Vu Meter For Power Amplifiers

One yet circuit with the known LM3915. It does not differ in a lot of points from other applications with same IC. The circuit accosted in those who they want a Vumeter that will be connected in the exit of power amplifier. It can be adapted his sensitivity so as to it works with amplifiers that have different out power, enough we change the R1 price of according to Table 1. The price that will be found if it does not correspond in standard resistance price, it should selected the next standard price or if you want the biggest precision it should you put resistances in series or parallel so that you achieve the price. Can use various types Leds round or square, so that you take the optical and aesthetic result that you want. With switch S1 selects if the Vumeter works as bar or dot. In the place ON [switch closed] the operation is bar and in place OFF [switch open] the Led operation is Dot. In the place Bar the consumption goes up, because work all the Led and it can reach until 150mA. For amplifier with two channels obviously it’s that it will be supposed made two same circuits, one for each channel. The voltage circuit supply is +12Volts. The reception of this supply should become from the existing power supply of power amplifier. Usually the power amplifiers work with supply bigger than +12Volts that it needs the circuit. For this reason added one stage of voltage +Vp attenuation in +12Volts. This becomes with the stage that is found in the discontinued line and with the IC2. That is a regulated stabilizer. The use of small heatsink is essential because the voltage differences of input - output are big so that we have growth of high levels of temperature. The use of R5 helps in the fall of voltage so that goes down the voltage in the IC2 input in lower levels. The calculation of this resistance price becomes empirically using the Ohm law. The voltage in the IC2 input should it is bigger than +16Volts. For example if the voltage of power amplifier is for example +50Volts it will be supposed we have voltage fall 50-16=34 Volts above in R5. For current with mean 50mA that it wants circuit [can reach until 150mA] the price of R5= V/I=34V/0.05A=680 ohms 2W. Perhaps it needs boost or decrease this price after tests. Because the resistor heat good is it placed in some distance from PCB. Her regulation process of IC2 with the TR1 good is it becomes first without exists the IC1. For this reason it should it’s placed in a base. If you have benefit possibility of continuous voltage +12V from some point of power amplifier, obvious is to leave out the R5, the IC2 and the components that are found in the discontinued line.

Calculation For R1
R1=Vout-10 in Kohms rounding in the nearest standart value

Example
For 200W / 8 ohms => (R1=40-10=30K [33K])
For 100 W/ 4 ohms => (R1=20-10=10K)

Part List
R1 = See Text and Picture1
R2 = 10 KΩ
R3 = 1.2 KΩ
R4 = 8.2 KΩ
R5 = See Text [2W]
R6 = 270 Ω
C1 = 100 μF/25V
C2,C3,C4 = 100 nF/100V MKT 5%
C5 = 22 μF/25V
D1 = 1N4002
D2…D8 = Led Green 5mm
D9 = Led Yellow 5mm
D10 = Led Red 5mm
S1 = 2 dip Switch
J1 = 3 pin 2.54 mm step connector

Sam Electronic Circuits

http://users.otenet.gr/~athsam/index.htm

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