Feeling stupid: how to solve an AoE exercise

[Moriambar]

Hi,
I'm currently studying a bit more seriously the AoE.
Here is the circuit that I'm looking at (third editio, page 77, right column)


It is supposed to generate an output pulse from a step in the input.
I get how it does that etc. In the next page, there is the following exercise, though:

Show that the output pulse width for the circuit is approximately T_pulse = 0.63 R₃C₁=0.63µs. A good starting point is to notice that C₁ is charging exponentially from -4.4V toward +5V, with the time constant as above

My take is this: first of all the voltage of the capacitor, albeit charging from a 5V power supply will eventually stop at ≈0.6V, due to the fact that Q₂ will start conducting, therefore stopping the charging of the capacitor.

Recollecting the memories of the RC capacitor charging circuit (AoE 3^rd ed, page 22), I see that the voltage I am searching for is or should be of the form V_out=V_power+Ae^-t/RC.

EDIT: to clarify I consider V_out to be the voltage at the base of Q₂, which is the right terminal of C₁.

Were V_power=5V.
I then try and find the A parameter, knowing that at t=0, V_out=-4.4V; thence I have -4.4V=5V+A=>A=-9.4V

so my equation should be V_out=5-9.4e^-t/RC.

I then solve for V_out=0.6V, and try and find t. I have

0.6V=5V-9.4V e^-t/RC => 4.4V=9.4V exp[...] => 4.4/9.4=e^-t/RC => ln(4.4/9.4)= -t/RC => t= -ln(4.4/9.4) RC but the number I get is ≈0.75 RC and not 0.63 RC.

Where and why am I gone wrong?

Cheers.

[iMo]

I then try and find the A parameter, knowing that at t=0, V_out=-4.4V; thence I have -4.4V=5V+A=>A=-9.4V

The Vout only goes from 0V (aprox) to Vpower, it cannot be -4.4V.

You should focus yourself on the voltage at the Q2 base.. The voltage there goes from -4.4 to 0.6V.

[Moriambar]

The Vout only goes from 0V (aprox) to Vpower, it cannot be -4.4V.

You should focus on the voltage at the Q2 base..

Ok this I don't get. The previous paragraph of the text explained that

[when a step is applied and Q₁ collector is forced to ground]
because of the voltage across C₁, this brings the base of Q₂ momentarily negative, to about -4.4V

I interpreted this in the following manner: since the ∆V across the capacitor is ≈4.4 V (left terminal at 5V, right one at 0.6), when the voltage of the first terminal (which is connected to Q₁'s collector) drops to ground, the other terminal shows to be at -∆V with respect to it, ie -4-4V.

Again what am I missing?

EDIT

You should focus yourself on the voltage at the Q2 base.. The voltage there goes from -4.4 to 0.6V.

How can Q₂'s base be at a different voltage than C₁'s right terminal? They're connected directly... so no ∆V should appear there, right?

[iMo]

The capacitor C1 in "idle mode" will have +5V at the left terminal and 0.6V at the right terminal.
When the Q1 opens, the right terminal of the C1 will be -5V + 0.6 = - 4.4V, and the left will be 0V. From that time t=0 and VQ2base=-4.4V (Q2 closed) and Vout=5V C1 starts to discharge itself, such in time "t=.." it will reach VQ2base= +0.6V which opens Q2 (and Vout drops to 0V).

[Moriambar]

The capacitor C1 in "idle mode" will have +5V at the left terminal and 0.6V at the right terminal.
When the Q1 opens, the right terminal of the C1 will be -5V + 0.6 = - 4.4V, and the left will be 0V. From that time t=0 and VQ2base=-4.4V (Q2 closed) and Vout=5V it starts to charge itself, such in time "t=.." it will reach +0.6V which opens Q2 (and Vout drops to 0V).

thanks for your reply and patience.
So are you saying that I should focus about the other part of the capacitor charging up from 0 to 5V instead of considering what the right side is doing?

I still am confused and not capable of calculating
Thanks

[Moriambar]

nvm, it's in the errata:
https://artofelectronics.net/errata/

[iMo]

When calculating the time "t" (while the Vout=5V) you have to consider the voltage rising from -4.4V to +0.6 at the base of the Q2.
The rising voltage follows "discharging the C1" via the base resistor R3 of the Q2.
The Q2's base-emitter is a diode basically. With negative voltage (like -4.4V) on the base with that wiring the diode is reverse polarized (and the Q2 is closed).

PS: a simulation..

[Moriambar]

When calculating the time "t" (while the Vout=5V) you have to consider the voltage rising from -4.4V to +0.6 at the base of the Q2.
The rising voltage follows "discharging the C1" via the base resistor R3 of the Q2.
The Q2's base-emitter is a diode basically. With negative voltage (like -4.4V) on the base with that wiring the diode is reverse polarized (and the Q2 is closed).

PS: a simulation..

yes thanks. Unfortunately the error was not mine, but it was in the errata. So I got the correct solution to start with, and thus I think I already understood it. Thanks for your reply anyway. and your patience too

[rstofer]

Keep the errata in front of you at all times.  There isn't a book published that doesn't have errors.  The more diligent you are with understanding discrepancies, the more you will need the errata.   BTW, there's no guarantee that the errata is all-encompassing.  There may by other errors.

Learn to use LTspice and simulate every circuit you possibly can.  It's one thing to know the math, it's another to see it simulated.  I would go further and suggest that a breadboard is another step in the learning process.  But that takes a scope.  Once again, I'm going to recommend the Analog Discovery 2 but I'm not going to duplicate what I have written over the last couple of weeks.

There are a couple of links in this message:

https://www.eevblog.com/forum/beginners/a-cost-effective-upgrade-from-a-6022bl-from-a-beginner/msg2990108/#msg2990108

[Moriambar]

Keep the errata in front of you at all times.  There isn't a book published that doesn't have errors.  The more diligent you are with understanding discrepancies, the more you will need the errata.   BTW, there's no guarantee that the errata is all-encompassing.  There may by other errors.

Learn to use LTspice and simulate every circuit you possibly can.  It's one thing to know the math, it's another to see it simulated.  I would go further and suggest that a breadboard is another step in the learning process.  But that takes a scope.  Once again, I'm going to recommend the Analog Discovery 2 but I'm not going to duplicate what I have written over the last couple of weeks.

There are a couple of links in this message:

https://www.eevblog.com/forum/beginners/a-cost-effective-upgrade-from-a-6022bl-from-a-beginner/msg2990108/#msg2990108

hi,
thanks for yout reply. I actually already have a bench and scope, and already worked a bit on many projects but still at an amateur level (mainly focused on i2c and arduino/attiny). Also I always used mosfets since I basically just had only the need for switches in my projects.
BJTs have always been incredibly complicated for me and so I decided to tackle them using basically the only book I have (AoE).

Regarding breadboarding I should be fine.

LTSpice is another beast and I actually couldn't understand how to make it work for the life of me. I can draw a circuit but then… that's it.
It does nothing.

[iMo]

Here is the .asc file with your example.
Double click the file and it opens in LTspice.
Click at the runner there and it opens the black result window as well.
Go to the schematics window and click at, for example, "out" wire with the cursor (a red "probe" appears when hovering over a wire) - you should get the waveform visible.
When in the black window you may zoom-in by framing a part of waveform with the cursor while pressing a mice button.

[rstofer]

Once the schematic is entered, the tricky bit might be getting the transient analysis command in place:

Edit->Spice Analysis then select the Transient tab and set the end time.  Close the dialog.  The example uses an end time of 1 second.  This is far too long and could be changed to something much smaller.  OR, just use the window approach mentioned to zoom in on the important parts of the waveform.

Click on the Running Man button to run the analysis.

Setting up the signal source is also an issue:  Right-click on the voltage source and select the PULSE radio button in the upper left.  Fill in the blanks and you are good to go.

If you get a trace color that is essentially invisible (like that lime green trace on a white background), right click on the signal name at the top of the plot window and select another color.

I don't use the default black background
Tools->Control Panel->Waveforms Tab->Color Scheme Button->"Click to edit Background color->Set RGB values to 255 for white background.

The reason I don't use the black background is that it uses too much toner if I print the screen image.  That, and I prefer the white background.

[rstofer]

hi,
thanks for yout reply. I actually already have a bench and scope, and already worked a bit on many projects but still at an amateur level (mainly focused on i2c and arduino/attiny). Also I always used mosfets since I basically just had only the need for switches in my projects.

Then you are good to go in terms of visualization.  You can always generate a square wave with your Arduino.

There are some example problems that aren't worth the time to breadboard them.  Others provide a lot of learning in a single example.  This circuit is one of them.  Change the capacitor value and see how the pulse changes.  Sure, it's easy to understand it mathematically, it's more important to know it really works using actual components.

[Moriambar]

hi,
thanks for yout reply. I actually already have a bench and scope, and already worked a bit on many projects but still at an amateur level (mainly focused on i2c and arduino/attiny). Also I always used mosfets since I basically just had only the need for switches in my projects.

Then you are good to go in terms of visualization.  You can always generate a square wave with your Arduino.

There are some example problems that aren't worth the time to breadboard them.  Others provide a lot of learning in a single example.  This circuit is one of them.  Change the capacitor value and see how the pulse changes.  Sure, it's easy to understand it mathematically, it's more important to know it really works using actual components.

Thanks, I usually generate a square wave with a 555, but still have to understand well how to get 50% duty cycle, but I plan to do it later.

I feel the AoE may not be the best book to learn on, but if one is really committed, the examples are never trivial, and they always make you think. Wonderful

[Renate]

Hey, that was fun!
The interesting part is that it is the exponential decay of the voltage across the resistor that we care about.
The cap could have been fed with a bias of 1000 V (but still P-P of 4.4 V err, I meant 5 V) and it would all work out the same.

On the 555 duty cycle:
Always remember, pin 3 (totem pole output) and pin 7 (open collector) are the same behaviour.
Want a 50%? Connect pin 3 through an R to 2 & 6 & cap to ground.