Switched by Transistors

A switch is an electrical or electromechanical component that can interrupt the current running through an electric circuit. An electromechanical device consists at least of two electrical contacts which can be either separated from each other, or touching each other. In the first case the switch is called to be "open" while it is "closed" in the second one. Ideally the resistance in the state "open" is infinite, while it is zero in the state "closed". At the previous video we have learned that the resistance of a bipolar junction transistor can be altered by attaching a voltage to the base pin. The resistance is high if 0V are attached to the base pin and it becomes low if those voltage increases above +0.6V at NPN types. A transistor can conceptually open or close electric circuits by altering the emitter collector resistance from it's minimum value to it's maximum value or vice versa. Let's have a closer look at the status "open": The most simple output circuit consists of the transistor and a load resistor - 1Megaohm are used here. The potentiometer at the input circuit is turned to it's minimum value, by what the emitter-base line gets bypassed, hence there are 0V attached to the base pin. The emitter-collector line and the load resistor are forming a voltage divider. The measured voltage drop at the load resistor is 0V, the total voltage is 12.15V and the voltage drop at the emitter-collector line is 6.07V - why? If you have watched my video about digital multimeters, you should remember the inner resistance of those devices. The multimeter used here has an internal resistance of 20Megaohm, by what there is a clear shift of the resistance ratio at the voltage divider formed by the transistor and the load as soon as the device is connected to the circuit. Hence the measured sum of the voltage drop at the load resistor and those at the transistor equals not the total voltage. The higher the resistance of the discovered device, the higher the resulting measuring error. Apparently the resistance of the transistor is clearly higher than the inner resistance of the multimeter. Even by using the multimeter itself as a 20Megaohm load resistor, no voltage drop can be detected. The condition "open" is met nearly perfectly by the used transistor while attaching 0V to the base pin. The second switching status to be discovered is "closed". By using a 1kiloohm load resistor and a LED as output circuit, we can easily observe the switching status of the transistor. While increasing the base voltage, the LED starts "glimming" around 0.6V, it is shining brightly at 0.65V and there is no more increase in luminosity above 0.67V. To turn the LED on, a base voltage of 0.67V has to be attached to the base pin. Let's discover the resistance of the transistors emitter-collector line. The voltage drop at the 1kiloohm load resistor is 9.85V, by what a current of 9.85mA is running through the circuit. The voltage drop at the transistor is 0.106V, by what the resulting resistance is 10.8Ohm. Let's replace the LED and the resistor by an electric motor. The base voltage is still 0.67V, but the motor doesn't turn - why? Well, the ohmic resistance of the electric motor is clearly lower than those of the LED and the 1kiloohm resistor, by what the voltage drop at the transistor increased above 11V (just remember the correlations in a voltage divider). The base voltage has to be increased above the 0.67V, to lower the resistance of the emitter-collector line. While it is around 0.74V, the engine is running with it's maximum idle speed. Now we get a resistance of just 1.3Ohm for the emitter-collector line. While remembering the functionality of an electric motor, you should know that the current running through the windings of the inductors increases clearly if the motor gets locked. Now we get 240mA for the current running through the circuit and approximately 7V for the voltage drop at the transistor. The resulting resistance is 30Ohm. Despite the constant base voltage, the resistance of the emitter-collector line increased. The emitter-collector line is no ohmic resistor - there is no linear correlation between current and voltage. The electric power dissipated by the transistor is 1.8W, by what the device is heating up noticeable. Like mentioned before, the resistance of a perfect switch is zero while it is closed, but what's the minimum value of a transistor? At the datasheet of the BD135 there is no minimum resistance listed, but the maximum continuous base current is 500mA at a base voltage of 1V. Thereby the emitter voltage is 2V, while the maximum continuous collector current is 1.5A, by what we get a resulting minimal resistance of 1.3Ohm. Considering those extreme values, 500mW of input power are required to safely turn on a load with a low resistance. Keep in mind that the resistance of a transistor get's low while it is turned on, but it is never zero! A voltage divider is rarely used at the input circuit of bipolar junction transistors. Normally there is just a single resistor switched in series to the base pin. Those resistor is forming the accordant voltage divider with the forward biased junction between base and emitter. The resistance value must meet the specifications of the transistor and the voltage output of the input signal. If a base current of 50mA is required to safely turn your transistor "on" and your input signal has a voltage output of 5V, the resulting resistance value is 100Ω when neglecting the resistance of the emitter base junction which is low. Keep the input power in mind which has to be provided by your signal source 220mW of power are dissipated by the resistor, 39mW by the transistors emitter-base line. If your input circuit doesn't provide such a high power, a Darlington transistor might be useful. Even less input power is required to keep the switching state of a field-effect transistor. A normally insulated MOSFET is "open" if the voltage drop between source and gate pin is 0V. The resistance is decreasing with increasing gate voltage, by what the characteristics are similar to those of a bipolar junction transistor, however the gate voltage is usually higher than the base voltage. The maximum gate voltage of the IRLZ24N, an n-channel MOSFET, is 16V, which would blow up all bipolar junction transistors. The minimum resistance of MOSFETs is often listed at the datasheet. The static drain-source resistance of the IRLZ24N is 60milliohm at a gate voltage of 10V and a drain current of 11A, respectively 105milliohm at a gate voltage of 4V and a drain current of 9A. Without any problem our electric motor can be switched by those transistor. There is no internal conductive path from the gate area to the source or the drain like inside of a bipolar junction transistor. The charge cumulated at the gate area can't flow off. To ensure that our N-channel MOSFET is turned "off" while the gate pin is not connected to an input circuit, a resistor has to be connected between gate and source, pulling the voltage level to ground. Those resistor is called pull-down resistor. If a resistor is connected between gate and the positive terminal of the supply voltage, our n-channel MOSFET is turned "on" while it is not connected to an input circuit. This is a so called pull-up resistor. Note that the supply voltage should not exceed the maximum gate voltage, or else a voltage divider must be used to lower the gate voltage caused by the pull-up resistor. There is a reason why I am not using an electric motor to demonstrate the functionality of the pull-up resistor. Some later we will see why. Always use high resistance devices as pull-up respectively pull-down resistors to keep the input power low, because now a current is running trough the resistor even while the switching status of the field-effect transistor is not altered. Switching transistors at their maximum frequency is tricky. Altering the state of a transistor always takes some time. When turning a transistor "on", the voltage drop at the emitter collector line drops from (nearly) the supply voltage of the circuit to (nearly) 0V. Vice versa the voltage is increasing from 0V to the value of the supply voltage while the transistor is turned "off". The maximum rate of change of a signal, hence the voltage drop is called slew rate. It is usually expressed in units of V/μs. Sometimes the rise time is listed at the datasheets of transistors instead of the slew rate. At the oscillograph shown here you can see the input voltage as yellow curve and the accordant output signal between the source drain line of an IRLZ24N as green curve. A 1kΩ resistor is switched in series to the gate pin simulating an input signal with a high inner resistance resulting in a worse slew rate of the MOSFET. There is always some kind of feedback between the signal source and the amplifying circuit like demonstrated at the previous video. Just remember the capacitance of a MOSFETs gate area! The power output of the input signal has to be sufficient to safely turn the transistor "on". Let's have a closer look at the power dissipated by the emitter collector line while the state alters from "on" to "off", meaning the resistance of the emitter collector line rises from it's minimum to its maximum value. While considering the correlations inside of a voltage divider, the power dissipated by the transistor can be calculated with knowing the total voltage and the constant resistance of the load resistor. The maximum collector current of the BD135 is 1.5A, the maximum emitter collector voltage is 45V and the maximum power dissipation is 8W. So it looks like there is no problem in switching a 30Ω load connected to a 40V power supply. While altering the switching state from on to off, the resistance of the transistor is around 10Ω for a short span of time. By inserting those values at the equation above we get 10W! The transistor is heating up clearly while operating with high switching frequencies. Finally some remarks on switching non ohmic loads. Here you can see the voltage progression between source and drain while switching an electric motor. As you can see, the voltage drop at the transistors source-drain line exceeds the supply voltage of 12V clearly. The peak voltage is above 25V! The maximum source-drain voltage of the transistor used for the switching operation must be designed for those voltage peaks. While turning on capacitive loads, the current climbs to it's peak value, so keep the maximum collector current in mind. In this spirit - be careful while switching your own circuits.