Bad Circuits - Test Your Knowledge

[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.

 

[berkeman]

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.

I'd like to take this offline via PM with NoTime for a bit, to see if I can better understand what he is saying. We'll report back with a resolution in the next couple days, I would guess. That will keep this thread going on the new problems more cleanly.

Thanks.

 

[ranger]

For circuit F [Zero-Crossing Counter], I'm asking myself - what if the input signal is noisy? We surely won't want false triggering of the counters when the input signal [nosily] oscillates about its zero crossing position. The way the circuit is now, its just a plain old comparator; a perfect candidate to swing its output as a function of the noise that may be present on the input signal on the non-inverting input. We need to give the circuit a good degree of noise immunity. For this I suggest that we use positive feedback to give the circuit hysteresis. We would effectively be building a schmitt trigger around the 741 by doing this.

I'll see if I can find more stuff wrong with F tomorrow.

 

[ranger]

Hey berkeman, is it too early to sum up the problems we were having with the current source op-amps?

 

[berkeman]

Hey berkeman, is it too early to sum up the problems we were having with the current source op-amps?

Not at all. Would you do that for us please? (You understand them, I'm pretty sure.) Please leave out the Zener power dissipation issues right now, though. NoTime and I are using SPICE simulations to resolve the disagreements via PMs.

BTW, I don't know if this came out in the Zener discussions or not, but 5.1V Zeners have the tempco closest to zero. So if you need to use a Zener for a voltage reference and want a good tempco, use a 5.1V Zener (biased at its specified Iz) and multiply it up or down if you need a different voltage.

 

[berkeman]

For circuit F [Zero-Crossing Counter], I'm asking myself - what if the input signal is noisy? We surely won't want false triggering of the counters when the input signal [nosily] oscillates about its zero crossing position. The way the circuit is now, its just a plain old comparator; a perfect candidate to swing its output as a function of the noise that may be present on the input signal on the non-inverting input. We need to give the circuit a good degree of noise immunity. For this I suggest that we use positive feedback to give the circuit hysteresis. We would effectively be building a schmitt trigger around the 741 by doing this.

I'll see if I can find more stuff wrong with F tomorrow.

That's definitely part of the problem. But even if there is no noise in the input signal, a comparator almost always will need some form of hysteresis (via attenuated positive feedback) around it. What you commonly run into, is that the act of the comparator or opamp switching its output generates enough of a transient in the chip's internal power supplies and internal circuitry to make apparent differential noise at the inputs. So even if you have a perfectly clean signal and great PCB layout and decoupling, the switchover point of a comparator without hysteresis will have multiple buzzing transitions. If you ever see a comparator shown in a circuit without hysteresis, that is a red flag to look further.

That having been said, in the last work project that I helped out with, we used comparators with zero hysteresis in the reciever of a network transceiver. I have to be a little careful what I say about this, but as a Quiz Question -- When do you think you could use a comparator with zero hysteresis when receiving an analog signal with the goal of accurately digitizing its zero crossings?


(EDIT -- clarified a bit about the output --> internal noise feedback issue)

 

[AlephZero]

More problems with the zero-crossing counter:

There are two power supply options shown, and each has a problem.

1. With the +15/-15 supply, the input to the 7493 will flip between about +13/-13V. Since the 7493 is a TTL device with 5V upply, this is not a good idea. The absolute max voltage for TTL inputs is usually +5.5V or +7V.

2. With the +5/0 supply, the comparator won't work at all because one op-amp input is grounded. The output from the opamp will probably be stuck close to +5V.

3. To drive TTL from an opamp with a +5/0 supply, you need an opamp where the output can swing right to to supply levels. The TTL logic levels are < 0.8V and > 2.0V. A 741 isn't the right part to use for this. You need an opamp that is designed to work on a 5V supply and the output can swing rail-to-rail.

 

[ranger]

That's definitely part of the problem. But even if there is no noise in the input signal, a comparator almost always will need some form of hysteresis (via attenuated positive feedback) around it. What you commonly run into, is that the act of the comparator or opamp switching its output generates enough of a transient in the chip's internal power supplies and internal circuitry to make apparent differential noise at the inputs. So even if you have a perfectly clean signal and great PCB layout and decoupling, the switchover point of a comparator without hysteresis will have multiple buzzing transitions. If you ever see a comparator shown in a circuit without hysteresis, that is a red flag to look further.

Another little circuit design tip that I have to put in (as H&H calls it) bag of tricks.

That having been said, in the last work project that I helped out with, we used comparators with zero hysteresis in the reciever of a network transceiver. I have to be a little careful what I say about this, but as a Quiz Question -- When do you think you could use a comparator with zero hysteresis when receiving an analog signal with the goal of accurately digitizing its zero crossings?

At first I can see no need for this (my lack of experience), but the more I think about it, there may be use after all. There has to be some transmission protocol out there that has a specification where it requires a very small input differential to be detected and pass as valid data. If we add hysteresis to our comparator, there is a chance that we may lose some data.

 

[ranger]

Voltage-controlled current source and 200mA "current source" - solutions

Wrapping up the possible problems with the circuits in post #74; circuits D and F.

Circuit D - Voltage-controlled current source
*The solution for this one is pretty simple. We need to swap the positions of the resistor and the load. D won't work because the opamp is supposed to be operating in a linear mode, but its two inputs are different voltages^[1]. 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.

Circuit F - 200mA "current source"
*Pushing 200mA of current through the zener may be a bad idea. The original configuration of circuit F does just this. As a fix for this problem, we could use the topology in the fixed circuit D. It would be best to have the zener be used as a voltage reference (to the non-inverting input) and have the load hooked up to the inverting input - circuit F fixed. This fixed version uses negative feedback, meaning that we would have the zener reference voltage at the inverting input - remember one of the golden rules of op-amps as outlined in the Art of Electronics - the op-amp's output will attempt to do what it can to bring the voltage differential between the two inputs to zero (provided we use negative feedback; see chapter 4 for more discussion on this). 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.

*Even with the fix mentioned above, we still have problems becuase we are dealing with real op-amps here i.e. non-ideal. 200mA is a lot of output current to ask from a general purpose op-amp.

*Next we have the issue of unbalanced input impedances. Remember that we have input bias current. Due to unmatched impedances on both op-amp inputs, we'll have some error introduced. In addition to matching the impedances, we could use an op-amp with very little input bias current (BJT input stage) or very little leakage current (FET input stage).

 

[sheldonstv]

whats wrong with this circuit?

 

[berkeman]

At first I can see no need for this (my lack of experience), but the more I think about it, there may be use after all. There has to be some transmission protocol out there that has a specification where it requires a very small input differential to be detected and pass as valid data. If we add hysteresis to our comparator, there is a chance that we may lose some data.

For the data detection in the network that I have in mind, it is important to detect the zero crossings, without any delay or offset that hysteresis would generate. The buzzing that can happen at the zero crossing detection by comparators without hysteresis is actually removed by subsequent digital signal processing in this case. The buzzing is at a much higher frequency than the data rate, so it's relatively easy to filter out with DSP. In a circuit like the Bad Circuit edge counter, extra edges of any frequency will give an incorrect answer.

 

[berkeman]

More problems with the zero-crossing counter:

There are two power supply options shown, and each has a problem.

1. With the +15/-15 supply, the input to the 7493 will flip between about +13/-13V. Since the 7493 is a TTL device with 5V upply, this is not a good idea. The absolute max voltage for TTL inputs is usually +5.5V or +7V.

2. With the +5/0 supply, the comparator won't work at all because one op-amp input is grounded. The output from the opamp will probably be stuck close to +5V.

3. To drive TTL from an opamp with a +5/0 supply, you need an opamp where the output can swing right to to supply levels. The TTL logic levels are < 0.8V and > 2.0V. A 741 isn't the right part to use for this. You need an opamp that is designed to work on a 5V supply and the output can swing rail-to-rail.

Good stuff, AlephZero. There's at least one more thing that is generally wrong with the circuit, having to do with the counters. Any ideas?

 

[berkeman]

whats wrong with this circuit?

I'm not very good with 555 circuits... Where is this one from?

 

[sheldonstv]

its one of my own design with a deliberate mistake on the schematic-not too hard to spot if you have a close look...

 

[ranger]

For circuit F [Zero-Crossing Counter]:

*Should we not use a comparator chip like the 311? We could build hysteresis around the 311. I'm suggesting this becuase of the limited slew rates of general purpose op-amps.*Would it also be a good idea to use a pair of input protection diodes on the input of the op-amp [thats getting the analog signal]? Just being careful that we don't exceed the maximum voltage differential allowable by the 741. But then again if we're dealing with relatively small signals there's no need for this.

 

[ranger]

Circuit G - SR Latch

It would be a idea to have both inputs grounded when we want to send in a LOW signal. Unconnected pins are usually a bad idea. If we are dealing with TTL inputs, an unconnected input is about 1.3V, but there is no current. Also there is no noise immunity in this case.
If we were using CMOS, we should also have the inputs grounded when we need a LOW.

 

[sheldonstv]

nobody know?re 555 circuit?

 

[dlgoff]

"nobody know?re 555 circuit?"

okay. It's hard to tell from you skematic but it looks like you have the output connected to the threshold and the discharge.

 

[sheldonstv]

that is one fault yes...there is one more if you have a careful look...

 

[berkeman]

that is one fault yes...there is one more if you have a careful look...

One thing that would help on your Bad Circuit post, would be if you could give a web pointer to the best web page for info/tutorial help on the 555 IC. I've played with them way back in school, but never really liked them much. But there are a fair number of posts here on the PF about 555 questions (mainly from students having to use them in labs), so a good web tutorial about them would be helpful. There is the datasheet that I could go find and read through, but it would be better if there were something like a "Circuit Ideas" web page for the 555 timer/monostable IC. Anybody have a good pointer? Thanks much!

 

[ranger]

Hey berkeman, no comments on the previous circuits you posted (post #134 and #135)?