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f4cepl4nt

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I have a few questions about transistors, so here we go:

How can you tell how much current must flow between base-emitter to allow current to flow collector-emitter??

How can you tell which connector on a transistor is the emitter?

When a transistor gets a small current running through the base-emitter, which then allows current to flow collector-emitter, does the current through the base-emitter get cut off? Or does it keep flowing in tiny amounts?

One last one: If a transistor has a current flowing base-emitter which then allows current to flow collector-emitter, if the base-emitter current is turned off (by a switch or something) does the collector-emitter current turn off too? Or does it just need a kick start to get going then nothing can stop it except a switch or broken circuit?


I'm very sorry if my questions are unclear, I'm not very good at describing things  ;D

Again, thanks for all the help, you guys are life savers (no not the candy  8) )

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using a multimeter but with some jiggling u can come to know about the terminals base,emitter ,collector.Emitter conducts most heavily.(Q2)

It is the base-emitter voltage which drives the transistor into saturation and makes the heavy current to flow from collector to emitter.(Q3)

It is the beta and internal mechanism of conduction in a transistor the answer to the
first question.

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How can you tell how much current must flow between base-emitter to allow current to flow collector-emitter??

A transistor has a range of DC current gain. It is different between two transistors with the same part number and is different at different collector currents. The transistor's datasheet explains it all.

How can you tell which connector on a transistor is the emitter?

By looking at the labelled picure of it on its datasheet.

When a transistor gets a small current running through the base-emitter, which then allows current to flow collector-emitter, does the current through the base-emitter get cut off? Or does it keep flowing in tiny amounts?

Only the input circuit to your transistor can cutoff the base-emitter current.

If a transistor has a current flowing base-emitter which then allows current to flow collector-emitter, if the base-emitter current is turned off (by a switch or something) does the collector-emitter current turn off too? Or does it just need a kick start to get going then nothing can stop it except a switch or broken circuit?

If you disconnect the base current, collector to base leakage current might cause the transistor to conduct. Usually the base of a transistor is shorted to its emitter with a resistor or another conducting transistor to make certain it is turned off.
The collector current follows changes in the base current, but is amplified. ;D
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  • 2 weeks later...

Hi Prateek,
I don't think a high beta (current gain) makes a transistor amplifier have a high voltage gain. A darlington transistor has an extremely high beta, and its gain is about the same as an ordinary transistor. A high beta transistor in a common-emitter stage has a higher input impedance.

A good amplifying transistor has a linear or flat beta, and a very low Re. The voltage gain is the collector resistor's value (and load in parallel) divided by the total of its Re plus emitter resistor's value. If you buffer its output so it isn't loaded down, and use bootsrapping or a constant current source as its collector resistor, its gain will be very high and linear. Here is a good explanation with measured results: http://www.dself.dsl.pipex.com/ampins/discrete/twoq.htm

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I think that high beta transistors are considered high gain because of the current multiplication. Generally, to get gain, you look to the collector and emitter resistors. It is always safer to play the voltage game. What you end up doing in the current gain scenario is to lower the change in base current rather than affect the change in collector current and thus the voltage. So with a constant change in input voltage, you have not gained anything with a high beta transistor.

Now let's consider the current game for a moment. If we had a current source as the signal, you would apply that to the base of a high beta transistor only to realize that yes you have the change in collector current, but oh no, you also have a greater input voltage, so your gain is not as high as you thought.

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Hi Audioguru. I think that feedback, zero time difference, is only there as a necessity. It let's current flow around the opamp. But of course why do you need that. It just so happens that when you put a resistor from the output to the  inverting input, your gain looks like it mimicks the resistors. I have done several things with opamps. I hook them up seemingly backwards with fairly reasonable results. But what negative feedback appears to be is just an adjustment to the signal at the input of the amplifier. And it's there because you don't want too high a gain. It is my opinion that the only reason they use opamps, which are simple circuits, is because they need high gain. How high? Well, let the designers be the judge of that.

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I have done several things with opamps. I hook them up seemingly backwards

Why?

what negative feedback appears to be is just an adjustment to the signal at the input of the amplifier. And it's there because you don't want too high a gain. It is my opinion that the only reason they use opamps, which are simple circuits, is because they need high gain.

An opamp circuit doesn't have to have high gain. Its negative feedback reduces its extremely high open-loop gain with two resistors to however much gain you want. The extra gain makes the opamp an error amplifier so that any non-linearity in its output is reduced to very, very close to zero.
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hi kevin!
we use negative feedback through an emitter resistor just to stabilize the gain and make the amplifier much more linear.
and to increase the range of i/p signal. ;D
hi audioguru!
do we have any disadvantage of a transistor with 2000 value of beta as amplifier or is it an excellent amplifier?what do u think? ::)

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Transistors with a beta as high as 900 is the highest I've seen and they work about the same as transistors with lower beta, except the input bias current is very low and the input impedance is high. BC549C for example.
Some opamps use "super-beta" transistors on their inputs for a very low input bias current and they have a very low breakdown voltage, requiring the use of clamp diodes between the inputs.

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Hi Prateek,
I don't think beta affects a transistor's voltage gain. It affects a transistor's input impedance and input bias current.
A common-emitter transistor's voltage gain is simply Rc/Re, where Rc is its collector resistor in parallel with its load. The voltage gain is higher at higher current (if the collector resistor remains the same then the supply voltage must be increased) because Re is decreased. ;D

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It doesn't matter if a transistor has a high current rating and a low beta, or a low current rating and a high beta. Beta changes with temperature and amount of current.

The beta of a 200mA 2N3904 drops to about 70% of its highest value at half of its current rating.
The beta of a 15A 2N3055 drops to about 14% of its highest value at half of its current rating.
You think the high current, low beta transistor keeps its beta more constant?
I don't think so.

post-1706-14279142327214_thumb.png

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hi kevin !
i too dont like non linearitities.
lower but linear beta is better for me.
hi audioguru!
that means
By changing emitter resistor we can change the gain of the transistor.
Whats the problem if we make extremely low Re?Do we have biasing problems?
or we may use high Rc?
Actually i wish to conduct a study on gain control of a transistor.How should i proceed and turn the knob for high gain?
;D

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Whats is the problem if we make an extremely low Re? Do we have biasing problems?

You can't change a transistor's built-in Re, but you can bypass its emitter resistor with a capacitor for high gain. Without collector to base negative feedback, a transistor without an emitter resistor is not stable with transistor changes from one to another, nor with temperature changes. A common emitter transistor needs an emitter resistor.

or we may use high Rc?

Then you need a high supply voltage or a low current for it to be biased correctly. The output impedance will be high.

Actually i wish to conduct a study on gain control of a transistor.How should i proceed and turn the knob for high gain?

Use an emitter resistor so it works with different transistors and is temperature-stable. Then add a pot in series with a capacitor and connect them across the emitter resistor for your gain control. With the pot at low resistance, the emitter resistor will be bypassed for high gain. Notice the high distortion at high output levels. With the pot at high resistance then the capacitor doesn't do anything, and the gain is about Rc/Re, with low distortion. ;D
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Guest SM2GXN

Hi audiguru!

Is there any known good and easy way to calculate the gain when Re is bypassed by an capacitor, I know how to calculate the capacitor.
As you said the gain without bypassing Re is simply Rc/Re but when it comes to bypassing Re  ::)

Bjorn

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Hi Bjorn,
Transistors have a built-in Re. It is calculated by dividing the transistor's hIE by its hFE. Both change when the transistor's current is changed.

Then the transistor's voltage gain is roughly the collector resistor (and the parallel load) divided by your calculated Re. Determining the voltage gain is rough because transistor amplifiers without negative feedback are not linear. On the collector voltage swing when the current is increasing, the gain is decreasing. A pure sine-wave input is distorted at the collector and looks like this on a 'scope:

post-1706-14279142327313_thumb.gif

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Guest SM2GXN

Hi audioguru!

Thank's for the reply.
What do think of this:
When the emiter resistor is bypassed the gain rise to very high values and the lowest frequency to be amplified depends mostly on the bypass cap, I'm thinking of audio now :)
As for the Ce I used to think of a divider which has a very low reactance for the frequency to be amplified at the bottom and in some way I'm back at the formula without bypass cap Rc/Re but Re is in parallel with the capacitor and thus making the reactance for the working frequency very low.
Without bypassing the emiter will follow the voltage change at the base, right?
Just my thoughts  ;D
Do you see my point?

Bjorn

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Hi Bjorn,
If you bypass the emitter resistor with a capacitor having a very low reactance at the working frequency, then the transistor will have high gain and distortion.
You can adjust the gain by using two emitter resistors in series, with only one of them bypassed. The two resistors could be a pot. ;D

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