Discussion in 'General Electronics Chat' started by electronics1, Sep 22, 2009.
Practically the circuit enables voltage on relay (thus turns it on) when Vin=5V, and disables it when Vin<Vbe~0.7V , where Vbe is base-emitter voltage drop, which is equal to voltage drop on a diode.
The transistor can be only in two states: cut-off for Vin<~0.7V and feed forward(if I remember correctly the name of the state) for Vin=5V.
The SOL relay resistivity in DC is calculated form its nominal values: RSOL=Vsol^2/Psol=12^2/24m=6Kohm. When Vin=5V the voltage drop on SOL relay is Vcc-Vcesat=12-0.2=11.8V, thus the current throuh the relay is Isol=11.8/6K~1.9mA. Then the base current is Ibase=Isol/B=1.9mA/110~0.173mA. The maximum Rb resitance in this case is Rb_max=(Vin-Vbe)/Ibase=(5-0.7)/0.173mA=248.95Kohm.
The diode role here is to get on itself the negative polarity voltage burst coming from the SOL relay (which is actually an inductor) the moment it is powered off. In this way the diode protects the transistor and power supply from such burst. During most time it is just in cut-off.
Is this explanation clear enough?
It is a relay with a driver transistor.
It could be a wide variety of transistors. Generally, you would want to select a transistor that was capable of twice the necessary collector current.
Rb = (voltage_of_signal_driving_the_base - 0.7v) / (Ic x 0.1). Not exact, but is close enough.
The relay is dissipating 24mW? In that case, current is P/E = .024/12 = 2mA - doesn't sound quite right for a relay coil.
If you mean the transistor's beta is 110 - you don't use the rated beta for transistor saturation. Use 1/10 collector current.
If it's the transistor that's dissipating the 24mW, then Ic is roughly 110mA, and Ib is roughly 10mA. So, (5v-0.7v)/.010a = 430 Ohms.
Many will work. 2N2222 would be a likely choice.
If the diode is not in the circuit, the relay coil will have a very high reverse voltage spike when the transistor turns off. This will likely destroy the transistor.