Bad Circuits - Test Your Knowledge

[berkeman]

Yes, I meant the last problem with circuit G.

Well to minimize such offset voltage, one could use a trimming network or simply use an op-amp with very small V_OS or a FET amplifier which has little input current. Another way I see to eliminate the input-offset voltage due to offset current is to make the the resistances of both the inverting and non-inverting input equal. In this way, both inputs will have negligible offset voltage due to offset current becuase they see the same equivalent resistance.

According to the datasheet for the 741, we should expect a typical offset voltage of about 1mV. I guess this also applies to a properly trimmed op-amp?

You've pretty much got [G] figured out, ranger. Good job. You might be able to trim out the remaining offset voltage, but I honestly don't know how stable that would be over temperature and aging. Usually you'll just try to do something that minimizes the offset, and be careful that your gain is not too high. The gain rolloff capacitor in the V- leg to ground is one way to get rid of the DC input offset voltage issue, if you can afford the loss of low-frequency gain, and time constant at circuit turn-on.

So how about the 2nd circuit...?

 

[dlgoff]

Well I didn't look at the specs for the op-amp on the last circuit, but it looks like the non-inverting input needs to be biased up a bit. Wouldn't the output be clipped on the low side?

 

[berkeman]

Well I didn't look at the specs for the op-amp on the last circuit, but it looks like the non-inverting input needs to be biased up a bit. Wouldn't the output be clipped on the low side?

Yes, absolutely. The LM358 is used when the inputs are going to be near ground, but it certainly can't drive down below ground when amplifying an AC signal. So you would need to do something like bias up the V+ input to half of the supply or something in order to keep the output signal centered between the top rail and ground.

So to summarize the answers for the two circuits in Post #62,

[G] As drawn, there will be an input DC offset that will peg the output of the x100 amp. And even with a fix to the input offset from the BJT stage, a V- side capacitor should be used to eliminate the input offset voltage of the opamp, to avoid about a 100mV output offset error.

[J] The output will clip at ground as drawn. The input needs to biased to half of the supply to keep the output nominally centered between the upper rail and ground.

 

[berkeman]

Okay, how about a couple last opamp circuits, and then we can switch gears. Here are two more from H&H Chapter 3 on opamps. I especially like the 2nd one [F], which was published in another textbook as an example of a "good" circuit idea...

[D] Voltage-controlled current source

[F] 200mA "current source"

 

[EugP]

Okay, how about a couple last opamp circuits, and then we can switch gears. Here are two more from H&H Chapter 3 on opamps. I especially like the 2nd one [F], which was published in another textbook as an example of a "good" circuit idea...

[D] Voltage-controlled current source

[F] 200mA "current source"

I'll give it a shot.
[D] I think it has something to do with R.

[F] Quick question about F. That "zener" thing in the middle with 9v next to it, does that mean there's a 9v voltage drop across it?

Sorry if I'm wrong, I'm still an undergrad with almost no real life experience.

 

[NoTime]

I'll give it a shot.
[D] I think it has something to do with R.

[F] Quick question about F. That "zener" thing in the middle with 9v next to it, does that mean there's a 9v voltage drop across it?

Sorry if I'm wrong, I'm still an undergrad with almost no real life experience.

Yes a zener is supposed to show a constant voltage.
It was intended as a 9v reference. Note that there is a maximum current that can flow thru the 240 ohm resistor before the voltage will drop below 9v.

In [D] there is something you can do with R that would fix the major issue.
What do you think it is?

 

[EugP]

Yes a zener is supposed to show a constant voltage.
It was intended as a 9v reference. Note that there is a maximum current that can flow thru the 240 ohm resistor before the voltage will drop below 9v.

In [D] there is something you can do with R that would fix the major issue.
What do you think it is?

For [D] I think the R and "load" should be switched.

I'm still thinking about [F] though, I've never seen that before.

 

[AlephZero]

D won't work because the opamp is supposed to be operating in a linear mode, but its two inputs are are different voltages. For example think about the case where the load has zero resistance. One input is at 0V, the other is at the "control voltage".

Swapping R and the load seems to fix that problem.

 

[AlephZero]

F: I'm assuming the op amp has a +15/-15 power supply, not +15/0.

If you took away the 240 resistor, this would "sort of" work, except that all the load current would flow through the zener. A typical small zener wouldn't take 200mA current and 1.8W dissipation though.

I suppose the 15V / 240R is meant to supply 25 mA through the zener which seems a sensible design value.

If that's identified the problem, somebody else can have a go at fixing it.

 

[ranger]

For circuit F, I think we need a resistor from the non-inverting input to ground to deal with offset voltage due to the small input bias current. Or are we intentionally leaving it this way to amplify the [offset] voltage difference to produce an output?

For circuit D, I also think that swapping the positions of the R and the load will work. With the circuit we'd want to fix it by having R connected to the inverting input to ground. And have the load connected to R and the output of the op-amp.

 

[berkeman]

Good job folks on circuit [D]. Yep, the reference resistor and the load were swapped. Doh!

On [F], if the opamp is doing its job and holding its V- input at ground, what is the voltage on top of the "9V" Zener (just calculate the resistor voltage divider first, ignoring the Zener diode). So what does this mean the current through the load really is?

It looks like they were trying to make an opamp circuit that would take a current 9V/45 Ohms and pass that current through the load. Maybe altering the 240 Ohm resistor value might make this circuit work, but it's dumb to brute force push extra current through the 9V Zener (and have to use a bigger, more expensive Zener part), when the opamp is supposed to be the amplifying device. Do you see a way to use the topology of circuit [D] instead (after being fixed with the swap that we talked about)?

 

[NoTime]

Maybe altering the 240 Ohm resistor value might make this circuit work

Not if a 200ma load current is desired.
200ma * 45 Ohms is a 9v drop.
At least they got the values right even if they wired it wrong.

 

[ranger]

Good job folks on circuit [D].
It looks like they were trying to make an opamp circuit that would take a current 9V/45 Ohms and pass that current through the load. Maybe altering the 240 Ohm resistor value might make this circuit work, but it's dumb to brute force push extra current through the 9V Zener (and have to use a bigger, more expensive Zener part), when the opamp is supposed to be the amplifying device. Do you see a way to use the topology of circuit [D] instead (after being fixed with the swap that we talked about)?

So our problem now is to not push 200mA thru the zener? If we used the fixed version of circuit D, we would apply the reference voltage of 9V to the non-inverting input. We'd of course need to bias the zener properly by choosing an appropriate resistor. Since the fixed version of circuit D uses negative feedback, we would have the zener voltage at the inverting input. So if we use R to be 45ohms, we'd have a current of 200mA thru the load. IN this configuration we don't have to worry about excess current thru the zener because we are using the resistor to get the appropriate zener current for voltage regulation.

 

[NoTime]

Looks like you have the right idea.

we would apply the reference voltage of 9V to the non-inverting input.
...
we would have the zener voltage at the inverting input.

Yes, the negative feedback voltage will approximate the reference voltage.
Why won't it be exact?

We'd of course need to bias the zener properly by choosing an appropriate resistor.

The choice of zener bias resistor (the 240 ohm) will affect the final load current.
Why?

 

[ranger]

Yes, the negative feedback voltage will approximate the reference voltage.
Why won't it be exact?

Because the feedback loop tends to subtract some of the output voltage proportional to Z*Vout; where Z is impedance and is obtained by applying the voltage divider to resistors on the inverting input (feedback network)?

The choice of zener bias resistor (the 240 ohm) will affect the final load current.
Why?

Does it have to do with the fact that the inverting and non-inverting inputs see different resistance? And becuase of the high bias current on the non-inverting input, we have offset voltage due to this?

 

[NoTime]

Because the feedback loop tends to subtract some of the output voltage proportional to Z*Vout; where Z is impedance and is obtained by applying the voltage divider to resistors on the inverting input (feedback network)?



Does it have to do with the fact that the inverting and non-inverting inputs see different resistance? And becuase of the high bias current on the non-inverting input, we have offset voltage due to this?

No. The answers have to do with the properties of real op-amps and zeners.

You are vaguely headed in the right direction with the first answer.

Your second answer makes me wonder just where did you put the 240 ohm resistor and zener?

 

[ranger]

No. The answers have to do with the properties of real op-amps and zeners.

You are vaguely headed in the right direction with the first answer.

Your second answer makes me wonder just where did you put the 240 ohm resistor and zener?

Well, let me recap [IIRC] the properties of real op-amps. They have offset voltage, bias and offset [input] current, output current limit, and gain degradation with increasing frequencies.

With regards as to why the inverting input will not see the exact zener voltage - the best answer I can come up with is that the feedback network subtracts some of the output voltage.

The 240ohm is connected to the 15V power supply, the zener to ground (anode:ground; cathode:non-inverting input), and the non-inverting input is hooked up to the zener to get the reference voltage. No?

 

[NoTime]

The 240ohm is connected to the 15V power supply, the zener to ground (anode:ground; cathode:inverting input), and the inverting input is hooked up to the zener to get the reference voltage. No?

Ouch! No!

Where did you lose this statement?

we would apply the reference voltage of 9V to the NON-inverting input.

Take another look at the revised circuit [D]

 

[ranger]

Ouch! No!

Where did you lose this statement?Take another look at the revised circuit [D]

That was a typo, sorry. I'll fix it in post #87 so others don't get confused.

Care to give anymore hints?

 

[NoTime]

That was a typo, sorry. I'll fix it in post #87 so others don't get confused.

Care to give anymore hints?

The statement might have been a typo on your part.
But, it was correct
And the reason I thought you had this figured out.

 

[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.