We found a nice looking 1970′s R-125 (Mediterranean) Hammond Organ for the house looking through Kijiji. It was up for 30 bucks but the lady was nice and said we could have it for free! The only problems were getting it home, repairing a key that was broken, and fixing a short in the bass pedals.

Well after a full day of loading and driving, we got it back home. Today I repaired the key by dismantling the top, removing the key, and then krazy gluing the plastic back together. This guide really helped alot.

I am still reading up on the bass pedal circuit, and giving it a nice polish right now. Thanks Sharie!

Other than that, a DW 5000 Accelerator (DW5000ad3) Single Bass drum pedal which is the most amazing pedal I have ever played, and two pairs of nice sticks. Vic firth 7a’s (for Anna) and 5a’s (for me). She loves drumming with the sticks and her meter is great (even on the Nar). Drum Drum Drum! now if Dad can catch up!

I was tagged for an upgrade to a Blackberry, and we went in to check them out. It turned out for 99 bucks, I could get a Blackberry Bold 9000, so I said yes plz.

Being able to surf and look at stuff, receive emails, and update web pages from anywhere rules! In the places I hang out, if you pull a net laptop out, it’s a conversation starter, but now, I’m just another loser addicted to playing with his phone in public. Hah!

Ap = Pl/Pin

where:
Pl is the load power, calculated by

Pl = Vl^2/Rl

and Pin is the input power, calculated by

Pin = Vin^2/Rin

Rin being the input resistance.

This is usually expressed in RMS, which is .707 times the peak voltage. If you measure AC voltage with an RMS voltmeter, this is the way to calculate load power. More often you are looking at the AC output voltage with an oscilloscope. In this case use

Pl = Vpp^2/8Rl

When the voltage gain is known, another equation that can be used is

Ap = Av^2(Rin/Rl)

Assume a common collector amplifier has an input resistance of 10k ohms, and a load resistance of 100 ohms. Voltage gain is approximately one for common-collector, so the power gain is

Ap = 1(10k/100)
Ap = 100

The AC and DC load line

During the positive half cycle of ac source voltage, the collector voltage swings from the Q-point towards saturation. During the negative half cycle, the collector voltage swings from the Q-point towards cutoff.

Maximum output voltage can be achieved when the Q-point is at the center of the AC load line. This differs from the DC load line in amplifiers (common emitter for example) because the DC and AC collector resistances are not equal. The DC collector resistance is simply the collector resistance, where the AC collector resistance is the collector resistor in parallel with the load resistance.

In the image of the AC load line for a CE amplifier, point a is the AC saturation point, and is calculated by

Icq + (Vceq/Rc)

and point b is the AC cutoff, calculated by

Vceq + IcqRc

DC quiescient power

Pdq = (Icq)(Vceq)

This is saying the power dissipation of a transistor with no signal input will just be the product of q-point Ic and Vce. Class A power amplifiers must maintain a quiescient current that is at least as large as the peak current requirement for the load current. The output power is

Pout = Vl(rms)Il(rms)

This formula can be used to determine the output power maximum.

Pout(max) = .5(Vceq)(Icq)

The Efficiency of an amplifier is the ratio of the signal power to the load, to the power supllied from the DC source.

%Eff(max) = (Pout/Pdc)100

Classifications of Power Amplifiers

There are a few classifications of power amplifiers, and they are based on the percentage of the input cycle that the amplifier operated in the linear region. In the previous examples, reaching cutoff or saturation was undesirable and resulted in clipping and distortion.

Amplifiers operating solely in the linear region are known as Class A amplifiers. They are usually mid-point biased to maximize the available gain. Any distortion or clipping is undesired. They are usually constructed in a common-emitter or common-source configuration. The amplifier conducts for the full 360 degrees of the input cycle, always in the linear region, and the output wave is 180 degrees out of phase with the input. Class A efficiency is usually around 25%.

Class B amplifiers have the q-point at cutoff. For this reason, they operate for 180 degrees of the input signal, and since Icq = 0 and Vce = Vce(cutoff), the transistor is not conducting until an AC signal is applied. Two transistors are usually used in class B amplifiers to create a push-pull configuration. Each transistor conducts for 180 degrees of the input signal, and the full signal is sent to the load. Class B amplifiers have a 79% maximum efficiency.

Bipolar junction transistors have the .7 (silicon) or .3 (germanium) volt drops that must be overcome, or the signal becomes distorted as it flips between the two transistors. This is known as crossover distortion. Diode biasing can be used to overcome it. The diodes compensate for the base-emitter voltage drops and produce a undistorted signal.

Class AB is a modified form of Class B push-pull operation when biasing resistors are used to put the push-pull stages into slight conduction, even when there is no input signal applied.

Basic Class C amplifiers are biased so they conduct for even less than 180 degrees of the input cycle. More power can be obtained, but the output is very distorted, and so Class C is used more often in RF applications. They are biased way below cutoff, and therefore much less heat is generated from this momentary conduction. A negative voltage is applied from the base, and the transistor conducts only when Vin exceeds this negative voltage and the voltage from the base to the emitter.

Power dissipation is very low for a class C, and can be found through

Pd(avg) = (time on/T)(Vce(sat)Ic(sat))

Remembering that the voltage drop across a transistor is around .2 volts, this will usually be a pretty small amount.

In tuned operation, a tank circuit containing an inductor and a capacitor set for resonance is used. This tank circuit would normally start out at a full wave form, and then slowly discharge with one pulse from the input. These circuits are tuned so each pulse from the input keeps the oscillation of the tank circuit going. Efficiency for Class C operation can approach 100%!

The transconductance curve for a FET is a comparison of values of voltage from gate to source (Vgs) to values of drain current (Id). When Vgs is closest to Vgs(off), which means the transistor is no longer conducting, then Id is at a minimum. When Vgs = 0, then Id is at it’s maximum, Idss. Idss is the drain current with the source shorted, which under normal conditions, is the highest amount the transistor will let flow.

Vgs(off) you get Id at a minimum
Vgs = 0 you get Idss (Id maximum)

The different types of JFET’s and MOSFET’s are detailed in my other posts so I won’t go into it too much here. Until further mentioned, I am going to assume a mid-point biased JFET transistor. This gives a bit of distortion due to the transconductance curve, but is acceptable for some applications. When less distortion is needed, D-Mosfets, which operate in either depletion or enhancement mode, the area around Idss can be a fairly linear and therefore a good spot to bias for amplification. For now, I’ll just stick to JFET’s.

The main advantage to FET amplifiers is their high input resistance. Since the gate to source junction is reverse biased, it has as much input resistance as a reverse biased diode.

Gain is still and can always be defined as the ratio of Vout/Vin. In the case of a FET amplifier, it can also be defined as

Av = Vds/Vgs

Gain can also be determined using transconductance (gm) measured in Siemens (S) times the value of the drain resistor (Rd). Remember that in the case of a loaded amplifier, the drain resistor is parallelled with the load resistance (Rd = RD || RL).

Av = gmRd

gm0 is a value given on datasheets, and represents the value of gm measured at Vgs = 0. From this, you can calculate values of gm for different values of Vgs.

gm = gm0 (1-(Vgs/Vgsoff))

There are three main configurations. Common-Source, Common-Drain, and Common-Gate.

In FET Common-Source amplifiers, the DC and AC share a common point at the source of the transistor, and share a lot of the characteristics of the Common-Emitter BJT. The signal at the output has a 180 degree phase shift from the input, and some voltage gain.

In FET Common- Drain amplifiers, the DC and AC share a common point at the drain, and can be compared to BJT Common-Collector amps. Vout is in phase with Vin, and the voltage gain is ~1. They are current amplifiers, and are also referred to as source-followers.

In FET Common-Gate amps, the gate is the common point for DC and AC, and can be compared to Common-Base BJT’s. They have a low input resistance. Common-Gates are mostly used for high frequency circuits, often as the first stage, and sometimes connected directly to an antenna.

JFET’s are constructed in two types. They can either be N-channel, or P-channel. In an N-channel JFET, There is a solid layer of N-type semiconductor, with two layers of P-type material attached to the sides. These two P-type materials are calledd the gate, and the two ends of the N-type material are called the source and the drain.

In diagrams, the drain is at the upper end, and the source is at the bottom end. Currrent in the drain circuit flows from the source to the drain.

The JFET is always operated with the gate-source junction reverse-biased. This reverse biasing of the gate-source junction with a negative gate voltage produces a depletion region in the p-n junction, which extends into the N-channel and increases the resistance between the source and the drain terminals.

In an example with two power supplies, one is attached from the drain to the sourceand is called Vdd, and is known as the drain circuit. The negative terminal is connected to ground, as well as to the source of the JFET. The positive end is connected to a series limiting resistor (Rs) and also to the source terminal of the JFET.

The gate supply (Vgg) is connected with the positive end to ground, and the negative end to the gate. This creates a negative gate voltage, which is needed for the reverse biasing of the gate source pn junction.

A greater value of Vgg narrows the channel, which increases the resistance of the JFET, and decreases drain current (Id).

Less Vgg widens the channel, which decreases resistance and increases drain current(Id).

Pinch-off voltage

The Pinch-off voltage (Vp) is the value of voltage from drain to source at which drain current (Id) becomes constant. In this area, known as the constant-current area, drain current will remain constant until it reaches breakdown. Once breakdown occurs, the JFET is being operated out of range and current will increase quite rapidly until it is destroyed.

Cutoff voltage

The value of voltage from the gate to the source that produces a drain current of approximately zero is called the cutoff voltage, or Vgs(off). For N-channel JFET’s, this will be a negative voltage, and this causes the delpetion region to become so large that current flow is stopped.

There is a relation between the pinch-off voltage and the cutoff voltage. Vgs(off) and Vp are always equal, but opposite in sign. That is, if Vgs(off) is -3 volts, then pinch-off voltage is 3 volts.

to be continued…

Links:

Ticalc.org

Backlighting

Starting with the forward region, once the biasing voltage source overcomes the barrier potential, the diode begins to allow electron flow. For a normal doped silicon diode, this is .7 volts. This is also known as the knee voltage, because once .7 volts is achieved, the voltage on the graph turns very sharply up, creating what looks like a knee in the line. Above the knee voltage, diode current increases very rapidly. Once the barrier potential is overcome, all that impedes the flow of current is the resistance of the P and N junctions. This is called the Bulk resistance of the diode and can be calculated from the sum of the resistance of the P and N junctions.

Another thing to consider in the forward region is the maximum DC forward current. This can be found on datasheets. Once this is achieved, the diode will probably be destroyed due to excessive heat. This is usually termed If(max) or Io. Diode datasheets also have a maximum power disspation rating.

When diodes are operated in the reverse region, you get a very small amount of leakage current, and there is a point when the diode will breakdown, due to an effect called avalanche. When so many electrons are being forced onto the diode, the energy that propels them is enough to force other electrons out of their valence band and across the P-N junction. This breakdown voltage is also put on the datasheet. Although some specialty diodes, like zener diodes are meant to be operated in this way, on a normal diode, avalanche is to be avoided.

Special purpose diodes

Rectifier diodes are constructed to allow current in only one direction. when used with an AC voltage source, this cuts off one side of the sine wave, and creates a pulsating DC wave. Say the circuit is connected so only the positive alternations are passed, once the sine wave reaches 0 volts, it remains there until the wave reaches 0 volts again, and then continues on passing a positive sine.

This arrangement of a diode in a circuit is known as a half-wave rectifier.

where:
Vp is your peak voltage
Vs is your voltage source
and .7v is the voltage drop across the diode (silicon).

If you arrange a diode this way on both ends of a AC voltage circuit and then combine them, you get what is known as a full-wave rectifier. Only the positive waves are passed, and they are 180 degrees out of phase with each other. The end effect is as the positive sine wave of the first signal drops to zero, the other side pulses and completes it’s sine, and so on. the result ends up looking like a regular sine wave, with the negative alternations flipped positive. Full-Wave rectifiers can be used in power supplies, where an AC signal is provided and a DC voltage is desired. Full wave rectifiers must use a center tapped transformer.

Since the negative alternations are simply dropped, normal full-wave rectifiers are wasteful. When designing a rectifier circuit, it is better to use a bridge rectifier. Bridge Rectifiers have two ways for the current to flow, so there is a path on each alternation. Most power supplies use this configuration. Since a center tap is not needed, the rectified voltage is twice what a full wave recifier would create.

In bridge rectifiers, another thing to consider is since you have two diodes dropping voltage on each path, the voltage is calculated by:

where:
Vp is your peak voltage
Vs is your voltage source
and 2(.7v) is two .7 voltage drops across the diodes (silicon).

If you connect a DC Voltmeter across the load, it will indicate the average value of the full wave signal, which is:

which is equivalent to

.636 * peak voltage.

The frequency of a full wave signal is double the input frequency, since a waveform completes it’s cycle as soon as it repeats. For a 60 hertz input:

Time = 1/Frequency

Time = 1/60

Time = 16.7ms

The rectified voltage has a period of

Time2 = 16.7ms/2

Time2 = 8.33ms

Frequency2 = 1/8.33ms

Frequency2 = 120 hertz.

Another way to put this simply is to say:

Fout = 2Fin

where:
Fin is Frequency In
Fout is Frequency Out

Another type of diode is the Zener diode. Most diodes are never operated in the breakdown region beacuse it would damage them. A zener is manufactured to be operated in the reverse region, and to have a specific voltage where it will begin to conduct. Zener’s are available in many different voltages. A zener diode is sometimes referred to as a zener voltage regulator becuase they can be used in parallel to allow a certain voltage to pass to the load, and then begin to conduct once the zener voltage is reached, therefore passing the remaining voltage through the zener and bypassing the load. A series resistor is always used in this configuration to limit current flow.

Maximum power through a zener diode is found by:

Zener Impedance can be found through:

The change in Zener voltage (^Vz) can be found by:

LED‘s or light emitting diodes are another specialty diode. As the electrons cross the P-N junction and fall into holes, they radiate energy. LED’s are constructed to show this as visible light. By using elements like arsenic and phosphorus, LED’s can be manufactured in red, green, yellow, blue, orange and even infrared. The exact voltage drop across LED’s depends on the color. The typical voltage drop is 1.5 to 2.5 volts for currents between 10 and 50 milliamps.

All diodes have an associated capacitance, due to the way they are constructed. The P and N regions can be thought of as the plates, and the depletion region is the dielectric. Varactor diodes are built to take advantage of this, and are used in tuning circuits where a variable resonant frequency is desired. As the voltage is varied, the depletion region expands and contracts, causing the capacitance to change. You can connect a varactor in parallel with an inductor to get a resonant circuit, and then vary the biasing voltage to achieve specific resonant frequencies.

Doping is the process of adding impurities to silicon (or other semiconductors) to alter the electrical characteristics of the semiconductor. If we think of a pure, or intrinsic piece of silicon, there is ideally no free electrons, and no free “holes” in the valence bands for electrons to go.

We add an atom with 3 valence electrons. Elements with 3 valence electrons are aluminum, boron, gallium, and indium. This creates a tri-valent bond and leaves on open “hole” for an electron to flow in and out of. Doping a semiconductor this way creates a P-type material. To remember, you can think of the “P” as positive. There is a deficiency of one electron, so the 3 valence atom added is known as a acceptor impurity element.

The other method of doping is to add an atom with 5 valence electrons to a piece of silicon. Elements with 5 valence electrons are arsenic, antimony and phosphorus. This creates a crystal with an extra electron that is free to move around and is known as a penta-valent bond. This is known as a N-type material and can likewise be remembered that the “N” is for negative. There is an extra electron, so the 5 valence atom is known as a donor impurity element.

As you can see, the amount of impurities added directly effects the electrical characteristics, and can be used to regulate the amount of electrons moving through the material. Simply having a P or N type material on their own might have some uses, but when the two are used together, a P-N junction is formed, and is the basis of many electronic devices used today.

When the two materials are put together, they repel each other. The free electrons spread out, and some of them diffuse across the junction. This is known as the depletion region. Each time an electron crosses over, it leaves a pentavalent ion, with a relative positive charge. This electron in turn falls into a hole in the P-type material, and causes a negatively charged trivalent ion. It has space for one electron, and when it is filled we can say that it has gained a relative negative charge. This region, with positively charged extra electron ions, and negatively charged electron deficient ions, creates a potential difference between them. This is known as barrier potential. The barrier potential is usually .7v for silicon, and varies for other types of semiconductors. The basic idea is exactly the same though.

The barrier potential must be overcome to allow electron flow in the P-N type material. Biasing a P-N junction is the process of adding a voltage source to either allow or prevent the flow of electrons.

When the N-type is negative with respect to the P-type material, the electrons easily flow from the power supply, to the junction, then from one side to the other. The N-type material constantly feeds the electrons to the P-type, and the electrons flow from the p-type back to the power supply. This is known as forward bias. The P-N junction is arranged in the circuit to allow electron flow.

Reverse biasing is done by changing the polarities of the voltage source, so the negative terminal is connected to the P-type, and the positive terminal is connected to the N_type material. This causes the depletion layer to widen, because the negative terminal attracts the free “holes” and the positive terminal attracts the free electrons. Current is not allowed to flow.

So far we have been looking at this in a perfect world, but in reality, there are a few holes in a N-type material and likewise there is a few extra electrons in a P-type material. These are known as the minority carriers, and are mostly caused by thermal energy, or heating of the P-N junction. Under normal operating temperatures, this amount is negligible. Datasheets are invaluable when seeking the maximum temperatures, voltages, and dissipation of power for any electronic device.

The lower of the two reactances is subtracted from the higher one. Say Xl is 100 ohms, and Xc is 75 ohms, the resulting reactance (Xnet) is 25 ohms, and since the resulting reactance is Xl, we say the circuit is acting inductively. Likewise, if Xc is greater than Xl, after we subtract one from the other, we say the circuit is acting capacitively.

It is easier to remember this by looking at vector diagrams. Current is once again our reference vector, with resistance in phase on the horizontal, inductance is plotted upward, and capacitance plotted downward. Whether we are talking voltages (Vl/Vr/Vc/Vt) or resistance (Xl/R/Xc/Z), in series circuits, the vectors look the same.

When applying the pythagorean theorem to series RLC, we simply subtract the reactances, and use the remainder (Xnet) in the formula. First subtract lesser reactance from the greater.

or

and then

same thing for reactances to calculate impedance.

Xl – Xc or Xc – Xl = Xnet

Z = √(R^2 + Xnet^2)

An example to calculate Impedance in Series RLC:

R = 100 ohms
Xl = 75 ohms
Xc = 60 ohms

75 – 60 = 15 ohms = Xnet

√(100^2 + 15^2) = 101.119 ohms = Z

Tan-1( 15 / 100 ) = 8.53 degrees = Phase angle

Say we had an applied AC voltage of 20 volts @ 100Hz . First calculate total current using Ohm’s law. I = V / Z.

20 / 101.119 = 197.787 mA

From there we can calculate the voltage drops across the components

Vr = .197787 * 100 = 19.7787 volts
Vl = .197787 * 75 = 14.834 volts
Vc = .197787 * 60 = 11.8672 volts

As you can see, there is a lot of voltage here, but Vl and Vc are canceling each other out. Working through it again, we can double check our work.

Vl – Vc = Vnet
14.834 – 11.8672 = 2.9668 volts

√(19.7787^2 + 2.9668^2) = 20 volts = Vt

Tan-1( 2.9668 / 19.7787 ) = 8.53 degrees. It checks out.

Parallel RLC circuits

Once again, the reactances cancel each other, the same way as in series, but in parallel circuits, voltage is our reference vector (since the same voltage flows though each branch) and is plotted horizontally, along with resistive voltage (in phase). Capacitive current leads the voltage by up to 90 degrees (ICE) and inductive current lags the voltage by up to -90 degrees.

An example of a parallel RLC circuit. Let’s keep our 20 volts @ 100Hz and use:

R = 55 ohms
Xl = 225 ohms
Xc = 125 ohms

Say for simplicity, that each component is on it’s own branch. In a parallel circuit, we want to calculate branch currents first.

Ir = Vt /R
Ir = 20 / 55
Ir = 363.636 mA

Ixl = Vt / Xl
Ixl = 20 / 225
Ixl = 88.8889 mA

Ixc = Vt / Xc
Ixc = 20 / 125
Ixc = 160 mA

We subtract Ixl from Ixc to get our net reactance current (Inet):

160 – 88.8889 = 71.1111 mA = Inet

find total current through the pythagorean theorem:

It = √(Ir^2 + Inet^2)
It = √(.363636^2 + .07111111^2)
It = 370.524 mA

use arctan to solve for phase angle:

Tan-1( .0711111 / .363636 ) = 11.0649 degrees = phase angle

and lets finish by calculating impedance from Ohm’s law.

Z = Vt / It
Z = 20 / .370524
Z = 53.9776 ohms

References:
Foundations of Electronics, by Russell L Meade