for Ie I did:

-Vz + Vbe + V_R_Load = 0
-Vz + Vbe + Ie*R_Load = 0
Ie = (Vz - Vbe) / R_Load
Ie = (12 V - 0.67 V ) / 100 Ω
Ie = 113.3 mA

Until here the values kinda match LTSpice's...

But then I did:

Ie = Ib (hfe + 1)
Ib = Ie / (hfe + 1)
Ib = 113.3 mA / (50 + 1) = 2.22 mA

This one I can't check because the LTSpice model of the transistor I'm using, almost for sure, won't have 50 of gain!

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#### WBahn

Joined Mar 31, 2012
24,970
This looks good -- and thank you for tracking your units this time.

Do you understand why your earlier approach wasn't working?

BTW: You are definitely improving!

#### BramLabs

Joined Nov 21, 2013
98
• PsySc0rpi0n

#### PsySc0rpi0n

Joined Mar 4, 2014
1,446
This looks good -- and thank you for tracking your units this time.

Do you understand why your earlier approach wasn't working?

BTW: You are definitely improving!
I understand that one of the reasons was that I was getting a voltage drop at R_Load greater than the supply voltage! But probably that is not the only reason!

Maybe this wouldn't help you a lot, but at least you'll have an information how to design zener with a supply voltage vary from x to y and then you can know how to choose Iz_min and Iz_max and choose the Iz between the range of Minimum and Maximum.

So basically my circuit will work on the Voltage supply that vary from 26 Volt - 30 Volt.
Well, maybe it doesn't help right now but it might help in the future! I think we're supposed to consider 10% of the Zener's maximum current value!

#### PsySc0rpi0n

Joined Mar 4, 2014
1,446
Ok, here are the rest of my calcs for checking!

Nodal equation for node 2

I_R1 = I_z + I_b
I_z = I_R1 - I_b
I_z = 12.77 mA - 2.22 mA
I_z = 10.55 mA

P_z = V_z*I_z
P_z = 12 V * 10.55 mA
P_z = 126.6 mW

For checking I wrote a net equation:

-V_z + V_be + V_out = 0 V
-12 V + 0.67 V+ I_e*R_load = 0 V
-12 V + 0.67 V + 113.3 mA*100Ω = 0 V
0 V = 0 V...

Lastly, one more check... We (I) have assumed that the transistor was working on the active zone! So, let's check if that's true:

We already know that Ib ≠ 0 A. We need to check if Vce > 0.5 V

Vce = Vin - Vout = 18 V- 11.33 V = 6.67 V. Also checks!

Looks like it's correct, no?

However I find it weird that it was not necessary to find the Zener's minimum current. Also I still have a question. For a voltage supply that may vary from 18 V up to 22 V, is the lower voltage, the worst case scenario?

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#### WBahn

Joined Mar 31, 2012
24,970
Your voltage supply has two limits. For low supply voltages you have to worry about the zener falling out of regulation (that's the 10% I_zmax criteria you are working from) while for high supply voltages you have to worry about the zener exceeding its operating limits, generally max power rating which is often via the I_zmax figure.

#### Bordodynov

Joined May 20, 2015
2,454
You so raged the saving a zener diode that have forgotten about transistor. I have considered your scheme and has got the power on transistor is 755 mW. Datasheet gives only 500 mW. Not aptly chosen transistor.
Good transistor ZTX690B. Gain of 400 at IC=1Amp.
Power Dissipation at Tamb=25°C Ptot=1W. Case E-Line TO92 Compatible.
Тransistor 2N2222A Gain of 150 at IC=115 mAmp.

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#### WBahn

Joined Mar 31, 2012
24,970
Or an alternative is to put in a collector resistor or diodes to shift some of the power to it. Which option to take depends mostly on how much the maximum current is that we need to be able to deliver to the load.

But the point is very valid. All of the parts (including the resistors) need to be evaluated to ensure that they are operating within their specified limits.

#### Jony130

Joined Feb 17, 2009
5,089

#### PsySc0rpi0n

Joined Mar 4, 2014
1,446
You so raged the saving a zener diode that have forgotten about transistor. I have considered your scheme and has got the power on transistor is 755 mW. Datasheet gives only 500 mW. Not aptly chosen transistor.
Good transistor ZTX690B. Gain of 400 at IC=1Amp.
Power Dissipation at Tamb=25°C Ptot=1W. Case E-Line TO92 Compatible.
Тransistor 2N2222A Gain of 150 at IC=115 mAmp.
@Bordodynov I understand what you're trying to say. But in an academic context, some of those details are neglected as, usually, teachers are trying to understand weather we understand the circuit behavior through math and calcs and not if the circuit is viable in real life!

Note: I'm not trying to excuse the fact that I have not done every steps like they should have been done. I'm trying to say that probably, what I have done was enough to teacher approve the problem solving.

Of course you're right. I should have evaluated all components maximum ratings if I was thinking about to build it in the breadboard, for instance!

Or an alternative is to put in a collector resistor or diodes to shift some of the power to it. Which option to take depends mostly on how much the maximum current is that we need to be able to deliver to the load.

But the point is very valid. All of the parts (including the resistors) need to be evaluated to ensure that they are operating within their specified limits.
I'll do the same for the other voltage and at the end I'll need to compare the zener's power dissipation to the given value of 250 mW!

#### PsySc0rpi0n

Joined Mar 4, 2014
1,446
Ok, so for the other value of the Power Supply: 22V I got:

I_R1 = 12.77 mA
I_E = 113.3 mA
I_B = 2.22 mA
I_z = 19.06 mA
P_z = 235.2 mW

So, for the Zener, it would be everything within the limits!

#### WBahn

Joined Mar 31, 2012
24,970
Ok, so for the other value of the Power Supply: 22V I got:

I_R1 = 12.77 mA
I_E = 113.3 mA
I_B = 2.22 mA
I_z = 19.06 mA
P_z = 235.2 mW

So, for the Zener, it would be everything within the limits!
How do you get I_R1 to be 12.77 mA when the supply voltage is 22 V?

Since your I_z is correct, I'll assume this was just a bit of sloppiness.

Can the transistor beta be higher than 50?

If so, would that increase or decrease the current in the zener?

What is the highest beta you can have and not violate the 250 mW limit?

There are a couple of fine points you need to keep in mind.

When looking for the minimum zener current, you need to account for all things that could reduce that current, such as

1) A supply voltage at its lowest limit.
2) An R1 value that is at its upper limit.
3) A transistor Vbe at its lower limit.
4) A transistor beta at its lower limit.
5) A load resistance at its lower limit.
6) A zener voltage at its upper limit.

When looking for the maximum zener current, you need to account for all things that could increase that current, such as

1) A supply voltage at its upper limit.
2) An R1 value that is at its lower limit.
3) A transistor Vbe at its upper limit.
4) A transistor beta at its upper limit.
5) A load resistance at its upper limit.
6) A zener voltage at its lower limit.

The second thing is that you want to avoid operating near a device's limits. A common rule of thumb is to not exceed 85% of the limit and a common, more conservative, rule is to not exceed 50%.

Not only does this significantly increase the life expectancy of electronic components, but it also makes almost all of the above considerations moot in almost all cases. If your nominal values result in 50% of the max power rating, then is it almost always assured that even if everything is against the worst limit possible that you won't violate the specs.

Now, being conservative limits your flexibility and could result in you not being able to meet some other spec. You deal with those situations as they arise.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,446
How do you get I_R1 to be 12.77 mA when the supply voltage is 22 V?
Of course not...

I_R1 = (22 V - 12 V) / 470 Ω = 21.28 mA

About beta, I think we use to consider it to be constant, regardless the V_ce or any temperature change!
But assuming it could change, if Beta increases, that will increase I_E, meaning more current on the load, which means that the zener would have less current through it!

#### WBahn

Joined Mar 31, 2012
24,970
Of course not...

I_R1 = (22 V - 12 V) / 470 Ω = 21.28 mA

About beta, I think we use to consider it to be constant, regardless the V_ce or any temperature change!
But assuming it could change, if Beta increases, that will increase I_E, meaning more current on the load, which means that the zener would have less current through it!
Think about this. If beta increases, that will not affect the current in the load at all. The current in the load is dictated by the zener voltage and the base-emitter voltage drop of the transistor. All the change in beta does is shift how much load current comes from the collector and how much comes from the base. If the beta goes up, then the base current will go down (and the collector current will increase by the same amount). Thus there will be more current through the zener (by the amount that is no longer going through the base).

#### PsySc0rpi0n

Joined Mar 4, 2014
1,446
Think about this. If beta increases, that will not affect the current in the load at all. The current in the load is dictated by the zener voltage and the base-emitter voltage drop of the transistor. All the change in beta does is shift how much load current comes from the collector and how much comes from the base. If the beta goes up, then the base current will go down (and the collector current will increase by the same amount). Thus there will be more current through the zener (by the amount that is no longer going through the base).
I can understand that, but I can't explain it if I look to I_E equation... I_E = I_B * (β + 1). So if β increases I would say that I_E would also increase. And the amount that I_E increased, would be the amount that I_z current would be deviated to the increased I_E... I'm not sure if I made myself clear!

But I definitely understand what you said... The voltage at that net loop is kinda constant, so also I_E will be!

#### WBahn

Joined Mar 31, 2012
24,970
I can understand that, but I can't explain it if I look to I_E equation... I_E = I_B * (β + 1). So if β increases I would say that I_E would also increase. And the amount that I_E increased, would be the amount that I_z current would be deviated to the increased I_E... I'm not sure if I made myself clear!

But I definitely understand what you said... The voltage at that net loop is kinda constant, so also I_E will be!
Where you are getting confused is that if beta changes, there are TWO other variables than can change in order to keep the equality true. Yes, I_E can go up or I_B can go down (with the other remaining constant), but it's also possible that both I_E and I_B could change. By increasing beta, all that you know is that the ratio of I_E/I_B has increased.