Amplifier Circuit Help - Basic

[bonkerz]

Why is the signal output inverted for this circuit?

I'm struggling to understand the description in Electronics for Dummies.

 

[bonkerz]

I think I have it.

So as the base current increases so does the collector current which drops a larger voltage over the collector resistor and the output voltage drops accordingly. So if the base current keeps increasing then the output voltage keeps dropping.

Still not 100% but this seems a bit further forward that I was earlier.

 

[(*steve*)]

Yep, you've got it.

take a look at the voltage across Re and see if you can spot a difference.

 

[bonkerz]

Yep, you've got it.

take a look at the voltage across Re and see if you can spot a difference.

The time lag on the graph?

The actual circuit in the Dummies book doesn't have a capacitor in parallel with Re.

I'm still at the early stages of understanding circuits. What I still haven't got a picture for is the actual movement of current through all parts of the circuit at each point in time. But I'll get there.

 

[(*steve*)]

No. And ignore any time lag.

also ignore the capacitor across Re (which serves to make Re appear much smaller at audio frequencies)

if the collector-emitter current is determined by the signal on the base, and the collector voltage is determined by this current and the collector resistor, what is the emitter voltage determined by?

It is determined by the collector-emitter current and the emitter resistor (and also the base-emitter current, but that's much smaller).

Note that while increasing CE current (caused by a positive going input) drags the collector to gnd (causing a negative going output), the reverse is true of the emitter. A positive going input signal pulls the emitter up to Vcc resulting in a signal which is also positive going.

There are a number of other significant differences, but this is one difference between common emitter amplifiers (essentially taking the output from the collector) and common collector amplifiers (essentially taking the output from the emitter).

 

[bonkerz]

No. And ignore any time lag.

also ignore the capacitor across Re (which serves to make Re appear much smaller at audio frequencies) OK

if the collector-emitter current is determined by the signal on the base, and the collector voltage is determined by this current and the collector resistor, what is the emitter voltage determined by?

It is determined by the collector-emitter current and the emitter resistor (and also the base-emitter current, but that's much smaller). Got that

Note that while increasing CE current (caused by a positive going input) drags the collector to gnd (causing a negative going output), the reverse is true of the emitter. A positive going input signal pulls the emitter up to Vcc resulting in a signal which is also positive going.

There are a number of other significant differences, but this is one difference between common emitter amplifiers (essentially taking the output from the collector) and common collector amplifiers (essentially taking the output from the emitter).

When you say 'collector' what exactly are you referring to? Just the top N part of the transistor?

Is the output being taken across signal output and gnd?

 

[(*steve*)]

When you say 'collector' what exactly are you referring to? Just the top N part of the transistor?

I am referring to the collector of the transistor, i.e. the connection which is neither the base nor the emitter

Is the output being taken across signal output and gnd?

Almost always.

 

[bonkerz]

I am referring to the collector of the transistor, i.e. the connection which is neither the base nor the emitter



Almost always.

You say it "drags the collector to gnd". Do you mean that you are reading the voltage across the collector part of the transistor only and nothing to do with the resistors? It might be a stupid question but I'm still a little bit confused with what's going on. I understand the current parts but reading the voltages is twisting my melon.

 

[(*steve*)]

think of the base-emitter as a resistor where the value is determined by the base.

Now you have three resistors in series, the collector resistor, the emitter resistor, and the transistor which is now a resistor which varies in value.

You can then use Ohms law to calculate the voltages at the various nodes. It is (or should be) clear that as the resistance of the transistor is decreased, the voltage across the transistor reduces. Since the voltage across the supply doesn't change, the voltage across the 2 resistors (Rc and Re) must increase. This also means that voltage at the collector must fall and the voltage at the emitter must rise.

The signal at the collector and the emitter are 180 degrees out of phase (one is upside-down compared with the other)

 

[bonkerz]

think of the base-emitter as a resistor where the value is determined by the base.

Now you have three resistors in series, the collector resistor, the emitter resistor, and the transistor which is now a resistor which varies in value.

You can then use Ohms law to calculate the voltages at the various nodes. It is (or should be) clear that as the resistance of the transistor is decreased, the voltage across the transistor reduces. Since the voltage across the supply doesn't change, the voltage across the 2 resistors (Rc and Re) must increase. This also means that voltage at the collector must fall and the voltage at the emitter must rise.

The signal at the collector and the emitter are 180 degrees out of phase (one is upside-down compared with the other)

Cheers Steve.

I was laying awake last night until about 5am turning this circuit over and over in my head and I've got it now.

One of the things I was struggling with was the idea of the transistor having a variable resistance dependant on the supply. Happy with that now.

The other thing was what voltage is. What I was struggling with was that if the transistor was in cut off mode then then how could you take a reading of an open circuit. I now understand that voltage is the potential difference between two points. So would I be correct in thinking that if connected one lead of a voltmeter to a 15v rail in one circuit and to a 5v rail in a different circuit I would get a 10v reading even though they two circuits are not connected?

I think I've grown a new area of the brain dedicated to a common emitter circuit.

 

[Rleo6965]

You will get 10volt reading if the 15v and 5v rail have same common ground.

I suggest that you must learn Ohm' Law first. This is very important for those who want to learn electronics. Then try to assemble simple circuit as your mentioned above circuit. Then adjust or vary input or resistors then use your multimeter to confirm the results. It's faster to learn this way.

 

[bonkerz]

You will get 10volt reading if the 15v and 5v rail have same common ground.

I suggest that you must learn Ohm' Law first. This is very important for those who want to learn electronics. Then try to assemble simple circuit as your mentioned above circuit. Then adjust or vary input or resistors then use your multimeter to confirm the results. It's faster to learn this way.

Thanks. That's my next step. I'm at the practical part of the book so now it's just getting all the tools and components.

 

[davenn]

......

One of the things I was struggling with was the idea of the transistor having a variable resistance dependant on the supply. Happy with that now.

No not of the supply voltage, but of the base current as the base current increases the resistance of the transistor ( that is the "variable resistance" between the collector and the emitter) decreases

Dave