Questions about Transistors and Voltage Bias

[Schappy]

Ive been reading a ton of tutorials on this site and on the web trying to grasp the inner workings of circuitry.
Its starting to come together but Im having trouble piecing some things together as far as transistors are concerned.

So before the transistor we set up a voltage bias with our resistors.
Also in combination with a capacitor these resistors set up a high pass filter.
They also set up a high input impedance meaning only a small amount of current will be used up by the transistor.

Now when it comes to the actual working of the transistor is where I get confused.
So what makes the gate "open" on the transistor. Is it the voltage difference between the gate and collector or gate and emitter? Or is it current related?

Once the gate is open current flows from the collector to the emitter. Does this increase the voltage difference between the gate and collector?

Ive been trying to follow the tutorial on this thread.  http://www.diystompboxes.com/smfforum/index.php?topic=47572.0

How to we utilize the transistor in this diagram?
Its seems that we are grounding the emitter so we lose the gain in current. Are we just utilizing a voltage gain?

[Schappy]

Let me shorten my question to make it more concise.

A voltage is applied to the gate that if high enough will open the gate and allow current to flow from the collector to the emitter.

So at least in the input stage transistor we are limiting current while maintaining voltage since we are grounding the emitter.

Is this correct?

[pjwhite]

A small current flowing from the base to the emitter causes a larger current to flow from the collector to the emitter.  There's also a semi-constant voltage drop to over come between the base and the emitter.  So, the voltage on the base with respect to the emitter has to be higher than this drop in order for current to flow into the base.  This voltage drop is on the order of  0.6 to 0.7 volts or so for silicon transistors, and 0.3 volts for germanium transistors.

[Schappy]

So how does this apply to the input transistor(FET) of an effects signal?

What are we accomplishing by using the transistor?

[pjwhite]

A FET works differently from a bipolar transistor.  And the pins are named differently -- on a FET you have a gate, drain and source instead of a base, collector and emitter.  The FET conducts current from the drain to the source when the voltage on the gate is greater than the voltage on the source.  This voltage is known as Vgs.  The amount of current from the drain to the source depends on the value of Vgs.  Higer Vgs = higer current flow.  No appreciable current flows into the gate.

[Schappy]

Ok, its making sense now.

So the drain is connected to the power source(9v) and the gate is connected to the input buffer. When the voltage between the gate and source is high enough current flows from
the drain to the source.

This is all controlled by a small current entering the gate.
We then use this larger current to boost or distort the signal. We now have a larger current but about the same voltage.

Is this correct?
Also Ive seen schematics where the source goes to ground.
Maybe another type of transistor was being used.

Ive heard people talk of using transistors for voltage gain and that is what is confusing me.

[R.G.]

You're still confused.

FETs (Field Effect Transistors) are not the same as Bipolar Transistors (NPN or PNP).

All FETs have a resistive channel from drain to source. They are further subdivided into Junction FETs (JFETs) and Metal Oxide Semiconductors (MOSFETs).

A JFET is always low resistance unless you do something to the gate to make it turn off. They are called "depletion mode" devices because you have to do something to *deplete* the conduction of the channel. In this respect only, they are similar to vacuum tubes, which conduct fully unless you turn them somewhat off.

Bipolars and MOSFETs are normally off, and you have to do something to the base or gate respectively to make them turn on. They are called enhancement mode devices because you have to *enhance* their conductivity to get something to happen.

All three work differently inside.

Let's take the bipolar. The bipolar is two diode junctions back to back. A DC voltage across this thing will reverse bias one of them, and so no current will flow. However, if you externally inject some *current* into the base (middle layer), the current poisons the ability of the reverse biased junction to resist current flow, so some current slips through in spite of the reverse biased junction. The current that slips through is roughly proportional to how much current you put into the base. The proportionality constant of collector flow to base current is the current gain of the transistor. The base and the third terminal, the emitter, are always no more than one silicon-diode drop apart, because making them more than that would require huge currents, and the transistor is already fully on long before you can pump that much current into the base.

The MOSFET is completely different. It has a channel through the silicon body from drain to source that has a very few charge carriers, not enough do support current flow. The gate is isolated from the channel by a layer of insulating glass. When you put a biggish voltage on the gate, it literally pulls more charge carriers from other places in the silicon and makes the area just below the gate more conductive because there are more carriers to carry current. So you enhance a MOSFET by putting a few volts on the gate to let current flow.

The JFET is different yet. It's channel is full of charge carriers and will flow some maximum current if you just hook a battery across it. The gate is an area of reverse polarity silicon that overlays the channel. If you ignore it, you have a silicon resistor. If you make the gate negative with respect to the channel, it does two things. First, the gate is isolated from the channel by the reverse bias, so no current flows in the gate. Second, the reverse bias field in the channel REPELS charge carriers off into the rest of the silicon; Unfortunately, there isn't much of anywhere to go, so the gate voltage effectively squeezes the channel closed, exactly like squeezing a garden hose pinches off water flow.

To work with either of these, you have to understand what's happening inside. If you mix up what device does what, you'll keep being confused.

It's always good to start with a tour through GEOFEX for most questions. I have somewhat detailed discussions of bipolar action and biasing MOSFETs in articles there.

[pjwhite]

Ok, its making sense now.

So the drain is connected to the power source(9v) and the gate is connected to the input buffer. When the voltage between the gate and source is high enough current flows from
the drain to the source.

This is all controlled by a small current entering the gate.

With a FET, the current into the gate is so small as to be effectively zero.  It's all about the _voltage_ on the gate, not the current.

We then use this larger current to boost or distort the signal. We now have a larger current but about the same voltage.

Is this correct?

Depending on the circuit configuration... yes.

Also Ive seen schematics where the source goes to ground.
Maybe another type of transistor was being used.

Ive heard people talk of using transistors for voltage gain and that is what is confusing me.

Bipolar transistors can be used for voltage gain by putting a resistor on the base and feeding the input voltage into this resistor.  The resistor becomes a voltage to current converter in effect.  A small voltage applied to the input becomes a small current into the base, which causes a larger current to flow between the collector and emitter, and with a collector resistor to V+ and the emitter grounded, the collector output develops a larger signal (inverted from the input signal).

Here's some info on FETs:
http://www.ibiblio.org/obp/electricCircuits/Semi/SEMI_5.html#xtocid155811

Sorry I don't have time to explain in more detail.  There's probably lots more info on this board, or you can Google for transistor tutorials...

[Schappy]

R.G.  and PJ Thanks for your time.

I will read on and continue to digest as much info as I can. Sometimes 1 response in a thread can save many hours of toil searching the net.

I have a few books ordered, one being Art of Electronics that I will read as soon as I get them.

If you could just tell me briefly what is going on with the transistor in this diagram it would clear things up greatly. http://www.diystompboxes.com/smfforum/index.php?topic=47572.0

In this diagram it appears that the current that has been amplified is going to ground. In a JFET, the current(with gate open) flows in only one direction right(toward the source)?

I just dont understand what we are accomplishing by using this transistor in this schematic.

[pjwhite]

(Circuit repeated here for clarity)


Let's start by looking at the DC characteristics of this circuit (no signal applied to the input).

I'll make a few simplifying assumptions, but the basic theory holds in any case. 

R1 and R2 form a voltage divider that biases the base of Q1 to approximately 0.8 volts ((9V / (430K + 43k)) * 43k.  Thanks to the sort-of constant BE voltage drop (assume 0.7 volts), we have 0.1 volts across R4 after subtracting: Vbias - Vbe.  We can compute the current through R4 since we know the voltage across it and its resistance:  0.1 / 390 = 0.00025 amps.  Assuming that the current through R4 is about equal to the current through R3 (since Ibe is much less than Ic), we can compute the voltage drop across R3:  10k * 0.00025 = 2.5 V.  Subtracting that from the power supply voltage (9V) we see that the idle DC voltage at the collector of Q1 will be about 6.5 volts.  The simplified formula for DC Vout is 9 - (R3 * (Vre / R4)).

Now, suppose we apply an AC signal of, let's say, 0.01 volts p-p to the input.  This will add to the base DC bias voltage and also appear at the emitter with the same relative amplitude.  So now we have a changing voltage across R4 with an amplitude equal to the input AC voltage, just with a different DC offset.  Using the formula above, we can see that for a 0.01 volt input signal, we get an output voltage swing of about 0.25 volts.  0.25 / 0.01 = a voltage gain of 25.

Let's look at that in more detail.  Our 0.01V peak-to-peak input is an AC voltage that swings between peaks of -0.005V and +0.005V.  Add that to the base bias voltage of 0.8 volts and we see that the base voltage will swing between 0.795V and 0.805V.  Subtract the Vbe voltage drop of 0.7 volts and we get a range of voltage on R4 of 0.095V to 0.105V.  Plug these values into the DC formula above and we see the output will swing from 6.56V to 6.31 volts.  After the DC offset is removed by the output capacitor, that works out to an AC voltage level of (6.56 - 6.31) = 0.25V peak-to-peak.

The simplified formula for AC signals is Vout = -Vin * R3 / R4.  Note the minus sign -- this indicates that this amplifier configuration is inverting.

If the input signal gets large enough, the output will start to clip when the output starts wanting to go higher that 9 volts or lower than 0.1V. 

[Schappy]

PJ

Thats exactly what I needed. Youve been a great help.

[Schappy]

So we are using the transistor to amplify the AC voltage. Since the transistor only needs a small amount of voltage(greater than 0.7 v) we attenuate the Power Supply voltage(DC) with a voltage divider using resistors.

After passing through the transistor we have both AC and DC voltage with our AC being increased to 0.25V.
We then filter out the DC with the capacitor.
Lastly R5 is a potentiometer that will bleed a certain amount of signal to ground altering the output.

[Paul Perry (Frostwave)]

Maybe this http://www.bcae1.com/transres.htm will help.
Explanation of transistor with interactive applet (vary bias in simple amplifier stage etc).

[RedHouse]

Maybe this http://www.bcae1.com/transres.htm will help.
Explanation of transistor with interactive applet (vary bias in simple amplifier stage etc).

That's very good Paul.

[Schappy]

Paul,
Thanks a lot thats a great site. I think that really solidifies my basic understanding of transistors.

If our signal we get from our collector is out of phase how does that affect our output signal?
Do we do anything to get it back in phase?

[Schappy]

I understand the math and purpose of each component but Im having a difficult time visualizing the actual movement of the electricity through the current and what happens at each step.

If anyone could give me a simple walkthrough or direct me to a place that will do so I would greatly appreciate it.

As a beginner it is quite difficult to piece together the functions of each component  and the theory behind it and apply it to an actual circuit.