The general rule of thumb for this condition, called saturation, is a gain of 10 to 20, where you've got a choice of exactly how optimistic you want to be. The most important is that, when the transistor is being operated with very low Vce (less than a volt, typically), its gain drops significantly. This is perfectly possible, but there are limits. In addition to this obvious issue, keep in mind that the power dissipated in the transistor is the product of the voltage (Vce) and the current (mostly collector current). This means that the voltage across the transistor (Vce) must be as low as possible, and certainly less than 1 volt. Put it this way - in order to get much current (and therefor much torque or power) from the motor, you need the voltage to be as close to 5 volts as possible. The circuit you're looking at is perfectly fine in principle, but it cannot be used for any but the smallest motors. The current \$I_ V_F\$, which is safe for the transistor.Īs a slight aside, I thought I should amplify Tony Stewart's comment. 1), the supply voltage is applied to the inductor L and it begins charging. So, when the transistor T is turned on (Fig. The key of intuitive understanding inductive circuits is to think of the inductor as of a "rechargeable current source". I have considered the similar but simpler configuration with an inductor (e.g., a relay coil) but it can be applied to the motor as well.ġ. I know this from my personal experience that is why, I have visualized these invisible electrical quantities in the pictures below by voltage bars (in red) and current loops (in green). To understand this trick, a beginner needs to imagine what the voltages are (magnitude and polarity), and where the currents flow (direction and path). Really it's just a convention so that all the signs work out. When talking to someone, current always flows from high potential ( ) to low potential (-), even though the electrons are going the opposite direction. The current for the motor comes from the 5V rail. Only about 330uA (closer probably to 270uA.I haven't typed any numbers into a calculator) will flow from Pin 9 into ground through the transistor. Very little current (uA or mA) can be used to control a motor (~200mA with that transistor from 5V through the motor). That is incorrect, see the next question. NOTE: Your question says "Pin 9 power source". Usually you'll also put a capacitor on the 5V rail to further absorb the spikes (otherwise you'll get noise everywhere). You'll also find these diodes across relay coils. The diode provides a path for this energy so that it can be dissipated into the 5V rail rather than going somewhere else less predictable. As the motor spins, the coil will be turned on and off with the commutator inside the motor which will cause voltage spikes. A motor consists of a coil of wire which is effectively an inductor (and an electromagnet). The diode in this configuration is called a "flyback" diode.
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