CIRCUIT DIAGRAM: February 2012

current divider rule

In electronics, a current divider is a simple linear circuit that produces an output current (I_X) that is a fraction of its input current (I_T). Current division refers to the splitting of current between the branches of the divider. The currents in the various branches of such a circuit will always divide in such a way as to minimize the total energy expended.
The formula describing a current divider is similar in form to that for the voltage divider. However, the ratio describing current division places the impedance of the unconsidered branches in the numerator, unlike voltage division where the considered impedance is in the numerator. This is because in current dividers, total energy expended is minimized, resulting in currents that go through paths of least impedance, therefore the inverse relationship with impedance. On the other hand, voltage divider is used to satisfy Kirchoff's Voltage Law. The voltage around a loop must sum up to zero, so the voltage drops must be divided evenly in a direct relationship with the impedance.
To be specific, if two or more impedances are in parallel, the current that enters the combination will be split between them in inverse proportion to their impedances (according to Ohm's law). It also follows that if the impedances have the same value the current is split equally.

Resistive divider

A general formula for the current I_X in a resistor R_X that is in parallel with a combination of other resistors of total resistance R_T is (see Figure 1):

$I_X = \frac{R_T}{R_X+R_T}I_T \$

where I_T is the total current entering the combined network of R_X in parallel with R_T. Notice that when R_T is composed of a parallel combination of resistors, say R₁, R₂, ... etc., then the reciprocal of each resistor must be added to find the total resistance R_T:

$\frac {1}{R_T} = \frac {1} {R_1} + \frac {1} {R_2} + \frac {1}{R_3} + ... \ .$

General case

Although the resistive divider is most common, the current divider may be made of frequency dependent impedances. In the general case the current I_X is given by:

$I_X = \frac{Z_T} {Z_X+Z_T}I_T \ ,$

Using Admittance

Instead of using impedances, the current divider rule can be applied just like the voltage divider rule if admittance (the inverse of impedance) is used.

$I_X = \frac{Y_X} {Y_{Total}}I_T$

Take care to note that Y_Total is a straightforward addition, not the sum of the inverses inverted (as you would do for a standard parallel resistive network). For Figure 1, the current I_X would be

$I_X = \frac{Y_X} {Y_{Total}}I_T = \frac{\frac{1}{R_X}} {\frac{1}{R_X} + \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}}I_T$

Example: RC combination

Figure 2: A low pass RC current divider

Figure 2 shows a simple current divider made up of a capacitor and a resistor. Using the formula above, the current in the resistor is given by:

$I_R = \frac {\frac{1}{j \omega C}} {R + \frac{1}{j \omega C} }I_T$

$= \frac {1} {1+j \omega CR} I_T \ ,$

where Z_C = 1/(jωC) is the impedance of the capacitor.
The product τ = CR is known as the time constant of the circuit, and the frequency for which ωCR = 1 is called the corner frequency of the circuit. Because the capacitor has zero impedance at high frequencies and infinite impedance at low frequencies, the current in the resistor remains at its DC value I_T for frequencies up to the corner frequency, whereupon it drops toward zero for higher frequencies as the capacitor effectively short-circuits the resistor. In other words, the current divider is a low pass filter for current in the resistor.

Loading effect

Figure 3: A current amplifier (gray box) driven by a Norton source (i_S, R_S) and with a resistor load R_L. Current divider in blue box at input (R_S,R_in) reduces the current gain, as does the current divider in green box at the output (R_out,R_L)

The gain of an amplifier generally depends on its source and load terminations. Current amplifiers and transconductance amplifiers are characterized by a short-circuit output condition, and current amplifiers and transresistance amplifiers are characterized using ideal infinite impedance current sources. When an amplifier is terminated by a finite, non-zero termination, and/or driven by a non-ideal source, the effective gain is reduced due to the loading effect at the output and/or the input, which can be understood in terms of current division.
Figure 3 shows a current amplifier example. The amplifier (gray box) has input resistance R_in and output resistance R_out and an ideal current gain A_i. With an ideal current driver (infinite Norton resistance) all the source current i_S becomes input current to the amplifier. However, for a Norton driver a current divider is formed at the input that reduces the input current to

$i_{i} = \frac {R_S} {R_S+R_{in}} i_S \ ,$

which clearly is less than i_S. Likewise, for a short circuit at the output, the amplifier delivers an output current i_o = A_i i_i to the short-circuit. However, when the load is a non-zero resistor R_L, the current delivered to the load is reduced by current division to the value:

$i_L = \frac {R_{out}} {R_{out}+R_{L}} A_i i_{i} \ .$

Combining these results, the ideal current gain A_i realized with an ideal driver and a short-circuit load is reduced to the loaded gain A_loaded:

$A_{loaded} =\frac {i_L} {i_S} = \frac {R_S} {R_S+R_{in}}$ $\frac {R_{out}} {R_{out}+R_{L}} A_i \ .$

The resistor ratios in the above expression are called the loading factors. For more discussion of loading in other amplifier types, see loading effect.

Unilateral versus bilateral amplifiers

Figure 4: Current amplifier as a bilateral two-port network; feedback through dependent voltage source of gain β V/V

Figure 3 and the associated discussion refers to a unilateral amplifier. In a more general case where the amplifier is represented by a two port, the input resistance of the amplifier depends on its load, and the output resistance on the source impedance. The loading factors in these cases must employ the true amplifier impedances including these bilateral effects. For example, taking the unilateral current amplifier of Figure 3, the corresponding bilateral two-port network is shown in Figure 4 based upon h-parameters.^[1] Carrying out the analysis for this circuit, the current gain with feedback A_fb is found to be

$A_{fb} = \frac {i_L}{i_S} = \frac {A_{loaded}} {1+ {\beta}(R_L/R_S) A_{loaded}} \ .$

That is, the ideal current gain A_i is reduced not only by the loading factors, but due to the bilateral nature of the two-port by an additional factor^[2] ( 1 + β (R_L / R_S ) A_loaded ), which is typical of negative feedback amplifier circuits. The factor β (R_L / R_S ) is the current feedback provided by the voltage feedback source of voltage gain β V/V. For instance, for an ideal current source with R_S = ∞ Ω, the voltage feedback has no influence, and for R_L = 0 Ω, there is zero load voltage, again disabling the feedback.

voltage divider rule (VDR)

What is the voltage divider rule?

The voltage divider rule is a simple way of determining the output voltage across one of two impedances connected in series. It is a useful tool for circuit analysis and design.

How is it used?

The voltage divider rule can be used with resistive, inductive, or capacitive circuit elements. It can also be used with AC or DC input sources. The equation for calculating the output voltage is different, however, depending on the type of circuit element. Following are the three general cases of two like elements connected in series:

Resistive divider:

The formula for determining the DC or AC output voltage of a resistive divider is:

Vout = Vin*R2/(R1+R2)
Example: In the following circuit, the output voltage would be: Vout = 9V*10K/(10K + 5K) = 6V

Inductive divider:

Inductive dividers can be used with AC input signals. A DC input voltage would split according to the relative resistances of the two inductors by using the resistive divider formula above. The formula for determining the AC output voltage of an inductive divider (provided the inductors are separate, i.e. not wound on the same core, and have no mutual inductance) is:

Vout = Vin*L2/(L1+L2)

Example: In the following circuit, the output voltage would be: Vout = 10VAC*50mH/(50mH + 100mH) = 3.33VAC. Note that the output voltage is not dependent on the input frequency. However, if the reactance of the inductors is not high at the frequency of operation (i.e. inductance not large enough), there will be a very large current drawn by the shunt element.

Capacitive divider:

Vout = Vin*C1/(C1+C2)

Example:

Pre amp Tone Control

1000W MOSFET CIRCUIT

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My only additional suggestion for you to think about it at length,
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Subwoofer filter Circuit

This week,I would like to presnt many amplifier and audio circuit.

I wrote this circuit and hold the PCB long since learned to write offline Protel99 PCB with new opportunities, but did not try to do more.
Circuit to solve new to the buffer input for the signal R + L and cutting area of the Muting issue and the cut-off point often fear for VR 4 floor hard so I divided into 2 layers using 2 the time that the cut frequency, if wanted. It has layers of fine cut to the same location.

I use ic tl072. And replace series capacitor 0.022 with 0.047 . That a soft voice that sounds good offline but not feeling heavy and dial around the back lot, try changing the capacitor. Beyond the traditional. Intended to adjust to extremely low frequency adjustment for the store before so I think that the device is used. Could differ. I sent in this condition before.

1500W INVERTER full schematics and pcb

Water lavel indicator

This circuit not only indicates the amount of water present in the overhead tank but also gives an alarm when the tank is full.
The circuit uses the widely available CD4066, bilateral switch CMOS IC to indicate the water level through LEDs.
When the water is empty the wires in the tank are open circuited and the 180K resistors pulls the switch low hence opening the switch and LEDs are OFF. As the water starts filling up, first the wire in the tank connected to S1 and the + supply are shorted by water. This closes the switch S1 and turns the LED1 ON. As the water continues to fill the tank, the LEDs2 , 3 and 4 light up gradually.
The no. of levels of indication can be increased to 8 if 2 CD4066 ICs are used in a similar fashion.
When the water is full, the base of the transistor BC148 is pulled high by the water and this saturates the transistor, turning the buzzer ON. The SPST switch has to be opened to turn the buzzer OFF.
Remember to turn the switch ON while pumping water otherwise the buzzer will not sound!