Just a quick question: Say we have an NPN common emitter amplfier. What is the point of having biasing resistors from the base to ground and to Vcc? We bias an NPN transistor negatively, so what is the point of having a resistor from base to Vcc?

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I'm no electronics god, i just talk too much.

 

Actually you bias an NPN transistor from the posititive rail!.

The reason for using a potential divider to bias the transistor is to swamp variations due to the gain spread of different transistors. You can just have a bias resistor from base to Vcc, but you have to select it's value for every single one you build. Another option is to place a bias resistor from base to collector, and the negative feedback sets the DC conditions and compensates for the differing transistor gains - however, the negative feedback also lowers your gain.

The classic circuit is four resistors, one in the collector, one in the emitter, and a potential divider on the base - this makes it as simple as possible to design.

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I don't know if there is a benefit of connecting a resistor from base to ground.

BUT, a voltage divider allows you to define a specific current and a specific voltage.

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-=: The best low-priced components to troubleshoot with are the speaker and the LED :=-

 

Zach, Nigel's description is very good. Basically the procedure is to set the base voltage with the voltage dividing resistors. Say we make it 1.6 Volt for example.

There will be about 0.6 V across the B-E junction so the emitter voltage will be about 1 V. If you had say a 1k emitter resistor to gnd, then the emitter current will be about 1V/1k = 1 mA. Most of the emitter current flows into the collector so the collector current is near enough to 1 mA also.

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Len

 

To add to what Nigel said, using two resistors in a voltage divider at the base not only stabilizes bias from one transistor to the next, but it also stabilizes bias over temperature too which is extremely useful.

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RadioRon

 

When I said resistor from base to ground I was thinking of having the transistor turn on at a certain point. I got to thinking about that later but it was too late, I was away from the computer :lol:

So its basically just to stabilize from one transistor to the next then? and temperature as ron added. I can see how variations in the transistor's gain would affect the circuit, but I don't see how placing the resistor from base to ground will keep the transistor's variation from changing the circuit too much.

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I'm no electronics god, i just talk too much.

 

Draw the circuit and do the maths!, it's really VERY simple.

1) The base voltage is simply calculated from the potential divider values.

2) The emitter voltage is 0.7V lower than the base voltage.

3) The emitter current can be calculated from the emitter voltage and the value of the emitter resistor.

4) The collector current is the emitter current MINUS the base current, but the base current is only very small and can generally be ignored. So for this simple example, collector current equals emitter current.

5) The collector voltage can be calculated from the collector current and the collector resistor.

Notice that the transistors gain appears nowhere (I ignored it in part 4).

If you would now like to try introducing it?, the emitter current is the sum of base current and collector current - try doing the sums for different gain values?. Notice that if the gain of the transistor is only 100, the base current is only 1% of the emitter current, and for a gain of 200 only 0.5% of the emitter current - these sorts of figures can safely be ignored if you are using 5% resistors!.

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Ahhh, I see, thanks for explaining that to me Nigel :lol:

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I'm no electronics god, i just talk too much.

 

One thing i always forget when i couple a signal to a potential divider biased transistor amplifer, is that the capacitor and the resistor to ground form a high pass passive filter, incase you're using low freq signals.

Megamox

 

You've also forgot a couple of other things as well!.

1) It's BOTH resistors in parallel that you need to consider (the two halves of the potential divider are effectively in parallel as far as AC is concerned).

2) Also in parallel is the input impedance of the transistor - not so simple to calculate as the two resistors!. The design values of the potential divider should swamp that fairly well though.

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Yes good point, how about the reactance of the coupling cap, that would be included too i think.

Megamox

 

The input impedance of a common-emitter transistor stage with its emitter AC-grounded could be as low as 100 ohms if its collector current is about 60mA. It is less than 2k ohms at a few mA.

Attached Images

transistor_input_impedance.png (25.9 KB, 761 views)

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Uncle $crooge

 

The reactance of the coupling capacitor is the other leg of the high-pass filter, and should be calculated to find the low frequency drop-off point.

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1.What is the role of f=1KHz and Vce = 10V if it changes with the signal frequency then what should be the design frequncy for input impedence?
2. In books they say that input impedence can be taken as hFE times emitter resistance then why to use this graph :?:
3. In some books coupling capacitor is shown as a elctrolytic capacitor how it can be :?:

 

The 1KHz is simply the frequency that graph was plotted at, it doesn't say it's frequncy dependent, although it will be to some extent. 1KHz is just the standard frequency for audio measurements.

Audioguru clearly pointed out the graph was for a transistor with a decoupled emitter resistor "common-emitter transistor stage with its emitter AC-grounded". So that formula doesn't apply.

It mostly will be, due to the fairly low impedances involved, what's the problem with that?.

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