# Example Transistor Circuits

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One of the most commonly used circuits is that of the VOLTAGE REGULATOR.

The simplest voltage regulator uses just a resistor and a zener diode. In the circuit diagram you can see a resistor (R1) and a zener diode (ZD1) connected across a power supply. The resistor is connected to the positive (+ve) supply wire and the zener diode anode is connected to the zero volt (ground) wire. At the junction of these two components the voltage is clamped by the zener diode to its specified voltage - in this case 5.6 volts.

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| This method is OK for low currents but the resistor becomes too hot if larger currents are needed. To cope with this problem we can add the NPN transistor (Q1). Now the transistor passes the current required at the output. What is the output voltage? |

It is easy to calculate. The voltage at Q1 base connection is 5.6 volts.

The voltage between base and emitter of a silicon transistor is always 0.6 volts if the transistor is "on".

So the voltage at the Q1 emitter (Vout) must be 5.6 - 0.6 = 5.0 volts. The output voltage will remain at a constant voltage of 5.0 volts provided that the input voltage from the supply is more than 6 volts (the zener voltage plus a little to compensate for that "lost" across the resistor).

In fact the input voltage can be swinging up and down between, say, 6 volts and 12 volts and the output voltage at Q1 emitter will still be a steady 5.0 volts.

The limiting factors are the amount of heat generated by R1, ZD1 and Q1 since all excess voltage must be shed as heat. The "wattage" ratings of the individual components must be calculated to suit:

1. The average input current (through R1 and ZD1) and

2. the output current (through Q1).

1. can be calculated from Ohms Law and

2. is decided by whatever the regulator is to supply voltage to.

Ohms Law **I = V/R** | V = Volts I = Amps if R = Ohms or I = mA if R = k½ |

Let's assume the following:

The circuit which this regulator is driving needs 5.0v at a current of 100mA.

A BC337 transistor is suitable since it can handle current up to 800mA.

Its gain at 100mA is listed as 100 (minimum) so it's easy to see that it will need at least 1mA into its base to allow 100mA to flow from collector to emitter.

For the zener diode let's choose a BZX55C5V6. This will need a minimum of 10mA of current to produce a stable voltage. So Q1 requires 1mA, ZD1 requires 10mA, making a total of 11mA through R1.

If the minimum supply voltage is, say, 7.8v then the minimum voltage across R1 is 2.2v.

Ohms law says that the resistance = V/I

= 2.2/11

= 0.2k resistance

= 200R

Suppose the maximum supply voltage might be 9.6v.

Then the maximum voltage across R1 will be 9.6 - 5.6 = 4.0v.

From Ohms Law, the current through R1 will now be V/R

= 4.0/200

= 0.02A

= 20mA

Watts = Volts x Amps | milliWatts = Volts x milliAmps |

Watts = Volts x Amps so the minimum Wattage of R1 must be

4.0 x 0.02 = 0.08W - not a lot!

A standard 0.25 Watt resistor will be more than adequate for R1. Let's check the zener diode rating under the worst conditions:

The voltage across ZD1 will still be 5.6v

The current in the worst case will be 20mA, assuming none goes through Q1.

So the Wattage of ZD1 must be at least 5.6 x 0.02 = 0.112W

= 112mW

A BZX zener diode will dissipate up to 500mW so the circuit is safe.

T0 provide **5 volts** at up to **100mA**, the final design will use:

R1 = 200 Ohms, 0.25W (220 Ohms will do)

ZD1 = BZX55C5V6

Q1 = BC337

| The first circuit can be improved upon to provide more current. All we need to do is to add a second transistor (which has a higher rating to handle the extra current) and change the zener diode to clamp at 6.2 volts in order to compensate for the b-e voltage of BOTH transistors. 6.2 - 0.6 - 0.6 = 5.0 volts |

If the first transistor provides 5.6 volts at 100mA (0.1 Amps) and the gain of the second transistor is 50 then it can provide **5.0 volts** at 0.1 x 50 = **5 Amps**. Be sure to use a transistor rated at 5 Amps or more! The final design can use:

R1 = 200 Ohms, 0.25W (220 Ohms will do)

ZD1 = BZX55C6V2

Q1 = BC337

Q2 = 2N3055 (on a suitable heat sink)

Note:

The combined gain of both transistors is 100 x 50 = 5000. We could not use just one transistor because no ordinary transistor capable of handling 5 Amps can have a gain of 5000. We could, however, use a "Darlington Pair" transistor which has two transistors (as above) in one package with just three wires.

| **Constant current generator used to charge a NiCad battery** This simple circuit can be used to provide a constant current, regardless of the input voltage. Again, the results are simple to calculate by using Ohm's Law. Suppose we have a NiCad battery of 9 volts and we need to charge it with a current of 40mA. |

We connect a transistor and a zener diode as shown. It's easy to see that the voltage across resistor R2 will be 5 volts. So to get 40 mA to flow through R2 its resistance must be ... what?

Ohms Law **I = V/R** | V = Volts I = Amps if R = Ohms or I = mA if R = k½ |

So 40 = 5/R

R = 5/40 k½

R = 0.125 k½

R = 125 ½

If 40 mA is flowing through R2 then most of that is coming via the collector-emitter junction of the transistor and must, therefore, be flowing through the battery. In practice, a little of the emitter current is also coming via R1 and the base-emitter junction. We can compensate for that by reducing the value of R so lets call it **R = 120 ½** which happens to be the nearest standard value. The current through R is now approximately 41mA and the current through the battery is approximately 40mA.

**What must the supply voltage be?**

The supply voltage must be at least the sum of the voltages across the battery (9v) and across R (5v) and across the c-e junction of Q1 (probably less than 1 volt). So the minimum supply voltage has to be 9 + 5 + 1 = 15 volts.

You can use this simple method to design a constant current charger for any NiCad battery. Just make sure that the transistor current rating (Ic) is higher than the value of the required charging current. The previous notes about the zener diode and resistor R1 still apply.

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