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Voltage Regulator Circuits For Model Trains

DC Voltage Regulator (NPN & Zener Regulator)

There are instances where it is handy to have a regulated DC voltage power supply for  use on your model railroad layout.  In a previous section I used a 9.1 volt Zener Diode to develop a 9.1 vdc regulator.  However, this type of circuit which has many model railroad applications is limited by the current carrying capability of the zener diode. In this section this approach has been modified to include the addition of an NPN transistor which enhances the current output of the regulator.   See the figure below for the Zener NPN Transistor Voltage Regulator Circuit Diagram.

Assume we are feeding the above circuit with a variable supply which is adjustable from zero to +12 vdc.

The diode being used is a BZX79B9V1  500 milliwatt Zener Diode. In order to find out the current limiting resistor R1 we must the perform the following.

Set Variable DC Supply to + 12 vdc.

Remove Transistor Q1 from the circuit (nothing attached to zener or R1.)

Max Zener Power from Datasheet = 500 milliwatts

Maximum Zener Current = Max power/Zener Voltage = 500 mw/9.1 = 55 ma

R1 = Voltage accross R1/Max Zener Current = (12-9.1)/55 ma = 55 ohms

Let’s use R1 = 56 ohms for our circuit.

Now lets discuss the NPN Transistor (Radio Shack Part Number TIP31)

Maximum Specifications VCE = 1.2 volts;  VBE= 1.8 volts; IC= 3 amps; Power = 40 watts

With input set to + 12vdc, base to emmitter voltage  = 0.6 vdc, and the circuit output = 9 – 0.6 = + 8.5 vdc

Izt = 5 ma which represents the minmum turn-on current for the BZX79B9V1 Zener ( Reference datasheet on internet).

Since Izt = 5 ma, then the load will have 55 ma – 5 ma =50 ma

Remember that the load associated with the Zener is the NPN Transistor, and we must use the DC current gain between the base and collector,( hFE = 10 to 50 as per datasheet on internet for Radio Shack  Tip31 NPN Transistor)

In this instance the collector current will vary as follows as the input voltage to the overall circuit is changed from zero to +12 vdc.

Zener output of 9.1 vdc appears when minimum zener turn-on current Izt is reached.

For hFE = 10, Collector Current = 10 x 50 ma = 500 ma

For hFE = 50, Collector Current = 50 x 50 ma = 2500 ma = 2.5A

The NPN will dissipate 2.5A x (12-9.1) = 2.5 x 2.9 = 7.25 watts which is well below the 40 watts maxium spec.

Negative DC Voltage Regulator (PNP & Zener Regulator Circuit)

The PNP Regulator will use a negative variable input power supply which will yield a negative regulated power supply. The Zener Diode must also be reversed.

Now we are feeding the above circuit with a variable supply which is adjustable from zero to – 12 vdc.

The diode being used is still a BZX79B9V1  500 milliwatt Zener Diode. In order to find out the current limiting resistor R1 we must the perform the following.

Set Variable DC Supply zero  to – 12 vdc.

Remove Transistor Q1 from the circuit (nothing attached to zener or R1.)

Max Zener Power from Datasheet = 500 milliwatts

Maximum Zener Current = Max power/Zener Voltage = 500 mw/9.1 = 55 ma

R1 = Voltage across R1/Max Zener Current = (12-9.1)/55 ma = 55 ohms

Now lets discuss the PNP Transistor (Radio Shack Part Number TIP42)

Maximum Specifications VCE = 1.5 volts;  VBE= 5.0 volts; IC= 6 amps; Power = 65 watts

With input set to – 12vdc, the zener voltage = – 9.1 vdc, and the circuit output = (-9.1 – 0.6) = negative  9.7  vdc

Izt = 5 ma which represents the minimum turn-on current for the BZX79B9V1 Zener ( Reference datasheet on internet).

Since Izt = 5 ma, then the load will have 55 ma – 5 ma =50 ma

Remember that the load associated with the Zener is the PNP Transistor, and we must use the DC current gain between the base and collector,( hFE = 15 to 75 as per datasheet on internet for Tip42)

In this instance the collector current will vary as follows as the input voltage to the overall circuit is changed from zero to -12 vdc.

Output of neg 9.1 vdc appears at zener when minimum zener turn-on current Izt is reached.

For hFE = 15, Collector Current = 15 x 50 ma = 750 ma

For hFE = 75, Collector Current = 75 x 50 ma = 3750 ma =3.75A

The PNP will dissipate 7.5A x (12-9.1) = 7.5 x 2.9 = 21.75 watts which is well below the 65 watts maxium spec.

If you wish to acquire a better understanding of Electronics Theory, I suggest you go to  the following  link: Electronics

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