Power supply 5 volts 5 A with a pass transistorBy using 7805 components, it can easily be made a series of excellent power supply output voltage regulation. However, the component 7805 can only be effectively distribute the current to 1 A only. 5 volt power supply is generally used to distribute many different applications, so it sometimes is not enough supply current 1A.
In this article presents the design electroniclab regulated 5 volt power supply that can supply enough current to 5 A, at least it is ever tested in the workshop electroniclab. Actually, this circuit could distribute up to 10 A or even more if the reader know tips to modify it.
The core of this circuit certainly is a basic circuit with a 7805 5 volt regulator. The difference is, the series pass transistor is added to the circuit consisting of transistors Q1 and 2 pieces of resistors R1 and R2. 7805 components in control regulates the output voltage, and a series pass transistor is essential to drain the remaining current to the load RL.
The transistors used are the MJ2955 PNP transistor. Transistor is known as silicon bipolar power transistors are often found in the market. Readers can in principle be replace it with other bipolar power transistor, provided with similar characteristics. From the datasheet, it is known that the category of transistor power transistor collector current Ic as 15A can be achieved with power dissipation that can reach 115 watts. Of course in designing a series of maximum limits should be aware of this, so it does not exceed the optimum can be achieved.
Note the series of images above. In the closed-loop current through the resistor R1, R2 and the emitter-base transistor Q1, can be formulated mathematically:
I1R1 = IeR2 + VBE (on) â € | â € | â € | (1)
For silicon transistors usually VBE (on) = 0.7 volts, the base-emitter voltage causes transistor starts working (ON). This voltage is known from the datasheet VBE (on) this can vary between 0.6 ~ 1.4 volts depending on which of the current Ic through the transistor. But for the simplification of the calculation, we set the course VBE (on) = 0.7 volts.
I1 is the current through the 7805 onwards will supply the load RL. With this series we will be setting a large current through 7805, for example if you set current I1 = 500 mA. So how to supply current to the load RL to 5A? Of course the rest would flow MJ2955 is passed through the transistor. Of formula (1) it is understandable that the current passing through R2 Ie will begin to flow only when the voltage across the resistor R1 flops greater than VBE (on) or mathematically:
I1R1> = VBE (on) â € |. (2)
If the magnitude of the above disubsitusikan to formula (2) of the R1 can be calculated that is required is:
R1 = VBE (0n) / I1 = 0.7/0.5 = 1.4 Ohm
How to set a large current I1 = 500 mA, may be whether more or less. If we trace back a little, we first want to make a power supply with Io = 5 A. In the above circuit, Io = Ic + IOA € ™. If we consider the IOA € ™ is quite small compared to Ic, then Ic = Io can be written. Of transistor known theory that Ic = Hfe Ib. MJ2955 datasheet from Hfe big unknown is 20 ~ 70. You can search for a transistor with Hfe = 50. If this is used, then that must be supplied base current is Ib = Ic / Hfe = 5/50 = 100 mA. With this calculation is not wrong if it is assumed to be masksimum current of 500 mA through R1. Because it will supply enough base current Ib (at 100 mA) required to supply current Ic transistor Q1 to 5 A.
Great resistance R2 can be calculated from Vin to Vout loop through the transistor Q1 is defined by:
Vin = IeR2 + VCE (on) + Vout â € |. (3)
Vin is the voltage output of the rectifier circuit made of a series transformer, diode bridge and capacitor elco. If for example Vin = 7 volts and the output voltage Vout = 5 volts, then the formula (3) can ditullis be:
+ 7 = IeR2 VCE (on) + 5
or
IeR2 + VCE (on) = 2 volts â € | .. (4)
This is the line or lines of work load transistor Q1. Assuming that Ie = Ic = 5 A and VCE (on) = 0 volts (ideal) when the transistor Q1 is working (ON), then it can be calculated of R2 = 2/5 = 0.4 Ohm. Finish â € |? of course not, because it must be specified wattage of the resistor is large. Of the general formula P = I2R power dissipation can be calculated on the resistor R2 is P = 52 (0.4) = 10 watts (minimum), then used the 0.4 Ohm 20 watt resistor to be safe.
Thus the sequence of this power supply design. Of course this design can be modified as needed. For tips last, With such a large current, temperature resistors and transistors will be so hot. Highly recommended to use a heatsink for the transistor Q1 and also R2 resitor. 7805 component should not require a heatsink, because the current through this component is relatively small. Elco capacitor C1 is the recommendation of the 7805 datasheet for more stable output voltage.
To the needs of larger currents, transistor Q1 can be replaced with a Darlington transistor, or a way to pass transistor cascade circuit into 2 or 3 level. In principle, the above calculations can also be applied to other power supply circuit design such as 12 volts or 24 volts.
