# Bipolar Transistor AC Amplifier

Discussion in 'General Electronics Discussion' started by RaevenCogan, Mar 13, 2011.

1. ### RaevenCogan

12
0
Mar 13, 2011
Hey guys This circuit recently came up in the book I'm reading and it has me a bit confused... To begin, I don't understand why the cathode of the capacitor is connected to the base of a NPN transistor because the base is P-Type silicon so shouldn't it connect to a positive element? Secondly, what's labeled as the "BIASED RESISTOR" is supposed to supply an output voltage of half the battery for the output to ride on but aren't resistors supposed to increase voltage while decreasing current Finally, what's labeled "LOAD RESISTOR" is said to make the output current a voltage. However, when was it ever "not" a voltage?

P.S. I'm sorry to bug you guys but I'm stuck

File size:
152.7 KB
Views:
1,747
2. ### davennModerator

13,866
1,958
Sep 5, 2009
NO .... a bias resistor simply supplies a small current to the base of the transistor
to just turn it on to a predetermined level. This is often done say, in amplifier circuits to allow the transistor to operate correctly within its linear range.

Now I'm not going to answer this totally accurately, Steve or Res. can chime in
Basically, the signal coming in from the left onto the + lead of the cap is going to be "more" positive ( during the positive parts of the signal cycle) than the very small + voltage on the base of the transistor. Not sure how to explain that one better

NO you get voltage drops across resistors the current stays the same

http://www.electronics-tutorials.ws/resistor/res_3.html

cheers
Dave

3. ### Resqueline

2,848
2
Jul 31, 2009
A bipolar transistor is a current-in current-out device. Ok, it has a voltage drop of approx 0.6V across b-e but that is just a coincidental & unwanted "side effect".
The output current is related to the input current and is largely voltage independent, hence the need to use a load resistor to "convert" it to a voltage (Ohms law).
The capacitor is the right way around if it's meant to be connected to a similar preceding amplifier stage. Then the anode would be at approx 5V and the cathode at 0.6V.
If it's to be connected to an unbiased signal source however (like a volume control for example) then it's the wrong way around.

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,497
2,839
Jan 21, 2010
I'll try to address a few things that davenn passed on

This is perhaps the first time I've heard anyone talk about the cathode of a capacitor outside discussions of the internal structure of polarised capacitors.

It's certainly not wrong, but applies only to polarised capacitors, and is a little tricky if you're talking about non-polarised electrolytic capacitors...

You need to look at the voltage levels involved. The capacitor exists to isolate any DC bias difference between the input of this stage and the output of the following one. If you presume that there are several similar stages like this connected together to get higher gain, then the output of this stage would be connected to the input of another similar stage.

When the transistor is off, the output is neat the supply rail. When it is fully turned on, it is some fraction of a volt (more likely a volt or so) above 0V. The base of the transistor is essentially fixed at 0.7V. It is clear that the output voltage can be considerable more positive, and possibly never more negative than the base. On that basis, the positive end of the capacitor should be connected to the source of the input signal.

If the input signal was from a source that had a negative bias, then you would have the capacitor the other way around.

It should be "bias resistor". It applies a bias to that which will thus be "biased"

...Either that or it has a strong opinion.

There are several models of a transistor, but a simple one is that it is a variable resistor
with the resistance determined by the base current.

The load resistance forms a potential divider with the "resistance" of the transistor. As the transistor's resistance changes, the voltage on the mid point of the voltage divider varies.

5. ### RaevenCogan

12
0
Mar 13, 2011
Thanks everyone for all the help!!!

I understand the circuit now thanks to all your help! Basically: The Bias Resistor supplies .6v (about) to the Base so that this voltage won't be taken from the input waveform. The Capacitor is hooked up the way it is because the input is more positive than the base it's connected to. The Load Resistor in conjunction with the Transistor forms a voltage divider of sorts for the output.

Correct me anywhere that I am wrong guys but I think I got this!

6. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,497
2,839
Jan 21, 2010
No, the bias resistor supplies current. The base of the transistor is never going to rise much above 0.6 to 0.7 volts above that of the emitter (which in this case is grounded). The resistor is chosen to give an "appropriate" quiescent base current.

The rest is essentially correct.

The "variable resistor" model of the transistor is not a very accurate model, but sufficient to understand the reason for the load resistor. You should not assume that the transistor will actually behave like a resistor.

7. ### RaevenCogan

12
0
Mar 13, 2011
But what I don't get is if resistors don't affect current then why use a bias resistor to supply the base current instead of a wire.

Sorry if I'm missing something Steve : /

8. ### Resqueline

2,848
2
Jul 31, 2009
Steve just said that the bias resistor sets the current going into the base. I = (Ubat - Ube) / R. Replacing it with a wire gives a divide-by-zero = infinite current = smoke.
Ube is a voltage drop inherent to the transistor technology. For Silicon transistors it's around 0.6V, but for the old Germanium technology it was around 0.2V.
This value is slightly current and temperature dependent. Silicon starts conducting microAmps at 0.5V and will conduct Amps at 1V.
It drops with 1.5mV for each degree C temperature rise, given a constant current.

9. ### RaevenCogan

12
0
Mar 13, 2011
Hey Res, are you basically saying that when current flows from the emitter to base the circuit would short if there was no load? In this case the bias resistor performs the job of load and also provides current going into the base for the signal to ride on?

Thanks for you response

10. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,497
2,839
Jan 21, 2010
This is essentially correct. Without a bias, only the positive half of the signal would be amplified. The bias current serves the purpose of allowing the negative half cycles to reduce the base current and therefore be amplified.

The base-emitter junction can be seen as a diode. If you connect the base directly to your positive supply, and ground the emitter (of an NPN transistor) then you are effectively shorting the power supply to earth via a diode -- not a good thing. The resistor limits the current. Because it's a diode (the BE junction) the resistor has almost the same voltage across it regardless of the current through it.

11. ### Resqueline

2,848
2
Jul 31, 2009
No. But if you replace "current flows from" with "power supply applied to", and "load" with "resistor" then it starts to make sense (to me at least).
I'm basically saying what has been said: a bipolar transistor is a current controlled current source that needs resistors to "convert" those currents into voltages.
If a base-emitter junction is connected directly to a power supply of 1.5V or more - then said junction will conduct an infinite current and will fry.
Hence the need for suitable resistors to reduce the currents to suitable levels (at the given design voltages). Please work with - and try to get the feel for - Ohms law.
The load resistor on the output sets the collector current. To keep the collector at half the supply voltage the base needs a current of approx 1/100th of this (= bias resistor).
Yes, the signal rides on the bias voltage/current level. Without a bias resistor only the signal peaks reaching above +0.5V would be amplified.

12. ### RaevenCogan

12
0
Mar 13, 2011
Thanks for all the help everyone! I understand now.