TIP2955 | Electronic Project Circuits

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A Simple Voltage Regulator +22V -22V by TIP3055,TIP2955

The TIP2955/TIP3055 transistors should be satisfactory for a power supply feeding a single amplifier. If two amplifiers are to be fed from a single power supply, these transistors should be changed to higher power devices such as the MJ2955/2N3055. Adequate heat-sinking must be provided for whichever transistors are used.

R9 and R10 can be replaced with a 10k preset potentiometer (or a 5k potentiometer in series with a 5k6 fixed resistor) to provide adjustment of the output voltage.

As with the capacitance multiplier circuit, Q2 and Q4 can be changed to a complimentary feedback pair arrangement, if required, to allow the use 2N3055s as the pass device in both halves of the supply (see the capacitance multiplier page for details). If this done, R9 and R10 must be made variable to allow the supply rails to be set to equal (but opposite) voltages.

Zener diodes of a different voltage rating can be used for ZD1 and ZD2, but the value of R9 and R10 will need to be adjusted to maintain the +/-22V output.

For different output voltages or a different Zener diode voltage, the output voltage can be calculated from the following equations:

+VOUT = ((R7 + R9) / R9) * (VZ + 0.6)

â€“VOUT = ((R8 + R10) / R10) * (VZ + 0.6)

where +VOUT and VOUT are the required supply rail voltages and VZ is the Zener voltage.

Read More Source:http://www.tcaas.btinternet.co.uk/jlhvoltreg.htm
Thank you.

Be the first to comment - What do you think? Posted by admin - December 25, 2007 at 5:41 pm

Categories: Power supply Tags: dc voltage regulator, dual power supply, TIP2955, TIP3055

7824 Single Power Supply 25V 2A with IC 7824 and TIP2955

This circuit uses a PNP Power Transistor TIP2955, you can use any other according to your current and voltage requirement. Look at R2 a 10 Ohm resistor, when the current in your load to the power supply is less than 70mA the voltage across R2 is less than 10E * 70mA = 700mV right.

The base emitter junction of Q1 will be biased or turned on around 700mV, less than 700mV the transistor just does nothing. When the current in your load goes over 70mA the voltage across R2 goes above 700mV and a small base current Ib flows from emitter to base of Q1 turning on the transistor. Now a collector current Ic flows from emitter to collector and then to your load supplying the excess demand. The Ic = Ib * hfe where hfe or beta is the DC gain value.

Some transistors will have only AC gain specified which is lower than DC gain. TIP2955 has a gain of 20 so for an Ib of 50mA the Ic will be 1 Amp which saves the regulator from heating up or shutting down as the main current flows thru the transistor. Q1 should be provided with a good heatsink.

Source:http://schematics.blogspot.com/2004/10/single-power-supply.html