Circuit diagram with SPICE
Switched to partsim Online Circuit Simulator with SPICE.
The circuit shows a NPN transistor and a power supply of 1 Volt DC. The transistor is connected in series with the power supply and 2 1K resistors.
The transistor blocks all DC voltage across its emitter to the collector. If a small current is supplied to the base of the transistor, current will flow across the terminals.
If an AC current is supplied to the base, this will result in a fluctuating current across the transistor terminals. Hopefully this will be of the same pattern of the input signal.
Resistor R2 could be a speaker of impedance 8 Ohms for example.
The input signal goes to the base of the transistor and the lower side of the resistor R1.
First question - won't the current from the battery flow into the input terminals since they are at a lower voltage than the power supply current?
Just thinking out aloud.
Switched to partsim Online Circuit Simulator with SPICE.
The circuit shows a NPN transistor and a power supply of 1 Volt DC. The transistor is connected in series with the power supply and 2 1K resistors.
The transistor blocks all DC voltage across its emitter to the collector. If a small current is supplied to the base of the transistor, current will flow across the terminals.
If an AC current is supplied to the base, this will result in a fluctuating current across the transistor terminals. Hopefully this will be of the same pattern of the input signal.
Resistor R2 could be a speaker of impedance 8 Ohms for example.
The input signal goes to the base of the transistor and the lower side of the resistor R1.
First question - won't the current from the battery flow into the input terminals since they are at a lower voltage than the power supply current?
Just thinking out aloud.
Attachments
The transistor blocks current flow as drawn.
To force current into the base, you first have to overcome the base-emitter volt drop of around 0.6 volts. No current will flow in the base until that threshold is reached. Once it is, current flow increases dramatically and non linearly.
Your question... no. Current can not flow from the battery to the input because the base-collector junction blocks it.
So if R2 were a speaker, then you would only hear the input signal as it passes through that 0.6 threshold. In other words, silence at low volume and huge distortion on the louder bits.
To force current into the base, you first have to overcome the base-emitter volt drop of around 0.6 volts. No current will flow in the base until that threshold is reached. Once it is, current flow increases dramatically and non linearly.
Your question... no. Current can not flow from the battery to the input because the base-collector junction blocks it.
So if R2 were a speaker, then you would only hear the input signal as it passes through that 0.6 threshold. In other words, silence at low volume and huge distortion on the louder bits.
So that's the reason for the quiescent current for a class A amplifier?
I am doing some reading as well : this helps: PNP Transistor Operation
In the meantime the simulator is asking for me to put in a ground.
Yes, you always need a ground as a point of reference in a simulation.
Lets take your basic one transistor amplifier as shown here and lets look what happens when we apply a 1 volt peak to peak input signal.
Look how the output does nothing until Vin reaches this magic 0.6 volt threshold. Until then the transistor is OFF. As Vin rises, Vout follows but it is 0.6 volts lower than Vin. This 0.6 volts is the base-emitter forward volt drop and is around this value for all silicon transistors.
The negative part of the input signal has no effect, it simply biases the transistor off.
The second and third picture shows a ramp waveform applied to the base. Look how the emitter follows 0.6 volts lower and how nothing happens until that 0.6 volt point is reached. The third is showing just the first part to make it easier to see.
Lets take your basic one transistor amplifier as shown here and lets look what happens when we apply a 1 volt peak to peak input signal.
Look how the output does nothing until Vin reaches this magic 0.6 volt threshold. Until then the transistor is OFF. As Vin rises, Vout follows but it is 0.6 volts lower than Vin. This 0.6 volts is the base-emitter forward volt drop and is around this value for all silicon transistors.
The negative part of the input signal has no effect, it simply biases the transistor off.
The second and third picture shows a ramp waveform applied to the base. Look how the emitter follows 0.6 volts lower and how nothing happens until that 0.6 volt point is reached. The third is showing just the first part to make it easier to see.
Attachments
I understand the diagrams, but the question is what is the other output terminal, and how does the voltage after the resister connected to the collector vary with the input signal? If the voltage is higher after the resistor connected to the collecter than at the input terminal, current will flow to the input terminal.
I'm not quite following your question 
You've no resistor connected to the collector in your (and my) diagram. The positive supply goes to the collector as drawn. The emitter is the one with the resistor.
The configuration as drawn is called a 'common collector' or more usually an 'emitter follower'. The voltage gain is always a little bit less than one (unity) but there can be tremendous current gain available. In other words a tiny current flowing in the base-emitter junction can control a large current flowing in the collector-emitter path. The output is always taken from the emitter in this configuration.
Does that help.
You've no resistor connected to the collector in your (and my) diagram. The positive supply goes to the collector as drawn. The emitter is the one with the resistor.
The configuration as drawn is called a 'common collector' or more usually an 'emitter follower'. The voltage gain is always a little bit less than one (unity) but there can be tremendous current gain available. In other words a tiny current flowing in the base-emitter junction can control a large current flowing in the collector-emitter path. The output is always taken from the emitter in this configuration.
Does that help.
You need to understand the way transistor works.
See a few good links under "Electronics theory" here:
USEFUL LINKS
I think you are missing some preliminary concepts, such as the difference between current and voltage, and how Ohm's law works.
So, may I suggest you start by studying and understanding how a voltage divider works? To understand this, you first need to understand Ohm's law, as well.
(A voltage divider is two resistors in series, the combination connected across some voltage source.)
A common-emitter transistor circuit can be visualised as a voltage divider, where the transistor is one of the two resistors, and the external collector or emitter resistor is the other.
So, once you've understood Ohm's law and the two-resistor voltage divider, it will become a lot easier to understand the transistor circuit.
Some online resources:
-Gnobuddy
Can anyone help me send me a SPICE circuit I can cut and paste into my SPICE sim I just installed?
gSpiceUI
Version 1.0.00 (23/09/2011)
What would help me is simply to see how the current and voltage varies at each junction in the circuit when the input is at selected values between -300 and +300 mA. That would clear up a lot of questions.
gSpiceUI
Version 1.0.00 (23/09/2011)
What would help me is simply to see how the current and voltage varies at each junction in the circuit when the input is at selected values between -300 and +300 mA. That would clear up a lot of questions.
Basic - I am trying to help you here. If you really want to understand this stuff, please go back to the beginning, use the links I gave you, and start learning from there. Even better, find someone who lives nearby (any electronics clubs there?), who can quickly show you the basics.
Trying one circuit simulator after the other won't help - you have to start by building a better foundation, so you can understand the more complicated concepts involved in this transistor circuit.
The question you just asked ("input is at selected values between -300 and +300 mA") is literally meaningless. It cannot be answered, because it makes no sense at all. The circuit in post #26 won't work. The circuit in post #24 won't work either. Looking at these, it is clear to me that you're not clear on basic concepts: voltage, current, voltage drop, Ohm's law, voltage dividers. That's why these transistor circuits are not making any sense to you yet.
There's nothing wrong with not understanding basic electricity: all of us started out from that same place.
You can do this: just start at the beginning, learn one step at a time, and very soon you will understand these transistor circuits.
-Gnobuddy
Trying one circuit simulator after the other won't help - you have to start by building a better foundation, so you can understand the more complicated concepts involved in this transistor circuit.
The question you just asked ("input is at selected values between -300 and +300 mA") is literally meaningless. It cannot be answered, because it makes no sense at all. The circuit in post #26 won't work. The circuit in post #24 won't work either. Looking at these, it is clear to me that you're not clear on basic concepts: voltage, current, voltage drop, Ohm's law, voltage dividers. That's why these transistor circuits are not making any sense to you yet.
There's nothing wrong with not understanding basic electricity: all of us started out from that same place.
You can do this: just start at the beginning, learn one step at a time, and very soon you will understand these transistor circuits.
-Gnobuddy
Another concept, which is vitally important to understand - Kirchhoff's laws:
- Kirchhoff's current law (KCL): The sum of currents entering any junction is equal to the sum of currents leaving that junction.
- Kirchhoff's voltage law (KVL): The sum of all the voltages around a loop is equal to zero.
Combination of those and the Ohm's law allows us designing or analysing the basic circuits in DC.
As soon as we start dealing with AC signals, capacitors and inductors, theory becomes much more complicated, but let's keep this area aside for now
- Kirchhoff's current law (KCL): The sum of currents entering any junction is equal to the sum of currents leaving that junction.
- Kirchhoff's voltage law (KVL): The sum of all the voltages around a loop is equal to zero.
Combination of those and the Ohm's law allows us designing or analysing the basic circuits in DC.
As soon as we start dealing with AC signals, capacitors and inductors, theory becomes much more complicated, but let's keep this area aside for now
Thanks Gnobuddy the tutorial here puts it really simply, explaining current, voltage and resistance in terms of water flow, hydraulic pressure and pipe diameter.
https://learn.sparkfun.com/tutorials/voltage-current-resistance-and-ohms-law
https://learn.sparkfun.com/tutorials/voltage-current-resistance-and-ohms-law
Yes, it's a good tutorial, which is why I suggested it to you in my post #31.
Ohm's law allows you to understand a circuit that only contains one single resistor. That was wonderful news back in 1827, when George Simon Ohm's work was first published. It is still essential knowledge, but, by itself, it does not let you go beyond one single resistor.
The other tutorial I linked in the same post (my post #31) is about voltage dividers. This will let you understand a circuit that has two (rather than one) series resistances. It is key to understanding those transistor circuits you were looking at.
Once you have Ohm's law and voltage dividers solidly understood, the doors to understanding DC electricity begin to open. Hint: the transistor itself acts as a sort of (special) resistor.
Vzhaichenko mentioned Kirchoff's laws; those are very helpful in understanding more complicated circuits. They are not yet essential at this point in the game; a voltage divider with two resistors actually incorporates the concept of Kirchoff's voltage law.
-Gnobuddy
This one seems to work.
Basic Audio Amplifier with one transistor, two resistors and a capacitor
https://www.youtube.com/watch?v=fegzhxbYblc
Basic Audio Amplifier with one transistor, two resistors and a capacitor
https://www.youtube.com/watch?v=fegzhxbYblc
AC and DC circuit
I have been going through some of the basic circuits here:
Circuit Simulator Applet
I understand how the transistor works. It can be compared to a dimmer switch for a light bulb circuit where the dimmer switch is operated by another smaller current, an electric motor or servo for example, except the operation of the transistor is not mechanical.
What I do no understand is the left hand side of the circuit, how two current sources, one DC and the other affect each other.
A simple circuit will hopefully clear this up : here goes:
This is the text file for the circuit: a simple DC circuit with two resistances.
$ 1 0.000005 10.200277308269968 50 5 50
v 384 144 384 208 0 0 40 5 0 0 0.5
r 256 144 256 192 0 100
w 384 144 384 96 0
w 384 96 256 96 0
w 256 96 256 144 0
w 384 256 256 256 0
w 256 256 256 192 0
r 384 208 384 256 0 100
Adding a second circuit on the left, which is a AC source with a 500 mA peak.
$ 1 0.000005 1.3241202019156524 50 5 50
v 384 144 384 208 0 0 40 9 0 0 0.5
r 256 144 256 192 0 100
w 384 144 384 96 0
w 384 96 256 96 0
w 256 96 256 144 0
w 384 256 256 256 0
w 256 256 256 192 0
r 384 208 384 256 0 100
v 112 176 112 224 0 1 40 0.5 0 0 0.5
w 112 224 112 256 0
w 112 176 112 144 0
w 208 144 208 192 0
w 208 192 256 192 0
r 112 256 256 256 0 100
s 112 144 208 144 0 0 false
p 112 256 112 320 0
This is the part I do not understand: The ac current and DC current keep on circulating without any change, as far as I can tell. Is this the way it works in reality? I would think some cancellation occurs.
Of course I could test it practically by connecting a DC voltage across a speaker when it is working. MIght burn out something unless I put a resister somewhere in the input.
I have been going through some of the basic circuits here:
Circuit Simulator Applet
I understand how the transistor works. It can be compared to a dimmer switch for a light bulb circuit where the dimmer switch is operated by another smaller current, an electric motor or servo for example, except the operation of the transistor is not mechanical.
What I do no understand is the left hand side of the circuit, how two current sources, one DC and the other affect each other.
A simple circuit will hopefully clear this up : here goes:
This is the text file for the circuit: a simple DC circuit with two resistances.
$ 1 0.000005 10.200277308269968 50 5 50
v 384 144 384 208 0 0 40 5 0 0 0.5
r 256 144 256 192 0 100
w 384 144 384 96 0
w 384 96 256 96 0
w 256 96 256 144 0
w 384 256 256 256 0
w 256 256 256 192 0
r 384 208 384 256 0 100
Adding a second circuit on the left, which is a AC source with a 500 mA peak.
$ 1 0.000005 1.3241202019156524 50 5 50
v 384 144 384 208 0 0 40 9 0 0 0.5
r 256 144 256 192 0 100
w 384 144 384 96 0
w 384 96 256 96 0
w 256 96 256 144 0
w 384 256 256 256 0
w 256 256 256 192 0
r 384 208 384 256 0 100
v 112 176 112 224 0 1 40 0.5 0 0 0.5
w 112 224 112 256 0
w 112 176 112 144 0
w 208 144 208 192 0
w 208 192 256 192 0
r 112 256 256 256 0 100
s 112 144 208 144 0 0 false
p 112 256 112 320 0
This is the part I do not understand: The ac current and DC current keep on circulating without any change, as far as I can tell. Is this the way it works in reality? I would think some cancellation occurs.
Of course I could test it practically by connecting a DC voltage across a speaker when it is working. MIght burn out something unless I put a resister somewhere in the input.
Attachments
Left hand circuit shows two series resistors across a DC voltage. The right hand diagram shows the same, but with a separate AC source simply using one of the conductors of the other circuit for its own connectivity. There is no ground, you have two independent voltage sources operating with no interaction. The AC source feeds into the left hand 100 ohm. The DC source is the same as in the left diagram.
Is it supposed to be like that ? That's how its drawn anyway.
Is it supposed to be like that ? That's how its drawn anyway.
You're right, excellent!
At some later point you'll find out that there is more to a transistor than just variable resistance, but that can wait a little.
Remember the part about voltage in a circuit being like pressure in water, or air?
Let's use that analogy. DC is like a constant, static pressure. For instance, the air around you has a steady pressure of one atmosphere.
AC air pressure exists too - we call it sound when it's between 20 Hz and 20,000 Hz.
Suppose you turn on your TV or click over to YouTube now, and turn on some music. The air around your head now has a DC component, AND an AC component. The DC component lets you breathe and stay alive; the AC component is the music you're hearing. At any one instant, the air pressure at your ears is the sum of the DC and AC pressures.
Now, there's nothing startling or hard to understand about that, is there? You've been listening to sounds in the earth's atmosphere all your life.
In exactly the same way, you can have a DC voltage and an AC voltage added together in a circuit. Nothing startling or hard to understand about that either, really. It's just unfamiliar.
I'll take the analogy a tiny bit further. You wouldn't survive long if you were listening to only AC air pressure: you need that DC pressure to be able to breathe!
Well, transistors "need" DC to be able to function, too, just like you do. Once they're happy and have the DC they need (it's called biasing), then they can handle a little AC added on top.
Analogies can only go so far, of course. But does that help make sense of your question about AC and DC co-existing in the same circuit?
-Gnobuddy
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