Bad Circuits - Test Your Knowledge

[berkeman]

Bad Circuits -- Test Your Knowledge

We were having a discussion in a different thread about the fun sections at the end of each chapter in "The Art of Electronics" by Horowitz and Hill, where they show simple schematics as examples of Bad Circuits, without any explanation of what is bad or wrong about them.

It was mentioned in that thread that sometimes it can be hard for students learning about circuits to determine what exactly is "bad" about the circuits shown, so this thread is meant to provide a place where we can discuss those circuits, and also to post other problematic circuits that folks have run across in their work or studies.

So if one of H&H's "Bad Circuits" is bothering you, or if you have run across any examples of errors in schematics that others can learn by, please post them here for others to see and discuss.

I'd like to keep this thread as tutorial as possible, so here are a couple requests:

-1- Please do not post homework problems in this thread. You all know where homework questions should be posted for tutorial help.

-2- If you can see the problem with a circuit easily, please hang back for a bit to let others discuss it, and maybe offer a small clue or two. Students will learn better if there is a discussion about the circuit, rather than hearing the answer early and outright.

That's it for now. I'll pull out my H&H copy and post something on Monday, unless somebody beats me to it.


EDIT -- BTW, we are posting small parts of Horowitz and Hill's textbook under the Fair Use portions of copyright law. They are being used for educational purposes only. Many of us here are fans of this book because of its practical approach to basic electronics, and use of real-world components throughout the course of the book. I encourage you to take a look at the book at your local technical library, or at your bookstore. It is now in it's 2nd edition:

https://www.amazon.com/dp/0521370957/?tag=pfamazon01-20

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[Maxwell]

What an awesome idea.

 

[berkeman]

What an awesome idea.

Yeah, we can thank @ranger for the idea. It came up in a homework help thread last week.

I'll post something here in a few minutes to kick things off.

 

[EugP]

This is a really good idea. Can't wait for it to start.

 

[berkeman]

Okay, here are a couple easy ones to start off with. These are from Chapter 9 ("Digital Meets Analog") of the first edition of H&H, page 450. There are at least two things wrong with each circuit. What are they, and what would be the best way to make things right?

Remember, if it's easy for you, hang back for a bit. We'll get to some harder ones soon enough.


EDIT -- BTW, let me know if this scan is hard to read. If it is, I'll figure out some way to enlarge the attachments in the future.

 

[EugP]

Could you please enlarge the picture? I can't make out the numbers.

 

[berkeman]

Could you please enlarge the picture? I can't make out the numbers.

Okay, I'm still figuring out how best to scan and post these. Here are (a) and (b) enlarged. How's that?

 

[Maxwell]

Well for (b) you need a resistor in front of the LED.

For (a), should the resistor be in series with the cap? I think so.

 

[goldy]

Hello there,
First of i would like to appreciate you for this wonderful idea.
And i think i can answer the second one..
I am not sure, but i think i am right, the two errors are:
1. There is always a resistor connected in serires with LED's because if the applied voltage is more than 1.5V than it get burnt.
2. There is no need to ground a LED.
please don't get angry if i am wrong, i am not sure about this.

 

[berkeman]

Maxwell and goldy are correct that at least a resistor is missing in (b), but that's about all that's correct so far. Keep on thinking them through...

Remember, H&H emphasizes real devices with real datasheet specifications and real-world signals...

 

[goldy]

and in (a)
there is no need to ground a capacitor.

 

[EugP]

Okay, I'm still figuring out how best to scan and post these. Here are (a) and (b) enlarged. How's that?

Perfect, thanks I appreciate it.
I have to be honest, I'm not very familiar with digital logic. I know the basics, but I don't have an in-depth knowledge of digital logic in real life problems and use in circuits.

If I may make a suggestion, I think that if someone posts an answer as to what is wrong with a circuit, they should explain why, just so the people who don't understand (like me ) could learn.

 

[goldy]

Hey EugP,
I am agree with you and I already have explained my first answer.
I am not quite sure with my second answer...

 

[ranger]

Berkeman, I'll give a crack at a) My reasoning is as follows:

As the circuit currently is, it has a RC time delay of 1s. This means that whatever was inputed will show up at the output with a time delay. If a high was the output of the first gate, the cap would take a great deal of time to charge up to that value (5RC). Will the signal of the first gate last that long? Then to discharge, will also take a long time. The issue with large time constants makes a difference because TTL logic uses a fairly narrow range for HIGH and LOW. What if only a short pulse was present at the the output of first gate? Then the final output could more that likely be in the "forbidden region". For cases with large time delays I'd use CMOS logic because of a more forgiving range of HIGH and LOW.

 

[Maxwell]

Yeah, I have a feeling that the RC time constant is too large.

 

[ranger]

and in (a)
there is no need to ground a capacitor.

Hi goldy,

I'm curious as to why you stated this. And also the same for the logic indicator circuit.

 

[goldy]

Hi goldy,

I'm curious as to why you stated this. And also the same for the logic indicator circuit.

Oh i am really very sorry that i have mentioned above, its just a misconception but i am still quite confused with the circuit (b), what is the need to ground the LED.

 

[ranger]

Oh i am really very sorry that i have mentioned above, its just a misconception but i am still quite confused with the circuit (b), what is the need to ground the LED.

[Circuit B]
What would happen if you left it floating?

Also if you didnt ground it, but connected to a voltage terminal of some odd polarity (+6V), would the circuit still function as it should?

 

[edmondng]

For circuit B;

i think for circuit b is should be +5V -> LED -> resistor -> output of logic. the way it has it configured now works (need resistor of course) but just bad practice because the output of the logic is driving the current to power the led.

I'll rather have a source driving the led, and grounding the other side. Much like a ucontroller, make a port be open drain and have the led connect to the source. toggle the port low to turn on led.

Is this right?

 

[chroot]

There are definitely some superficial lessons to be learned here. The second circuit (the one with the LED) will indeed fail catastrophically if you do not put a current-limiting resistor in series with the LED. As soon as you apply power to it, it'll burn itself up.

However, there are actually some much deeper lessons to be learned here, too. Slapping a resistor into the second circuit will prevent it from blowing itself up, but the circuit still won't work well at all. Anyone know why?

- Warren

 

[chroot]

Oh i am really very sorry that i have mentioned above, its just a misconception but i am still quite confused with the circuit (b), what is the need to ground the LED.

Generally speaking, any device with two terminals needs to have both connected in order for it to work at all. In the case of an LED, light is produced by passing current through it. If you only connect one terminal, you obviously have no path for current, so it cannot possibly light up.

- Warren

 

[chroot]

I'll rather have a source driving the led, and grounding the other side. Much like a ucontroller, make a port be open drain and have the led connect to the source. toggle the port low to turn on led.

It doesn't particularly matter if the gate is sourcing or sinking the current -- the current is still going through one of the gate's transistors, and thus is still bad practice.

- Warren

 

[ranger]

However, there are actually some much deeper lessons to be learned here, too. Slapping a resistor into the second circuit will prevent it from blowing itself up, but the circuit still won't work well at all. Anyone know why?

- Warren

Okay, let's see. According to the schematic, the gate is also fanning-out to other logic gates; hence more current. And according to your post #22, this is a bad practice. If I were to build this, I'd use a buffer right after the first gate. And to add to this, TTL based chips have a limited fan-out range when compared to CMOS. Solves the problem? Was that even a problem in the first place?

 

[chroot]

Solves the problem? Was that even a problem in the first place?

Well, can you tell me why fanout matters?

- Warren

 

[edmondng]

It doesn't particularly matter if the gate is sourcing or sinking the current -- the current is still going through one of the gate's transistors, and thus is still bad practice.

- Warren

how about putting a diode after the gate? then it always flow one way only. buffer amp isn't that bad but might as well use an opto isolator

 

[ranger]

Well, can you tell me why fanout matters?

- Warren

Well according to the schematic, the TTL circuit is also driving other gates in addition to the LED. Because TTL logic gates have relatively low input impedance [when compared to CMOS] it would obviously have a more limited range. Once this limit is met, any attempt to drive more logic inputs will cause voltage levels to fall, and with digital logic, that's no good. Hence the need for a buffer (or opto-isolator). I'm of course basing this response because I saw "to other gates" written on the schematic.

 

[chroot]

how about putting a diode after the gate? then it always flow one way only. buffer amp isn't that bad but might as well use an opto isolator

The LED is itself a diode -- current only goes through it in one direction -- so adding another diode would accomplish nothing. I'm also not sure why you bring up opto-isolators. The "sending" half of an opto-isolator is just an LED!

- Warren

 

[chroot]

Well according to the schematic, the TTL circuit is also driving other gates in addition to the LED. Because TTL logic gates have relatively low input impedance [when compared to CMOS] it would obviously have a more limited range. Once this limit is met, any attempt to drive more logic inputs will cause voltage levels to fall, and with digital logic, that's no good. Hence the need for a buffer (or opto-isolator). I'm of course basing this response because I saw "to other gates" written on the schematic.

You're overthinking the distinction between CMOS and TTL here. Fanout matters because gates, when conducting, act like current sources. They source (or sink) current into their load capacitances. This takes time.

If you increase the load capacitance, it will take more time for the gate to charge or discharge it, slowing down your circuit's maximum clock frequency. Similarly, decreasing the gate's output current capability will also slow down your circuit. Connecting an LED this way serves to rob the gate of output current it could be using to charge its load capacitance. This means that the LED will drastically reduce your circuit's maximum clock frequency.

If you're building an essentially DC circuit on a breadboard in an electronics lab, you will never notice this effect -- at all. If you're designing something that you intend to run at even a couple of kilohertz, though, this current-robbing LED will kill your design.

- Warren

 

[berkeman]

Berkeman, I'll give a crack at a) My reasoning is as follows:

As the circuit currently is, it has a RC time delay of 1s. This means that whatever was inputed will show up at the output with a time delay. If a high was the output of the first gate, the cap would take a great deal of time to charge up to that value (5RC). Will the signal of the first gate last that long? Then to discharge, will also take a long time. The issue with large time constants makes a difference because TTL logic uses a fairly narrow range for HIGH and LOW. What if only a short pulse was present at the the output of first gate? Then the final output could more that likely be in the "forbidden region". For cases with large time delays I'd use CMOS logic because of a more forgiving range of HIGH and LOW.

Interesting. I'm not sure H&H did this on purpose, but they might have as a subtle point in (a). When a schematic has 0.001 as a capacitor value, that will mean C=0.001uF. Often in the notes for a schematic, they will say something like "Values shown for inductors are in mH and for capacitors are in uF, unless otherwise noted." I personally always show the prefix and units on caps and inductors on schmatics, but that's up to personal style.

So, the RC time constant of (a) is more like 1ms, but there are still several problems with the circuit. Major hint on (a) -- the input to a digital gate usually has a fast transition time, usually measured in nanoseconds (ns). What issues might arise (there are at least 2 issues) with an input signal that is taking more like 1ms to ramp up or down?


EDIT -- holy smokes! I took too long to type my response...now I'm way behind!

 

[chroot]

What issues might arise (there are at least 2 issues) with an input signal that is taking more like 1ms to ramp up or down?

EDIT -- holy smokes! I took too long to type my response...now I'm way behind!

And now we're on to the third-order problems...

- Warren