Bad Circuits - Test Your Knowledge

[ranger]

So I guess we can wrap up circuits D and F?

 

[NoTime]

So I guess we can wrap up circuits D and F?

So what is your final solution?
Say where the 45 ohm resistor gets connected also.

You could also try to figure out what the op-amp parameter is that you left out of the list before.

 

[ranger]

So what is your final solution?
Say where the 45 ohm resistor gets connected also.

You could also try to figure out what the op-amp parameter is that you left out of the list before.

Well my final solution is this:
https://www.physicsforums.com/showpost.php?p=1392945&postcount=83

I'm still unsure as to whether the answers to the questions you asked in post #84 are in my post #85 and #87.

I believe the final op-amp property I left out was slew rate.

 

[NoTime]

Well my final solution is this:
https://www.physicsforums.com/showpost.php?p=1392945&postcount=83

After your statement that this solution involves connecting the 240 ohm and zener to the inverting (-) input of the op amp. I have to say your solution is wrong and it's back to the drawing board for you.

I'm still unsure as to whether the answers to the questions you asked in post #84 are in my post #85 and #87.

I believe the final op-amp property I left out was slew rate.

Slew rate is the same concept as "gain degradation with increasing frequencies".
However, one of the words in your original statement does relate to my question, but not as presented.

The question on the zener biasing is still unanswered.

 

[ranger]

After your statement that this solution involves connecting the 240 ohm and zener to the inverting (-) input of the op amp. I have to say your solution is wrong and it's back to the drawing board for you.

I indicated that it was a typo, and the linked post (#83) states that I have the zener to the non-inverting input. You're beginning to confuse me

 

[NoTime]

I indicated that it was a typo, and the linked post (#83) states that I have the zener to the non-inverting input. You're beginning to confuse me

I, apparently, am confused.

Ok, but Berkman will have to provide any additional challenges, since I don't have the book.

I would still be interested if you can come up with answers to my additional questions

 

[ranger]

Okay, I'll take sometime to think about them. But others should definitely step in and make an attempt to avoid a stall here. The questions that NoTime is talking about can be found here:
https://www.physicsforums.com/showpost.php?p=1392983&postcount=84
My attempt is here:
https://www.physicsforums.com/showpost.php?p=1393025&postcount=85

Just in case anyone got lost, NoTime's questions are in response to the fix for circuit F:
https://www.physicsforums.com/showpost.php?p=1392945&postcount=83

 

[AlephZero]

The attachment is what I think Ranger meant (and I agree with it).

Re the 240R resistor and zener, the zener isn't an ideal regulator so the voltage across it depends on the current through it. The 240R gives a current of 25ma which presumably is the recommended current sheet given on the data sheet for the part number being used.

After changing the circuit, the only function of the zener is as a voltage reference, it doesn't need to be a high current device taking the full 200ma load. 25ma seems a bit high for a low power zener. I would have thought about 5ma would be enough. Changing the 240R to 1.2K would set the current to 5ma.

 

[berkeman]

Well, you folks have been busy over the weekend! I'm not sure I'm 100% in synch with where you are on the latest problem, but I think there were two last things that NoTime and ranger were discussing.

** I think that NoTime may be referring to the limited Iout of most opamps (is that right?), and that would be my fault for not mentioning that I was assuming that we were carrying along the high-current opamp shown in the original 200mA Bad Circuit. Most opamps are not going to give you that kind of output current. The issue of unbalanced input impedances was mentioned by ranger -- where the input bias offset current will generate an error term in the output. This is hard to avoid in this simple implementation, but could be mitigated if the opamp had fairly low input bias current offset specs.

** The second issue raised by NoTime is a very important one, and I'll add an extra twist to it. Look at a datasheet for Zener diodes, like this one:

http://www.vishay.com/docs/85763/bzx84v.pdf

What current is the Zener voltage measured at for this small-signal Zener family? What should the resistor in our circuit be, if we were using this Zener series? What would be the effect of chosing a smaller or larger resistor than this?

And for my extra-credit Quiz Question -- how does the temperature coefficient of Zener diodes vary with Vz? What value of Zener diode generally has the lowest temperature coefficient?

 

[ranger]

** The second issue raised by NoTime is a very important one, and I'll add an extra twist to it. Look at a datasheet for Zener diodes, like this one:

http://www.vishay.com/docs/85763/bzx84v.pdf

What current is the Zener voltage measured at for this small-signal Zener family? What should the resistor in our circuit be, if we were using this Zener series? What would be the effect of chosing a smaller or larger resistor than this?

And for my extra-credit Quiz Question -- how does the temperature coefficient of Zener diodes vary with Vz? What value of Zener diode generally has the lowest temperature coefficient?

Well if we want to get the 9V regulation, we could go with the BZX84C9V1-V, which has a range of 8.5V-9.6V. it seems that they are indicating that they got that range by using a test current of 5mA. We are way over that limit here (62mA). We would need to use a bigger resistor to trim the current down to 5mA.

I have to run. I'll see if I can get to the quiz question later.

 

[AlephZero]

Well if we want to get the 9V regulation, we could go with the BZX84C9V1-V, which has a range of 8.5V-9.6V. it seems that they are indicating that they got that range by using a test current of 5mA. We are way over that limit here (62mA). We would need to use a bigger resistor to trim the current down to 5mA.

You calculated the zener current as 15V/240R = 62 mA? Oops...

You forgot the voltage drop across the resistor is (15-9) = 6V which gives the current as 6V/240R =25mA. The BZX series would handle 25mA (the power dissipation is 9x25 = 225mW < 300 mW absolute max) but it's a waste of power and it would change the reference voltage a bit (see the data sheet for dynamic resistance)

See my previous post #98. I admit I didn't look up the zener current in a data sheet for post #98, but my estimate of 5ma current was about right

Re the supplementary question, there's a nice graph near the end of the data sheet that shows the answer. Making sense of the temp coefficients in the tables is a little bit harder to do.

 

[ranger]

You calculated the zener current as 15V/240R = 62 mA? Oops...

You forgot the voltage drop across the resistor is (15-9) = 6V which gives the current as 6V/240R =25mA.

...

Ouch! That was a stupid mistake on my part.

I have to run. I'll see if I can get to the quiz question later.

I was in a rush to get to my school's library to renew my copy of The Art of Electronics. But they won't let me; they took it away. Now I have to wait till the new semester starts (about 3 weeks).

 

[NoTime]

You calculated the zener current as 15V/240R = 62 mA? Oops...

You forgot the voltage drop across the resistor is (15-9) = 6V which gives the current as 6V/240R =25mA. The BZX series would handle 25mA (the power dissipation is 9x25 = 225mW < 300 mW absolute max) but it's a waste of power and it would change the reference voltage a bit (see the data sheet for dynamic resistance)

See my previous post #98. I admit I didn't look up the zener current in a data sheet for post #98, but my estimate of 5ma current was about right

Re the supplementary question, there's a nice graph near the end of the data sheet that shows the answer. Making sense of the temp coefficients in the tables is a little bit harder to do.

Closer, but

I'll give you one equation.
The zener current equation is I_z = (V_source - V_z) / (R_bias + R_z)
R_bias is the 240 ohm resistor.

Zener power disapation is not V_z * I_z.
What is the correct formula?

The zener voltage V_z is not V_ref (junction of 240 ohm and zener).
What is the equation for V_ref?

Note: Berkmans datasheet shows production min/max values or tolerance. An additional consideration but not a direct part of the question.

 

[NoTime]

** I think that NoTime may be referring to the limited Iout of most opamps (is that right?), and that would be my fault for not mentioning that I was assuming that we were carrying along the high-current opamp shown in the original 200mA Bad Circuit. Most opamps are not going to give you that kind of output current. The issue of unbalanced input impedances was mentioned by ranger -- where the input bias offset current will generate an error term in the output. This is hard to avoid in this simple implementation, but could be mitigated if the opamp had fairly low input bias current offset specs.

All true.
I was simply going after open loop gain of the op-amp.
Just one more error term in the output, but one that often seems to be forgotten.

 

[AlephZero]

NoTime, which circuit are you talking about here? The original one, or my modified one (post #98)?

One of my former bosses used to say "If you already know something is a dumb idea, don't waste time calculating how dumb it is to 3 decimal places".

I think we agree what was wrong with the original circuit, so I'm talking about the modified one.

Closer, but I'll give you one equation.
The zener current equation is I_z = (V_source - V_z) / (R_bias + R_z)
R_bias is the 240 ohm resistor.

That's the small-signal equation for using the Zener as a regulator. The data sheet give the dynamic resistances R_z at two reference currents. The values only apply for small changes around those reference currents.

In this circuit the Zener isn't a regulator, it's providing a constant reference voltage.

If you take the design point I_z = 5ma, all you can say from the data sheet is that V_z is within the tolerance band around the nominal 9.1V value. R_z doesn't come into the calculation of R_bias.

If you look at the graph on page 4, you see that R_z is a strongly nonlinear function of I_z and decreases as I_z increases. There is no easy way to calculate the change in V_z if the current changes from 5mA to about 25mA. Using the data sheet value of R_z = 6R at I_z = 5ma in the "small signal" equation won't give the right answer.

Zener power disapation is not V_z * I_z.
What is the correct formula?

Sorry - I don't see what else it could be (assuming I_z is the total current through the diode). Sure, in the original circuit I_z = 25mA (approx!) + 200mA but we already know that's a bad design.

The zener voltage V_z is not V_ref (junction of 240 ohm and zener).
What is the equation for V_ref?

I'm not sure what you are getting at here - unless my previous comments have already answered it.

 

[NoTime]

NoTime, which circuit are you talking about here? The original one, or my modified one (post #98)?

That would be the modified one you posted in 98.
I'm confused as to what I might have said that make you think otherwise.
I'd actually appreciate it if you would point that out.

One of my former bosses used to say "If you already know something is a dumb idea, don't waste time calculating how dumb it is to 3 decimal places".

I like this

That's the small-signal equation for using the Zener as a regulator. The data sheet give the dynamic resistances R_z at two reference currents. The values only apply for small changes around those reference currents.

In this circuit the Zener isn't a regulator, it's providing a constant reference voltage. If you take the design point I_z = 5ma, all you can say from the data sheet is that V_z is within the tolerance band around the nominal 9.1V value. R_z doesn't come into the calculation of R_bias.

The distinction between regulator and reference is only intent and not construction. Selection of different values of R_bias (what I was asking about) causes the reference current to change.

If you look at the graph on page 4, you see that R_z is a strongly nonlinear function of I_z and decreases as I_z increases. There is no easy way to calculate the change in V_z if the current changes from 5mA to about 25mA. Using the data sheet value of R_z = 6R at I_z = 5ma in the "small signal" equation won't give the right answer.

Frankly, the dynamic aspect of R_z is news to me.
Never say you have nothing more to learn about something.
It has been a while since I actually worked with these and I remember getting specs for R_z at a nominal I_z.
Also, it does explain a rule of thumb I was given for dealing with R_z.

If you look at the chart 6 R_z is going to vary about 10 ohms for a current range fo 20 to 50 ma. So using a nominal value of 25 ohms for R_z will give reasonably accurate results if I_z is in this range.

Sorry - I don't see what else it could be (assuming I_z is the total current through the diode).

The power dissapated by the zener is not 9v * 25ma = 225mW.
Why?

 

[ranger]

The distinction between regulator and reference is only intent and not construction. Selection of different values of R_bias (what I was asking about) causes the reference current to change.

The only formula that I'm aware of for finding the zener impedance is:
Z_Z = \frac{\Delta V_Z}{\Delta I_Z}
And this demonstrates that the zener impedance is only for small signal variations. We have the current held at a constant 25mA in our circuit.

The power dissapated by the zener is not 9v * 25ma = 225mW.
Why?

I can't see why it wouldn't be this. Can you give a couple of hints?

 

[AlephZero]

That would be the modified one you posted in 98.
I'm confused as to what I might have said that make you think otherwise.
I'd actually appreciate it if you would point that out.

There was nothing specific that you said, but I wasn't sure about some of your comments (like your power dissipation question, below) so I thought it was worth checking we were both on the same page. If we WERE talking about two different circuits, the discussion probably wouldn't lead anywhere useful till we both realized that!

The power dissapated by the zener is not 9v * 25ma = 225mW.
Why?

Like Ranger, I'm stuck on that one.

 

[NoTime]

We have the current held at a constant 25mA in our circuit.

Perhaps rewording the question will help.
If you were given a 9v zener.
How would you select the appropriate bias resistor?

I can't see why it wouldn't be this. Can you give a couple of hints?

Try making up an equivalent circuit using a voltage source, resistor (and diode if you wish).

 

[ranger]

Perhaps rewording the question will help.
If you were given a 9v zener.
How would you select the appropriate bias resistor?

Given that we want I_Z to be 5mA, voltage source of 15V, and a zener voltage of 9V for the specific I_Z:

R = \frac{(V_S - V_Z)}{I_Z}
R = \frac{(15 - 9)}{0.005} = 1.2k \Omega

I don't see how this helps.

Try making up an equivalent circuit using a voltage source, resistor (and diode if you wish).

If we use a regular pn junction diode, the power dissipated is still I_F*V_F. Translate this to zener, we get I_Z*V_Z

It would be good if you could give a straight up answer to all your questions in your next reply. We've been on this circuit way too long.

 

[berkeman]

We've been on this circuit way too long.

Yeah, sorry about that. I've been swamped at work the last few days. I'll post a new set of circuits later today (hopefully).

 

[NoTime]

Given that we want I_Z to be 5mA, voltage source of 15V, and a zener voltage of 9V for the specific I_Z:

R = \frac{(V_S - V_Z)}{I_Z}
R = \frac{(15 - 9)}{0.005} = 1.2k \Omega

I don't see how this helps.

Not really a good choice. The 240 ohm is better.
The people that designed this circuit did know what they were doing even if the connections got printed wrong.
One of the problems I think is that graph is plotted log log.
Makes a nice straight line but hides the fact that the R_z approximates a constant at higher currents.
Given that the regulation of high current (like in this case) rails is low this will translate into your voltage reference being more noisy.
You can also use the bias resistor to correct for some of your op-amp output error terms.

If we use a regular pn junction diode, the power dissipated is still I_F*V_F. Translate this to zener, we get I_Z*V_Z

What is the power dissipation in a battery?
Last I checked a voltage source doesn't dissipate power, only it's internal resistance does.
The equivalent circuit of the zener is basically a battery and resistor.

Power dissipated in the 9v zener with a 240 ohm bias is I_z^2 * R_z or about 16mw.

 

[berkeman]

The equivalent circuit of the zener is basically a battery and resistor.

Power dissipated in the 9v zener with a 240 ohm bias is I_z^2 * R_z or about 16mw.

Interesting. Let me think this through. If I have a 10V voltage source with negligible output resistance, and connect to a 5V zener through a 500 Ohm resistor, then I'll have an Iz of 10mA. The voltage source is supplying 10V*10mA = 100mW to the external circuit of resistor + Zener. The resistor is dissipating 5V*10mA = 50mW, so it would seem that the Zener is also dissipating Vz*Iz = 50mW. If there is a voltage drop across the Zener and a current flowing through it, it seems like the powers of the external supplies and dissipative elements would force the dissipation of the Zener to be Vz*Iz, and not dependent on the dynamic Rz value.

 

[AlephZero]

If there is a voltage drop across the Zener and a current flowing through it, it seems like the powers of the external supplies and dissipative elements would force the dissipation of the Zener to be Vz*Iz, and not dependent on the dynamic Rz value.

I agree - unless NoTime can explain how a Zener dissipation of Iz^2*Rz is consistent with conservation of energy.

 

[NoTime]

Interesting. Let me think this through. If I have a 10V voltage source with negligible output resistance, and connect to a 5V zener through a 500 Ohm resistor, then I'll have an Iz of 10mA. The voltage source is supplying 10V*10mA = 100mW to the external circuit of resistor + Zener. The resistor is dissipating 5V*10mA = 50mW, so it would seem that the Zener is also dissipating Vz*Iz = 50mW. If there is a voltage drop across the Zener and a current flowing through it, it seems like the powers of the external supplies and dissipative elements would force the dissipation of the Zener to be Vz*Iz, and not dependent on the dynamic Rz value.

Well that is my point.
The current flowing thru the zener is not (10v- 5v)/500 ohms.
It is (10v - (5v + (I_z * R_z)) / (500 + R_z)

 

[berkeman]

Well that is my point.
The current flowing thru the zener is not (10v- 5v)/500 ohms.
It is (10v - (5v + (I_z * R_z)) / (500 + R_z)

The current flowing through the Zener has to equal the current flowing through the 500 Ohm resistor in my example, since they are in series. And the voltage across the Zener has to be 5V in my example, since it is a 5V Zener, and 5V is being dropped across the series resistor.

 

[berkeman]

Okay, I think we've had enough fun with the current source circuits (independent of the Zener diode discussion that may still go on independently here for a bit).

Here are a couple more Bad Circuits, this time from the Digital Meets Analog chapter (again, I'm using the old 1st Edition of H&H).

[F] Zero-Crossing Counter -- lots of issues

[G] SR Latch -- pretty easy, but how should it be fixed?

 

[NoTime]

The current flowing through the Zener has to equal the current flowing through the 500 Ohm resistor in my example, since they are in series. And the voltage across the Zener has to be 5V in my example, since it is a 5V Zener, and 5V is being dropped across the series resistor.

Is it 5v? I think not.
Take two equal 10 voltage sources, connect them in parallel.
No current flows.
Take a 10v voltage source a 1 ohm resistor and a 9v voltage source.
The 1 ohm resistor has a 1 volt potential difference across it, will have a current of 1 amp and will dissipate 1 watt.

This is the same situation represented by the zener circuit.
I just looks a little different.

 

[berkeman]

Is it 5v? I think not.
Take two equal 10 voltage sources, connect them in parallel.
No current flows.
Take a 10v voltage source a 1 ohm resistor and a 9v voltage source.
The 1 ohm resistor has a 1 volt potential difference across it, will have a current of 1 amp and will dissipate 1 watt.

This is the same situation represented by the zener circuit.
I just looks a little different.

I think you're carrying the voltage source analogy for a passive Zener diode component a bit to far. How about if you replace the passive reverse-biased silicon Zener diode with 8 passive forward biased silicon diodes, each with a Vf of 5V/8 = 0.625V. What is the difference in DC V and I behavior between the two circuits? I think you'd agree that the total power dissipation of the 8*diodes in the Supply+Resistor+8*diodes circuit is 50mW, right?

 

[NoTime]

I think you're carrying the voltage source analogy for a passive Zener diode component a bit to far. How about if you replace the passive reverse-biased silicon Zener diode with 8 passive forward biased silicon diodes, each with a Vf of 5V/8 = 0.625V. What is the difference in DC V and I behavior between the two circuits? I think you'd agree that the total power dissipation of the 8*diodes in the Supply+Resistor+8*diodes circuit is 50mW, right?

The only distinction between the straight 1 ohm resistor, I exampled, and the zener setup is that in the zener case the 1 ohm resistor is the sum of r_bias + R_z.
Also V_ref occurs at the junction of R_bias and R_z.

I don't see what is so difficult with this concept
And yes, it works this way in practice.