This circuit is the first one of the series that I'm about to upload the next days, based on the BJT Transistor Theory TEDIE Book. I need your participation into this work. I need you to subscribe tot he forum or start a conversation in Tweeter and tell me your suggestions for the next circuits or what an amateur really wants to learn regarding transistors. After all, this is the point of the TEDIE Book. The next book will be full with practical circuit examples using transistors, based on your suggestions.
This is the most typical application of transistors in digital designs. A DC Load Switching Transistor is used to control the power of a load such as a solenoid, a motor, a lamp etc. IC Chips are usually not capable to provide enough power for such loads in terms of both voltage and current. Take for example a DC relay. A typical 12V PCB relay requires about 20mA of current for its solenoid. A typical 5V relay requires about 45mA. As you can see, a microcontroller or a CMOS chip cannot provide the required power for these relays.
The solution is a transistor driver. The purpose of this transistor driver is to amplify the CHIP's (microcontroller for example) current or voltage (or both) and provide enough power for the load (relay).
For this application, the best transistor connection is a Common Emitter amplifier with Fixed Bias, as explained in the BJT Transistor Theory TEDIE Book in page 17.
First of, the circuit. It can't be any simpler:
We want to control a 24V relay from the output of a microcontroller (or any other low-voltage CHIP). For the sake of simplicity we substitute the microcontroller output with a 5V voltage source and a pushbutton (PB). So, how does a designer thinks in order to analyze and build this switch?
Step 1: Learn your load
I = V / R => I = 24 / 1440 => I = 0.0166 A = 16.6mA
Some datasheet provide only the solenoid power. No problem! We can calculate the current using the voltage (24V) and the power of the solenoid. Our relay consumes 400mW of power:
I = P / V => I = 400 / 24 => I = 16.6 mA
Step 2: Select the transistor
There are thousands of different transistors to select from. So, we need to narrow our selection range. Let's enumerate first the parameters that our transistor has to satisfy:
Still the selection range is huge. Here comes experience to give a solution. There are typical low-cost transistor used for such low-current low-voltage switching applications like BC337, BC338, BC548, BC549, BC550 and the list goes on. You can have a stock of cheap switching NPN transistors and use them for your circuits. We can safely choose a BC337 for this application. Let's take a quick look at the transistor's characteristics (click to enlarge):
First, the collector current. We want to provide 16.1mA. generally, always overpower this current requirement. A 20% is a safe number, so we want our transistor to be able to provide about 20mA. As you see, the BC337 is able to provide up to 800mA (IC) which is more than enough. As for the voltage, we will power the transistor with 24 volts. Again overpower this value by 20%. The BC337 can switch up to 45 volts (VCEO) which again more than enough.
One note: There are 2 values for the VCE, the VCES and the VCEO. Both of these parameters refer to the maximum voltage across the collector-emitter leads with the emitter being grounded. The difference is that VCES is measured when the emitter and the base are short-circuited (when the base is driven to ground like the emitter) and VCEO is measured when the base is open. It is obvious that we use the VCEO parameter for our calculations.
Step 3: Calculate the base resistor RB
We can now do some simple math to calculate the base resistor. To do so, we start from the current that we need to provide. The relay requires at least 16.6mA but since our transistor can provide up to 800mA, we can safely double the provided current. So we will run our calculations for 35mA collector current.
In chapter 2.2.1 page 17 of the BJT Transistor Theory TEDIE Book we find this formula:
IC = IB x hFE
Solving for IB we can calculate the base current as a result of the collector current:
IB = IC / hFE
So, the base current is calculated by dividing the collector current by the current gain hybrid parameter. We've talked about this parameter in the first chapter of the theory book. Each transistor has the hFE range indicated in the datasheet. Here is the datasheet entry for the BC337 (click to enlarge):
For IC=100mA the current gain varies from 100 to 630! If you've read the theory book, you already know that the hybrid parameters are not stable. For our calculations we will use the worst case scenario which corresponds to the lowest current gain, and that will be 100. So:
IB = IC / hFE => IB = 35 / 100 => IB = 0.35 mA
So we will calculate our resistor so that it provides at least 0.35mA (350uA) into the base of the transistor. Referring again to the theory book page 17, we find this formula:
IB = (VCC - VBE) / RB
Solving for RB
RB = (VCC - VBE) / IB
From the theory (page 6) we know that VBE = 0.7 Volts for a silicon transistor like the BC337. As for VCC in our example this is the voltage provided by the microcontroller and it is 5 volts. Putting it all together we have:
RB = (VCC - VBE) / IB = (5 - 0.7) / 0.35 => RB = 12.285 Ohms
PAY EXTREME ATTENTION TO THE UNITS!!! We use the number 0.35 mA for IB but we should have used a number measured in Amperes instead. So, it should rather be 0.00035 Amperes. But having all these zeroes in a number is quite confusing and not recommended. It is OK to use a sub-division like mA or uA, but do NOT forget to multiply (or divide) the result by the proper power of 10. In our example we use mA, so the end result must be multiplied by 1000!
So, from the previous results we come up with a 12 KOhms base resistor. And that's about it!
What about the power dissipation on the transistor?
Good question. How much heat is called the transistor to dissipate during the ON cycle? Well, practically the power being dissipated is zero. Referring to chapter 5.5 "Calculating the Power Dissipation of the Transistor" in the theory book, we dig this formula:
PD = VCEQ ICQ
When the switch is OFF, the DC current through the base is zero, so the current through the collector (ICQ) is zero as well, so the power dissipation is zero. The transistor operates in the Cut-Off Area (refer to theory book 3.1.2 "Region 2: The Cut-Off Area").
When the transistor is ON, most (if not all) the voltage is dropped on the load which is our relay solenoid. That being said, the voltage across the collector-emitter leads is very close to zero. So the power dissipation is again close to zero since VCEQ is very small. In that case, the transistor operates in the Saturation Area (refer to theory book 3.1.1 "Region 1: The Saturation Area").
So, when the transistor operates as a DC switch, the Q point operates either in the Cut-Off Area (OFF) or in the Saturation Area (ON). The power dissipation in both cases is a negligible value for collector currents less than 100mA.
Go to the next page (click here) to see this circuit prototyped on a breadboard for measurements and calculations.
Coming up next:
Within the next days i will upload:
- A PNP low-activated switch
- A video presentation
So stay tuned