Voltage Regulators - Problem 3

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,755
It's probably quite reasonable, except good luck finding a 350 Ω resistor. How much power is being dissipated in Rz? What wattage rating would you choose for this resistor?
Yeah, probably I won't find such a resistor! And is there any database where I can check the usual commercial resistor values? I mean, how do you know, if not by experience, which resistors you can find on the market?

P_Rz = 350 Ω * (18.29 mA)² = 117 mW.

Another option is to use the geometric mean of sqrt((160 Ω)(540 Ω) = 293.9 Ω => 300 Ω
I don't know that method! Never heard about that! I already saw @Bordodynov suggestion which also mentions that method!
 

WBahn

Joined Mar 31, 2012
29,978
Well, so for the larger resistor I should use a lower value... I got that! A lower value will grant a larger current which means that it's larger than the minimum acceptable current for the zener to hold his working mode (zener mode).

Also I saw it later that the power that I have calculated was for the Rz and not for the zener. I mentioned it after I saw @Bordodynov 's post that I think I missed!
Yeah, our posts crossed.

But I presume that you mean that I need to choose a resistor capable of dissipating more than 1/4 W... Probably 1/2 W but I don't know if these resistors has a specific name or not!
Yep. Resistors generally come in fractional watt sizes and, above 1 W (for a ways) follow a 1-2-5 sequence. So you can find 1/8 W, 1/4 W, 1/2 W, 1 W, 2 W, 5 W, 10 W, 20 W and so on.
There are numerous exceptions to this, so while you can generally count on being able to source these wattage ratings (and they are generally cheaper), you can often find ratings that are a closer match to your needs.

I also don't know that E24 sequence... What constraints are you referring to? Like, what current should I have if I couldn't use 1/2 W resistors?
The E24 sequence is simply the standard 5% resistor series.

http://www.logwell.com/tech/components/resistor_values.html

And, yes, that's what I'm talking about. If you have to use 1/4 W resistors and want to limit the power to half that to avoid stressing them, which resistor would you choose?

You're doing good -- and thanks for tracking your units so much better. At some point you will see how it really pays off.
 

WBahn

Joined Mar 31, 2012
29,978
Yeah, probably I won't find such a resistor! And is there any database where I can check the usual commercial resistor values? I mean, how do you know, if not by experience, which resistors you can find on the market?

P_Rz = 350 Ω * (18.29 mA)² = 117 mW.



I don't know that method! Never heard about that! I already saw @Bordodynov suggestion which also mentions that method!
The arithmetic mean, (A+B)/2, and the geometric mean, sqrt(A·B), are two ways of finding an intermediate value. Some problems strongly call out for one or the other, other problems don't and so you can pick. Problems in which the parameters are multiplicative argue for using the geometric mean.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,755
Hum, didn't know about that...
But I'm still not clarified about what those sequences are all about! What means a 1-2-5 sequence?

Also didn't know about the E??? stuff! I knew that there were several precision values for resistors but I didn't know they had a name for each!

And, yes, that's what I'm talking about. If you have to use 1/4 W resistors and want to limit the power to half that to avoid stressing them, which resistor would you choose?
So, if I have (must) to use a 1/4 W resistor, I presume I should keep the current constant, so I would have to change the resistor value. I would use the P = R*I² to find it, like:

0.25 = R*(18.29 mA)²
R = 747.3 Ω

What will change, I guess, is the voltage drop across the resistor, no?
 

WBahn

Joined Mar 31, 2012
29,978
Hum, didn't know about that...
But I'm still not clarified about what those sequences are all about! What means a 1-2-5 sequence?
A 1-2-5 sequence is merely an approximation to a logarithmic axis (think log graph paper).

You see this commonly on electronic test equipment such as oscilloscopes and volt meters. Each value is roughly twice the value of the one before it.

2 = 2 * 1
5 = 2.5 * 2
10 = 2 * 5

So, if I have (must) to use a 1/4 W resistor, I presume I should keep the current constant, so I would have to change the resistor value. I would use the P = R*I² to find it, like:

0.25 = R*(18.29 mA)²
R = 747.3 Ω

What will change, I guess, is the voltage drop across the resistor, no?
You're starting to go off the deep end a bit. Any value resistor you choose will have a current that varies with Vin. Just as before there was a range of resistance values that Rz had to be within in order to keep the Zener diode in spec, now you have a range of resistance values that will keep the power dissipation in Rz in spec. You have to pick a value of Rz that satisfies both range limitations.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,755
...

You're starting to go off the deep end a bit. Any value resistor you choose will have a current that varies with Vin. Just as before there was a range of resistance values that Rz had to be within in order to keep the Zener diode in spec, now you have a range of resistance values that will keep the power dissipation in Rz in spec. You have to pick a value of Rz that satisfies both range limitations.
Ah ok, I think I get it!

So, between 160 Ω and let's say 500 Ω, I'll need to come uo with a resistor that brings down the power dissipation on it, right?

I'm not sure if there is a specific way of doing this but I'm thinking about trial and error.
So, if I choose, let's say 470 Ω, I'll get:

P = R*I²
0.25 W= 470 Ω * I²
I = √(0.25 / 470) = 23.1 mA (min zener current was 4.46 mA)

So, I get 23.1 mA through Rz, and ignoring Ib, I'll also get 23.1 mA through the zener. Now, the power at the zener must be also re-evaluated.

Pz = 5.6 V * 23.1 mA = 129.2 mW

So, I think this could work in terms of power dissipation in both components.
But one other question arises.

I have been calculating currents for 250 mW of power dissipation. However, I'm not yet being very conservative because I'm working on the edge, in term of the resistor power dissipation.

So, wouldn't it be more reasonable to calculate the resistor value for a lower power dissipation limit? Like for 180 mW or so???
 

WBahn

Joined Mar 31, 2012
29,978
Ah ok, I think I get it!

So, between 160 Ω and let's say 500 Ω, I'll need to come uo with a resistor that brings down the power dissipation on it, right?

I'm not sure if there is a specific way of doing this but I'm thinking about trial and error.
No need to do trial and error. Set up the mathematical inequality that captures the design goal:

The power in the resistor should not exceed 1/2 of the resistor's 1/4 W power rating:

P < 50%·(1/4 W)
P = I²·Rz
I = (Vin - Vz)/Rz

Bring it together, we have

(Vin - Vz)²/Rz² < 50%*(1/4 W) = 0.125 W

Rz² > 0.125 W / (Vin - Vz)²

Rz > sqrt(0.125 W) / (Vin - Vz)

The maximum power in Rz will occur when there is the maximum current in it, which will occur when Vin is at it's highest value of 12 V.

So

Rz > sqrt(0.125 W) / (Vin - Vz)

Oops -- in trying to carry this out I can't get the units to work out since I will end up with √(A/V) or 1/√Ω instead of Ω. So, looking back, I see that I just made a mental algebra mistake and that, "bringing it altogether" should have yielded:

(Vin - Vz)²/Rz < 50%*(1/4 W) = 0.125 W

Very obvious, but easy to miss when you are worried more about getting in the keyboard symbols. I'm leaving this in to show how tracking units enables you to detect and track down the kinds of errors that we all make on an all-too-frequent basis.

So, going from this point again:

Rz > (Vin - Vz)² / 0.125 W

Rz = (12 V - 5.6 V)² / 0.125 W

Now the units work out since we have V² divided by V·A giving us V/A = Ω

Rz > 327.7 Ω

So, if I choose, let's say 470 Ω, I'll get:

P = R*I²
0.25 W= 470 Ω * I²
I = √(0.25 / 470) = 23.1 mA (min zener current was 4.46 mA)
UNITS!!

I = √(0.25 W / 470 Ω)

Track the units: 1 W = 1 V·A, Ω = 1 V/A, so 1 W/Ω = A²

So, I get 23.1 mA through Rz, and ignoring Ib, I'll also get 23.1 mA through the zener. Now, the power at the zener must be also re-evaluated.

Pz = 5.6 V * 23.1 mA = 129.2 mW

So, I think this could work in terms of power dissipation in both components.
But one other question arises.

I have been calculating currents for 250 mW of power dissipation. However, I'm not yet being very conservative because I'm working on the edge, in term of the resistor power dissipation.

So, wouldn't it be more reasonable to calculate the resistor value for a lower power dissipation limit? Like for 180 mW or so???
Yes, which is why I said to choose a value for Rz that limits the power dissipation to half of the 250 mW rating (in other words, no more than 125 mW).
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,755
Ok, I have just burned a zener I had here... And it was the only one! I need to buy more! I was trying to check this circuit on the breadboard but I have shorted the collector and the base of T1 when placing the scope's probes. I hope I have not damaged any scopes's channels inputs...

Edited:
I think I found another zener... Need to check if it has the same zener voltage!
 

WBahn

Joined Mar 31, 2012
29,978
Ok, I have just burned a zener I had here... And it was the only one! I need to buy more! I was trying to check this circuit on the breadboard but I have shorted the collector and the base of T1 when placing the scope's probes. I hope I have not damaged any scopes's channels inputs...
It's very, very unlikely that you damaged the scope. You only exposed it to, at most, about 12 V. You might have had some excess current flowing through the scope's probe tip, but it's doubtful you did any real damage (it can almost surely handle a lot more current than that zener can, because it is very low resistance).

Putting a base resistor between the zener and the transistor might be a good idea. With the kind of base current you are looking with that Darlington circuit, it can be pretty high. In fact, if you used the same size resistor that you use for Rz you will be looking at just a couple of millivolts dropped across it, but it might prevent the kind of mistake you just made (though note that that mistake could also smoke your transistor pretty easily). One way to mitigate both is to put two base resistors in series and treat the junction between them as your transistor base for voltage measurement purposes.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,755
Putting a base resistor between the zener and the transistor might be a good idea. With the kind of base current you are looking with that Darlington circuit, it can be pretty high. In fact, if you used the same size resistor that you use for Rz you will be looking at just a couple of millivolts dropped across it, but it might prevent the kind of mistake you just made (though note that that mistake could also smoke your transistor pretty easily). One way to mitigate both is to put two base resistors in series and treat the junction between them as your transistor base for voltage measurement purposes.
A base resistor, you mean this:
aac-.png

I don't know if is there any way of having 100% sure of I damaged somehow the scope, but at least the channel is measuring correctly the supply voltage of around 12.73 V...

The scope is a Rigol MSO 1104Z.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,755
That's what I'm talking about. And it sounds like your scope is fine.
A little bit of OFFTOPIC...

Do you know anything about this scope?

I was trying to use the MATH option, using 2 channels to measure the difference between the collector and emitter voltage but I can only see the plot line. I can't see the result number anywhere in the screens. Do you have any idea if this should be like this? Not showing the result in numbers somewhere on the screen?
 

WBahn

Joined Mar 31, 2012
29,978
I don't know about the Rigol, but most of these scopes in the Math mode requires that both channels be set for the same gain and that the individual signals have to individually be within each channels range. This if you have 9.10 V and 9.15 V and you want to display that 50 mV as a fairly large signal, you may be out of luck. Try using a voltage divider to make, say, a 5 V signal and a 6 V signal and then try to get the 1 V difference to display using Math mode. Then you will know that you know how to do it.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,755
I don't know about the Rigol, but most of these scopes in the Math mode requires that both channels be set for the same gain and that the individual signals have to individually be within each channels range. This if you have 9.10 V and 9.15 V and you want to display that 50 mV as a fairly large signal, you may be out of luck. Try using a voltage divider to make, say, a 5 V signal and a 6 V signal and then try to get the 1 V difference to display using Math mode. Then you will know that you know how to do it.

Ok, no need of that... I got the answer in the old IRC channels... A friend of mine said me that I should set the source of measurements as MATH and not any of the channels. This way I can place the value of the difference in the screen just as any measurement made with channels!

OFFTOPIC OFF!

I'm trying to figure out why I can't measure anything in the load resistor!
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,755
...

Oops -- in trying to carry this out I can't get the units to work out since I will end up with √(A/V) or 1/√Ω instead of Ω. So, looking back, I see that I just made a mental algebra mistake and that, "bringing it altogether" should have yielded:

(Vin - Vz)²/Rz < 50%*(1/4 W) = 0.125 W

Very obvious, but easy to miss when you are worried more about getting in the keyboard symbols. I'm leaving this in to show how tracking units enables you to detect and track down the kinds of errors that we all make on an all-too-frequent basis.

So, going from this point again:

Rz > (Vin - Vz)² / 0.125 W

Rz = (12 V - 5.6 V)² / 0.125 W

Now the units work out since we have V² divided by V·A giving us V/A = Ω

Rz > 327.7 Ω



UNITS!!

I = √(0.25 W / 470 Ω)

Track the units: 1 W = 1 V·A, Ω = 1 V/A, so 1 W/Ω = A²



Yes, which is why I said to choose a value for Rz that limits the power dissipation to half of the 250 mW rating (in other words, no more than 125 mW).
Shouldn't the underlined line be:
(Vin - Vz)²/Rz² < 0.125 W
Rz² > (Vin - Vz)² / 0.125 W
Rz = (Vin - Vz) / √0.125 W
Rz = (12 V - 5.6 V) / 0.354 W
Rz = 18.1 Ω

But, for some reason this returns a value of 18.1 Ω. Not sure why
 

WBahn

Joined Mar 31, 2012
29,978
Track the units. What do the units work out to be? Don't just tack the units you think the answer should have onto the end. Track the units and see what they actually turn out to be.

You have sqrt (0.125 W) in one line but then have that as 0.354 W on the next line. But the W is INSIDE the square root. So when you take the square root you have to take the square root of it, too.

So your fourth line should be

Rz = (12 V - 5.6 V) / (0.354 √W)

Since 1 W = 1 V·A, we have that 1 √W = 1 √(V·A) = 1 √V·√A

So the units on the answer are

Rz = 18.1 √V / √A = 18.1 √Ω

That's what the units work out to be. So now we know one thing -- that this answer is WRONG because the units don't work out to Ω.

This allows us to look back through our work and figure out where we made our mistake, which is what I did in the post when my answer didn't work out to the correct units. I left the earlier wrong work there specifically as an example of how to use units to catch mistakes -- but it ONLY works if you NEVER, EVER just tack on the units that you WANT a result to have!

If you take a moment to think about it, you'll recall that the power in a resistor is

P = I·V = I²·R = V²/R

Then you can spot where you made your mistake and work forward from there -- which is also a huge advantage of leaving the numbers out of the work until the very end.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,755
Well, I never thought about that that way! I never thought to do the math work on units too!

Anyway, I understand what you just said! But mathematically, where am I wrong? I mean, if that were just numbers with no electrical meaning, that math was correct, no?

I mean, if I have:

x² = y² / z
√(x²) = √(y²) / √z
x = y / √z

this is correct, no?
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,755
Ok, forget it... That was just some unexplainable stupidity...

I was doing some messed up math...

I < P
(Vin - Vz) / R < V*I

This is non-sense!

Forget it!
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,755
Ok, I'm using 2 base resistors and for the 330 Ω that we have just calculated, I'm using a 220 Ω + 100 Ω because I don't have any 330 Ω resistor!

The 2 base resistors are 1kΩ. But I don't know why, I'm not getting anything at the Load resistor!

The following values are average measure from scope!
Vc1 = Vin = 12.1 V
Ve1 = 5.9 V
Vb1 = 7.27 V

Vbe1 = -1.90 V ????????

Vc2 = 12.1 V
Ve2 = weird... It's a sin wave looking plot on the scope... If I put my finger on T2 this wave's positive peak goes a bit higher! If I remove my finger, after a few seconds the plot line looks almost flat but it's still possible to see some ripple/sine waveform and it says 5.9 V.
Vb2 = 6.70 V

Vbe2 = 0.37V, this one measured by the voltmeter! It's easier! ???????????

Vzener = 6.83 V ????? higher potential at the zener cathode (voltmeter red probe) and lower potential at the anode (black probe, GND)!


The question marked values are weird in my point of view! I guess! Both Vbe should be around 0.6V or 0.7 V and Zener voltage should be around 5.6 V.
 
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