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Electronics => Beginners => Topic started by: DiDBoGDaN on November 18, 2023, 09:48:14 pm
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Hi, my name is Bohdan, I'm completely new to electronics and I'm currently reading this great book The Art of Electronics by Paul Horowitz, Winfield Hill. I recently started a section on transistors and I have a few questions here. It is about "Generating a short pulse from a step-shaped input signal" (image attached). The book says, "A +5 V positive input step brings Q1 into saturation (note the values of R1 and R2), forcing its collector to ground; because of the voltage across C1, this brings the base of Q2 momentarily negative, to about −4.4 V.". Can you tell me why the voltage drops to -4.4V and not 0 or stays at +0.6V? I understand , that initially the left side of the capacitor is charged to +5 V and the right side to +0.6 V due to the PN junction of the base-emitter. And when the transistor Q1 opens, the left side begins to discharge to almost 0 V. But why does the right side go to -4.4V... How the voltage even lower than 0? I just can't figure it out.Thanks for your reply. [attachimg=1]
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Sorry, don't know what's wrong with previous attachment
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Capacitors temporarily "resist" a change of the voltage across them. So if the voltage on the one side goes up or down the other side tries to do the same. The collector of the first transistor goes down by almost 5 Volts The other side tries to go down by 5 Volts. As it was at 0.6 that means it goes down to -4.4. As the capacitor discharges the voltage across it decreases and the base of the second transistor heads back to 0.6.
(When you come to study inductors the thing to remember is that they temporarily "resist" change of current through them.)
It is preferable not to use the word "open" as it can have opposite meanings. You have used "open" as in an open door that lets people through. But in electronics most people consider an "open circuit" to mean that there is a break or a switch's contacts are not bridged and no current can flow. The terms "on" and "off" are less anbiguous when discussing switching circuits.
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Thanks for the answer! That is, if I understood correctly, the capacitor tries to maintain the initial potential difference (5-0.6=+4.4V), so when its left side instantly discharges to 0, it tries to keep the difference and discharges the right side to -4.4V. I may be wrong here, but my logic is as follows: in order to instantly discharge to -4.4V, a fairly high current occurs, that is, a large amount of current flows through the base of transistor 2, due to which it turns on and works until the right side is charged again to the level of +0.6V .
By the way, there is another question: does the current go from - to + or from + to -, about which conductivity is usually said: hole or electronic, because I noticed that in different literature they write differently. For example, at the beginning they say that current is a flow of electrons, electrons "flow" from - to +. And then, using the example of transistors, they show that electrons go from + to -.
And yes, thanks for the advice on vocabulary.
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Forget about the hole/electron flow. If you want to work and design with electronics, it's irrelevant. Voltage and current are just defined as they are, and everyone works with them that way. It's only interesting to physicists.
There is no immediate discharge involved here, it's simply a voltage/potential shift on both sides of the cap. The discharge comes later through the resistor.
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Thanks for the answer! That is, if I understood correctly, the capacitor tries to maintain the initial potential difference (5-0.6=+4.4V), so when its left side instantly discharges to 0, it tries to keep the difference and discharges the right side to -4.4V. I may be wrong here, but my logic is as follows: in order to instantly discharge to -4.4V, a fairly high current occurs, that is, a large amount of current flows through the base of transistor 2, due to which it turns on and works until the right side is charged again to the level of +0.6V .
No, your understanding is not correct.
When TRI collector goes down to close to 0 Volts, almost no current flows initially through the capacitor. The capacitor tries to maintain the voltage difference between its two sides. As the voltage on TR2 base goes below 0.6 Volts, no current flows into its base and that transistor turns off. There is now almost 10 Volts across the 10k resistor. Current flows through it into the capacitor, slowly raising the voltage at the junction of the capacitor, resistor and TR2 base until the voltage reaches 0.6 and current instead flows into TR2 base instead of into the capacitor.
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Okay, I'll think that the current flows from + to -. I'm just trying to understand everything to the smallest detail. I finally realised that initially transistor 2 is on, and a current of about ((5-0.6)/10,000=0.44 mA) is flowing through its base, and when transistor 1 turns on, a voltage shift occurs, which causes transistor 2 to turn off, and then the right side of the capacitor is starting to charge. So I know about the phase shift between the current and voltage across the capacitor, but I can't understand what a voltage/potential shift is and a Google search doesn't show any results for this query. So I'm wondering again why this is, what it is, how it works, how the charge from +0.6V drops to -4.4V, why there are 10 volts across the 10kΩ resistor (5--4.4V=10V). I don't understand the shift. I think of a charge as a certain number of electrons on one side of a capacitor, and in my head, for the voltage to drop from +5V to 0 or from +0.6V to -4.4V, some of the electrons must "leak" out of the capacitor. Perhaps these are too simple or stupid questions for someone, but I'm just starting out, I don't have enough experience but I have a lot of desire to understand everything.
I understand that an inductor resists a change in current because the current that flowed created a magnetic field, so when the current supply is switched off, the magnetic field gradually decreases, generating a current inside the inductor and a reverse EMF.
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Phase shift between capacitor current and voltage can be useful when considering sine wave signals. But not when analysing pulse circuits which have voltage steps with fast transitions.
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What you have here is a differentiator circuit. Since voltage across a capacitor needs time to change (via current charge or discharge), a sudden change on one side of a capacitor affects immediately also the other side.
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thank you
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Nobody seems to be seeing the obvious. C1 and R3 form a differentiator.
The differentiator CR time constant sets the "dt" .
www.electronics-tutorials.ws/rc/rc-differentiator.html (http://www.electronics-tutorials.ws/rc/rc-differentiator.html)
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Thank you, this article was really useful)