domingo, 25 de julio de 2010

Miniature Antenna Tower for Radio

Miniature Antenna Tower for Radio


The project uses a 4-element Yagi-Uda and some blinking lights at the top for transmission to create a miniature antenna tower project.


The Yagi-Uda antenna is made of copper clad steel wire with 14 AWG for the vertical parts of the structure while the zigzag structure was made by wrapping around 18 AWG copper wire around s flat strip of steel. The coil is then stretched out so that it will resemble the zigzag structure of a real antenna tower. The 3 vertical pieces and 3 zigzag pieces were soldered together after being made followed by the thrust bearing plate and rotator plate that are made from small bits of PC board and some small coils of copper wire.
The circuit board used was made thin to fit between the box and the base tight. The flasher circuit was built around 555 and soldered to the box with a battery connector taken from a dead 9V battery. The active components include the triode, NPN BJT, bridge rectifier, P-channel MOSFET, diode, and pentode. The passive components comprise of iron core transformer, resistor, crystal, battery/DC power source, ground connection, antenna, electrolytic capacitor, inductor, and momentary/morse key switch.

C.I: 17.557.095

Speaker Box Audio Amp

Design Philosophy
I did this design after getting a new computer. After buying the processor, monitor, and printer, I wasn't willing to spring for a set of speakers too. After going "soundless", I decided to add speakers. Of course, this was the perfect excuse warm up the soldering iron to try out a new design. This original design is a variation on a well-known design, examples of which can be found in a great many texts. My variation was to add a second differential stage to replace the usual common emitter-plus-constant current source. Doing so opens up a second inverse feedback path. The signal path being a series connection of the differential amp inverting inputs, and the feedback being the non-inverting inputs. To this end, the constant current source does not include bypass capacitors to render it a DC-only stage. The other feedback path is the usual one: output to the non-inverting input of the first differential stage. The idea is to add lots of inverse feedback in order to linearize the output transistors, as the non-linearity of these devices makes them rather poor analog amplifiers.

The other design criterion was the use of components readily available at a hobbist's store: "Radio Shack". No "exotic" components that would require a special order are used here. To this end, the output stage is a semi-complementary design, although this is not optimum so far as fidelity is concerned. (Radio Shack doesn't offer complimentary pairs of power BJTs.) Since the amp is intended to work into the "wide band" (60Hz -- 18KHz) speakers that this store sells, the gain is deliberately kept down. With the design given here, the max output is 6.0W. Even here, since these speakers are rated at 3.0W, you could blow speakers if the volume is turned up too high. Limiting output is accomplished with the initial voltage amp (Q1). This stage has a gain of 7.5, so that with a 1.0VRMS input, the output voltage becomes slightly less than 7.5VRMS. The gain of the circuit from the output end of the volume control to the speaker is unity, due to the large negative feedback. You could drive the power output as high as 10W, as there is plenty of "head room" in the power supply design. (Again, this was done to improve the fidelity as much as possible with a BJT-based design.) If you push the output, you will need better heat sinking on the finals than what's described here.
The selection of actual components is not critical. Any small signal "transistors anonymous" will work here. The only exception is Q8 and Q10. These two must be a 2N3904 and a 2N3906, as these are complimentary pairs. These transistors set the characteristics for the entire final, and must be balanced as to characteristics in order to keep DC out of the speakers, and for symmetrical amplification. As for the differential pairs (Q2, Q3, Q5, Q6) type isn't so important, however, matched pairs is important. It's a good idea to get a couple dozen of each, PNP and NPN, and select pairs that match closely in VBE and hFE. Matching pairs in the differential stages helps hold down DC offset, as this design uses DC coupling in the interest of good low frequency response. Speakers tend to object to any significant amount of DC by burning out the voice coils. It also compromises fidelity in that DC on the voice coils restricts the movement of the cone.
The 1800pF capacitor connected to the drain of Q1 was included since one of these amps showed a weak (10mVP-P) oscillation at 700KHz. If you include it, be certain to make the connection with the least possible lead length, as it is an RF component. It will have no effect on the audio quality since its cutoff frequency is about 58KHz.
Construction: Amplifier
The entire unit is designed to fit in a plastic "project box" cabinet of 7 X 6 X 3 inches. The "wide band" speakers are 3¾ inches in diameter, and fit in a cutout of 37/16 inches. The circuit board is: 6¾ X 2¼ double-sided, copper clad. The circuit is built "dead bug" style, simply by making all ground connections by soldering to the unetched copper, the rest of the wiring being made above the circuit board. If additional mechanical reinforcement is needed, this is provided by soldering 2.2MEG resistors to the copper ground plane for use as a "tie point". Connecting such large resistors between the circuit and ground doesn't affect the operation since the impedance at that point is going to be several orders in magnitude smaller. To fit the circuit on the board, both sides are used. The preamp module being constructed on one side, with the final being built on the opposite side.
The heat sink for the power transistors is made from two pieces of steel or aluminum 1 X 1 (in.) angle stock. Simply measure off a 2¼ length, and cut it in half. This will naturally provide the necesary gap to separate these heat sinks electrically. Sand off enough copper from one end of the circuit board to insure that the heat sinks aren't being shorted out. These are attached to the circuit board with one screw and nut. (Also, be sure to clear the copper away from the screw heads on the preamp side of the circuit board, otherwise, you will short out the finals.) In the middle of one flange of the angle stock, bore a hole slightly smaller than the diameter of the 2N3053s. This hole should be enlarged by careful reaming, so that the transistors fit snugly. This is essential for good heat transfer out of the transistor and into the heat sink. Once this has been done, give the transistor a thin coat of silicone heat sink grease and press into place. These can be set aside until you need them.
Once the circuit is completed, and you've checked for wiring errors, the initial test is done with a 10 ohm/10W resistor connected to the output in place of the speaker. It is essential that this circuit, at no time, ever be run with no load. The Q3 DC return is solely through the load. If there is no load, this differential stage becomes severely unbalanced, and that will propagate to the finals, and probably blow one or both finals. The first thing to check is the DC voltage across the 10 ohm dummy load. This should be less than 0.5VDC. If there is more DC than that, then check each differential stage to see where that DC offset is coming from. This is most likely due to mismatched transistors (you did select for balance didn't you?) or finals too far out of match. When I did this construction, I didn't bother matching 2N3053s as the characteristics of the finals are determined largely by their Darlington counter parts. I didn't have any trouble from these components, but you can't rule the possibility out either. If the DC balance is acceptable, then you can either test with the dummy load, a signal generator, and an oscilloscope. You should see a faithful reproduction of the input wave form. If all goes well, then you can connect the speaker. When making this connection, it's a very good idea to connect the speaker return to the same point where the connection to the DC ground is made. This prevents the possibility of setting up an audio frequency ground loop. This, in turn, can cause instability that's next to impossible to fix. That means do not try to "one wire" the speaker if you decided to build the amp in a metal project box. Adjust the bias for 10 -- 15mA of no-signal idling current. You should then be good to go.
Construction: Power Supply
The power supply is simply a straight forward symmetrical plus/minus supply constructed from a transformer with a balanced secondary and a bridge rectifier module. Even though the transformer was a self-wound unit, any transformer with a 25.2VCT secondary at 2.0A will be a good substitute. Such transformers oughht to be available as VT "heater transformers" for high power "ham" RF amplifiers. It is essential that the primary side of the transformer be connected exactly as shown. This PS uses a MOV for over voltage protection. These things have a nasty habit of not turning off once they turn on. The fuse needs to be connected between this MOV and the mains. Don't attempt to "go cheap (or lazy)" and fail to include it. A 2.0A, "slow blow" type works well here. It will stand up to the initial surge as the filter capacitors charge up, and will prevent damage or worse should the MOV be triggered. This supply is somewhat over rated for the application, and is heavily bled for good voltage regulation. There is still some 120Hz hum that could be eliminated with a more involved PS circuit. However, as this wasn't intended as a "premium" amp, that wasn't necessary. The hum is most noticeable when running with no input signal. While actually playing sound, it's no longer noticeable. If you want to substitute a better supply, then that's OK, too.

These units worked quite well, and do sound good, despite the inherent limitations of BJTs as analog amplifiers. There are a couple of improvements possible. The first would be the use of a full-complimentary final. If this is done, then any complimentary BJTs with a PC of 5.0W or more will work. Again, the most important part of this will remain the input Darlingtons. Even better would be the substitution of complimentary (N-Channel/P-Channel) VMOSFET power transistors. These, of course, don't require Darlington pairs in order to operate. If VMOS power transistors are used, it's important to include a 100 ohm resistor in series with the gate terminal, mounted close to the transistor body with the shortest lead length. This will help prevent the possibility of RF oscillation. (Unlike the BJT, the VFET is a high frequency device, and the inadvertant construction of VHF oscillators with these is quite easy.)
Parts List: Amplifier
CR1 -- 6: 1N914
Q1: 2N3819
NPN: 2N3904 (or other small signal NPN)
PNP: 2N3906 (or other small signal PNP)
Q8: 2N3904 (Critical)
Q10: 2N3906 (Critical)
Q9, 11: 2N3053
R<>: 1.0K/¼W miniature pot.
Parts List: Power Supply
C1 -- 2: 20000uF/35WVDC
CR1 -- 4: 50VPRV/5.0A Bridge Module
CR5: Green LED Panel Mount
F1: 2.0A, Slow Blow Fuse
T1: PRI: 120VRMS
SEC: 25.2VCT/2.0A

C.I: 17.557.095

El transistor sin polarizar

El transistor esta compuesto por tres zonas de dopado, como se ve en la figura:

La zona superior es el "Colector", la zona central es la "Base" y la zona inferior es el "Emisor". El Emisor está muy impurificado, la Base tiene una impurificación muy baja, mientras que el Colector posee una impurificación intermedia.
En este ejemplo concreto el transistor es un dispositivo npn, aunque también podría ser un pnp.
En principio es similar a dos diodosUn transistor es similar a dos diodos, el transistor tiene dos uniones: una entre el emisor y la base y la otra entre la base y el colector. El emisor y la base forman uno de los diodos, mientras que el colector y la base forman el otro. Estos diodos son denominados: "Diodo de emisor" (el de la izquierda en este caso) y "Diodo de colector" (el de la derecha en este caso).

Antes y después de la difusión

Vamos a hacer un estudio del transistor npn, primeramente cuando está sin polarizar (sin pilas y en circuito abierto) se produce una "Difusión" (como un gas en una botella), donde los electrones cruzan de la zona n a la zona p, se difunden, encuentran un hueco y se recombinan. Esto hace que en las uniones entre las zonas n y p se creen iones positivos y negativos.

Esta difusión y recombinación se da hasta llegar al equilibrio, hasta conseguir una barrera de potencial de 0,7 V (para el Si). Se crean 2 z.c.e., una en la unión E-B (WE) y otra en la unión C-B.

C.I: 17.557.095


los masconocidos


El transistor bipolar es el comun de los transistores, y como los transistores, puede ser de germanio o silicio.

existen dos tipos de transistores: el NPN y el PNP, y la direccion de flujo de la corriente en cada caso, lo indica la flecha que se ve en el grafico de cada tipo de transistor

C.I: 17.557.095


ok, here it is, as promised, AA's Circuit Splinter No.2! but first, a little warning- i am not an electrical engineer, nor do i claim to resemble anything close to one, here or on television. please take everything i post as just what it is- the off kilter ramblings of a misinformed diy enthusiast- nothing more, nothing less. i'm trying to learn just like you.


we are continuing our limited look into DC coupled buffers and amplifiers. in this case, a JFET/BJT Feedback Amplifier. this amplifier configuration is known as what is called a Sziklai Pair, otherwise known as a compound transistor, which is similar to a darlington pair, and sometimes called a compound darlington. this wikipedia article goes into the sziklai pair a bit more, such as how they are often used in discrete solid state output stages.

here is a great article as well from

and finally, a jensen transformers app-note for a piezo buffer amp-

basically, what we have done that is different here from the last edition of Circuit Splinters(in my limited understanding), is to reduce the amount of feedback from one stage to the next. previously we had an infinite amount of feedback, which gave us a voltage gain of 0. by inserting R5 into the picture, we raise the resistance, and therefore lower the amount of feedback. in doing this, the gain of the stage rises. as you can see below, the gain of this stage is equal to the ratio of R3 to R5. so, lower R3, or raise R5, and the gain goes up. the circuit below has R3-470, R5-4700, so 4700/470 = a gain of 10. the great thing about this kind of stage is that you have a high input impedance from the FET, a buffered output, and you get voltage gain instead of just current gain as in a classic buffer. remember that R2 may have to be adjusted to 1/2 your supply voltage on the drain of Q1 for the proper amount of headroom.

i think that this circuit tends to sound best when used in a "clean" gain type of stage, with just a slight amount of breakup. never the less, many different adjustments can be made. instead of R3, put a large bypass cap, and move R3 to the emitter of Q2 for more gain. try different bypass caps for different frequency responses. bypass Q2's emitter. if you adjust your bias circuit, you can use a bjt for Q1. try a Ge pnp for Q2. see what happens when you put a small cap from the source of Q1 to the collector of Q2.

here is an example of an RIAA circuit that uses this configuration. you can see what has been done to tailor the frequency response of the stage for the RIAA curve using RC filters in the feedback section.
(click to enlarge)

here's a version from John Linsley Hood that uses a bootstrapped input-
(click to enlarge)

here's another one that uses a bjt for Q1, and has an interesting baxandall-style tone control based on feedback as well. note the bias configuration for the BJT input transistor.
(click to enlarge)

as always, if anyone out there has any real knowledge they can share, feel free! the sky's the limit. go nuts. have fun. take it easy!

C.I: 17.557.095

Frequency Response for MOSFET/BJT

The frequency response of a BJT or MOSFET can be found using nearly the exact same process, with the only variations being caused by a single resistor and simple naming conventions that differ between the two devices.
Before we start let's think a little bit about what we're doing:
Our goal is going to be to find the pole(s) of the circuit.
Okay? What is a pole and why do I care where it is?
A pole is a frequency at which the gain of the device rolls off. (remember that when it rolls off , it will be at the -3dB frequency with a slope of -20dB/decade)
We care because if the gain of a device rolls off at a certain frequency, then we won't be able to amplify a signal above that frequency very well because the gain will be decreasing by 20dB/decade.
The procedure is nearly identical whether we are using a BJT of a MOSFET, but we will work each of them side by side just in case there might be any confusion, and we'll follow these steps as we go through. (we will also use some values that came from the output file when running a simulation of this circuit in Cadence (or PSPICE) )MOSFET
1. Take a look at one of the circuits and see what you notice, how about the MOSFET. This step is just to help us with our knowledge understanding of the circuit.
- At a glance it just looks just like another MOSFET right? Sure is, but let's take a look at a few things just for kicks. Notice that it is using a bypass capacitor at the source so we don't have to worry about R_s (at when working with high frequency). Since the capacitor C_s bypasses R_s to ground, you should notice that this is a common-source amplifier. You could notice the Values for R_1 and R_2 and start to think about what the Gate voltage is and how that may affect the circuit.
2. We are talking about frequency response so that means we are probably going to want to draw the small signal equivalent circuit.
Remember that the capacitors C_1 and C_2 will act like short circuits at high frequencies so we will ignore them, but we will have to account for some of the capacitance internal to the device.
Both devices have internal capacitances that are very similar. As you can see from the small signal models for a MOSFET (above) and BJT (below), the only significant difference is that the BJT has an additional resistance Rpi between the Base and Emitter.
Most of the analysis we will do is based on the small signal model. Note that small signal models are not typically used in PSPICE so this picture may look a bit odd, especially the controlled source but for our purpose it is good to have a visual reference. To start we will point out what everything is. Cgs is an internal capacitance betwe
en the gate and source. The
values for Cgs was similar to one the a PSPICE simulation may give. CM1 and CM2 are Miller capacitances which we will find values for later. ro is a Norton equivalent resistance that makes the model more ideal. And just pretend that the G2 looks more like a voltage controlled current source and that their gains are gm*Vgs and gm*Vpi. For the BJT CM1 and CM2 are both Miller capacitances, Cpi is similar to Cgs and Rpi the additional component used for BJTs but not MOSFETs. The other part should look familiar from the other figures.
We will find the device gain, overall gain, equivalent input and output capacitances, and the input and output poles. The process for both is essentially the same.
Device Gain: This is the gain from the control source to the output so we are looking for Vout/Vgs (or Vout/Vpi for a BJT). We will ignore CM2 for this process. Notice the resistances ro, RD, and RL are in parallel. Vout should be given by that equivalent resistance times the current though it which is gm*Vgs from the control source. So the equation for device gain is,
V_{out} / V_{gs} = gm*(r_o//R_D//R_L) (MOSFET)
V_{out} / V_{\pi} = gm*(r_o//R_C//R_L) (BJT)
Overall Gain: This will be the gain from the source (Vs) to the output (Vout). We already know what Vout/Vgs is so if we find Vgs/Vs, we can multiply them to get Vout/Vs = (Vout/Vgs) * (Vgs/Vs). Vgs/Vs is a simple voltage divider. Hopefully you can see this from the small signal model (remember that we are ignoring the capacitors for now but they will play a part later). The equations we will get for Vgs/Vs and the overall gain are.
V_{gs} / V_s = \frac{ (R_1//R_2)}{(R_1//R_2) + R_s} (MOSFET)
Overall Gain: V_{out} / V_s = \frac{ (R_1//R_2)}{(R_1//R_2) + R_s} * gm(r_o//R_D//R_L) (MOSFET)
V_{gs} / V_s = \frac{ (R_1//R_2//r_\pi)}{(R_1//R_2//r_\pi) + R_s} (BJT)
Overall Gain: V_{out} / V_s = \frac{ (R_1//R_2//r_\pi)}{(R_1//R_2//r_\pi) + R_s} * gm(r_o//R_C//R_L) (BJT)
Now we will find the input and output poles. For this we will need to look at the capacitances and use a formula to find the Miller capacitances, CM1 and CM2. Any explanation for the miller capacitance will have to wait for another post or check out your Electronics Book, Wikipedia, Google, etc. but we will need to use a couple of special equations. Overall we will need to find the input resistance and input capacitance for the input pole and the output resistance and output capacitance for the output pole.
Each pole will be at a frequency w=1/RC where the R and C are the equivalent R and C at that point, so to find the input pole, we will need to find the input resistance and the input capacitance. These are found by looking into the input (the left side of the small signal model). The voltage source will act like a short so we see Rs in parallel with R1//R2 for the MOSFET (the BJT will have Rpi in parallel also). The input capacitance will be Cgs in parallel with CM1 (the BJT will be the same). The output resistance and capacitance are found the same way only looking in from the output (the right side of the small signal model).
\omega_{IN} = \frac{1}{R_{IN}C_{IN}} \omega_{OUT} = \frac{1}{R_{OUT}C_{OUT}} (MOSFET or BJT)
So the input pole will be: (MOSFET)
R_{IN} = R_S//R_1//R_2 = 950 R_{OUT} = r_o//R_D//R_L =
C_{IN} = C_{gs} + C_{M1} = C_{OUT} = C_{M2} =
\omega_{IN} = \omega_{OUT} =
and the output pole will be: (MOSFET)
R_{IN} = R_S//R_1//R_2//r_\pi = R_{OUT} = r_o//R_D//R_L =
C_{IN} = C_{BE} + C_{M1} = C_{OUT} = C_{M2} =

C.I: 17.557.095

Diseño de un amplificador con BJT:

  • a gain of 10 una ganancia de 10
  • 50 K Ω or higher input impedance 50 K Ω o mayor impedancia de entrada
  • 50 Ω or lower output impedance 50 Ω o menor impedancia de salida
  • Low freq response : 20 hz Baja Frecuencia de respuesta: 20 Hz
  • High freq response : 100khz. Alta Frecuencia de respuesta: 100kHz.

Required: Requerido:

Design an amplifier using BJTs Diseñar un amplificador con BJT
The amplifier must have: El amplificador debe tener:
  • a gain of 10 una ganancia de 10
  • 50 K Ω or higher input impedance 50 K Ω o mayor impedancia de entrada
  • 50 Ω or lower output impedance 50 Ω o menor impedancia de salida
  • Low freq response: 20 hz Baja Frecuencia de respuesta: 20 Hz
  • High freq response: 100khz Alta Frecuencia de respuesta: 100kHz

1) Use Standard Resistor Values 1) Utilice los valores estándar de resistencia

We are going to use two stages for the amplifier , a common emitter stage followed by a common collector stage. Vamos a utilizar dos etapas para el amplificador, una etapa en emisor común seguida por una etapa de colector común.
Assuming β = 300, Vbe = 0.7 V, Vcc = 15 V Suponiendo que β = 300, VBE = 0,7 V, Vcc = 15 V

Common Collector Stage Etapa colector común

Let Rc2 = 2.2K Ω, Re2 = 2.2K Ω, VceQ = 7.5 V Que Tc2 = 2,2 K Ω, RE2 = 2,2 K Ω, VceQ = 7,5 V
Let R3//R4 = Rth = 60 Ω Que R3 / / R4 Rth = = 60 Ω
Solving for R3 & R4 Resolviendo para R3 y R4
R3 = 188K Ω , R4 = 88K Ω R3 = 188K Ω, R4 = 88K Ω

Using standard resistor values Utilizando los valores estándar de resistencia

R3 = 220K Ω , R4 = 100K Ω R3 = 220K Ω, R4 = 100K Ω
Rth(new) = 68.75K Ω Rth (nuevo) = 68.75K Ω

The Ac Equivalent Circuit El circuito equivalente de ac

Rout: omitting the output resistance of the first stage Ruta: la omisión de la resistencia de salida de la primera etapa
Which is less than 50 Ω ¿Cuál es inferior a 50 Ω

2) Common Emitter Stage 2) Etapa en emisor común

This stage must have a gain = 10, and input impedance ≥ 50KΩ Esta etapa debe tener una ganancia = 10, y la impedancia de entrada ≥ 50KΩ
A rough approximation for the gain ≈ Rc1/Re1 Una aproximación a la ganancia ≈ Rc1/Re1
Let Rc1= 10KΩ, Re1 = 820Ω , VceQ = 7.5 V Que Rc1 = 10KΩ, Te1 = 820Ω, VceQ = 7,5 V

The Ac Equivalent Circuit El circuito equivalente de ac

We want Rin ≥ 50KΩ Queremos Rin ≥ 50KΩ
Rin = Rth // Rib Rin = Rth / / Rib
50K = Rth // 258K 50K = Rth / / 258k

Solving for Rth Despejando Rth

Rth = 62KΩ Rth = 62KΩ
Solving for R3 & R4 Resolviendo para R3 y R4
R1 = 658K Ω , R2 = 68K Ω R1 = 658K Ω, R2 = 68 K Ω

Using standard resistor values Utilizando los valores estándar de resistencia

R1 = 470K Ω + 220K Ω = 690K Ω , R2 = 68K Ω R1 = 470K + 220K Ω Ω = 690K Ω, R2 = 68 K Ω
Rout = Rc1 = 10K Ω Ruta = = 10K Ω Rc1

3) Frequency Response 3) Frecuencia de respuesta

Fl = 20 Hz , Fh = 100KHz Fl = 20 Hz, Fh = 100KHz

Using standard capacitor values Utilizando los valores estándar de condensadores

C1 = 0.16 μF, C2 = 0.12 μF, C3 = 0.033 μF C1 = 0,16 uF, C2 = 0,12 mF, C3 = 0,033 mF

4) Simulation 4) Simulación

Which shows that VceQ 1 = 8.125 – 0.566 = 7.559 V Lo que demuestra que VceQ 1 = 8,125-0,566 = 7,559 V
VceQ 2 = 11.354 – 3.658 = 7.696 V VceQ 2 = 11,354 a 3,658 = 7,696 V

C.I: 17.557.095