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We are working on a circuit in which we are controlling two way solenoid valve (24V, 10W). We are using 3.3V based controller. The solenoid valve is connected to the controller via TIP 122 (NPN Darlington) transistor.

When the solenoid valve was connected in the collector end of the TIP 122, it was working fine. When the valve was connected to the emitter of the transistor, it was not switching ON.

Why there is discrepancy, when connecting the valve on the emitter end. In the end the transistor has to turn-ON and has to allow current to pass between collector and emitter end.

Answer: When connecting load across emitter and the ground: The VBE required to maintain the forward bias will go away as voltage will drop across the load. If VBE is set at 0.7 V, and then as soon as the voltage across load becomes 0.7V transistor will no longer act as a switch. As their will not be potential difference across VBE. Th

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    \$\begingroup\$ Why do you imagine that the emitter side should work? Explain your thinking. \$\endgroup\$
    – jonk
    Commented May 31, 2020 at 17:58
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    \$\begingroup\$ Useful search term : "emitter follower". \$\endgroup\$
    – user16324
    Commented May 31, 2020 at 18:03
  • \$\begingroup\$ The transistor is controlled by the voltage between its terminals (particularly, between emitter and base). It doesn't know or care where you think Ground/Zero Volts is. \$\endgroup\$
    – Peter Bennett
    Commented May 31, 2020 at 19:20
  • \$\begingroup\$ Experimentally while connecting the solenoid valve to emitter side of the TIP122, it is not working. \$\endgroup\$
    – TapasX
    Commented Jun 1, 2020 at 2:14

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Let's consider what the difference between the two topologies is...

1. Transistor switch (common-emitter stage). When you connect the load (solenoid valve in this case) between the collector and positive rail, and apply input voltage (through a resistor) to the base, the base-emitter voltage will increase because the emitter is firmly fixed (grounded). The transistor begins decreasing its collector-emitter "resistance"... and accordingly, its collector-emitter voltage. Finally both its "resistance" and voltage become (almost) zero (as they say, the transistor saturates or, more figuratively, behaves as a piece of wire)... and the whole supply voltage is applied to the load.

2. Emitter follower (common-collector stage). If you connect the load between the emitter and ground, when applying the input voltage the transistor begins conducting and its collector current will create an increasing voltage drop across the emitter resistor. Figuratively speaking, the transistor will "lift" itself its emitter voltage until makes it (almost) equal to the input voltage... and stops. So the maximum final voltage applied to the load is (almost) equal to the input voltage. The name of this phenomenon is "following negative feedback".

Now you can continue to think in this direction to decide what to choose...

Just a note: I have written "almost" but in the case of TIP (Darlington transistor) there is a significant voltage drop (roughly 1.4 V) across the base and emitter since actually there are two junctions in series. You can estimate how much voltage will remain for the solenoid...

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  • \$\begingroup\$ Understanding from your answer. 1. For the first case voltage across collector and emitter will be applied across the load. 2. For the second case the voltage of the base emitter voltage will be applied across the load (which is very less), Hence point 1, make sense, which is load should be connected across the collector end of the transistor. \$\endgroup\$
    – TapasX
    Commented Jun 1, 2020 at 2:18
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    \$\begingroup\$ The solenoid has low coil resistance; so, it should be connected in series to the transistor - for the first case, between the collector and 24 V; for the second case, between the emitter and ground. Then: 1. For the first case, the whole 24 V voltage will be applied across the load. 2. For the second case, only the difference between the input voltage and base-emitter voltage will be applied across the load (which is only 3.3 - 1.4 = 1.9 V). Hence point 1 makes sense, which is load should be connected between the collector and 24 V supply rail. I have added more explanations to my answer. \$\endgroup\$
    – Circuit fantasist
    Commented Jun 1, 2020 at 4:12
  • \$\begingroup\$ Load should be connected between the collector and 24 V supply rail, this should be true for while giving PWM input to the base to the TIP 122 (Transistor) via resistor. \$\endgroup\$
    – TapasX
    Commented Jun 1, 2020 at 5:42
  • \$\begingroup\$ A controller output (port) should be used to drive the transistor. Also, connect a backward biased diode in parallel to the coil (against the self induced emf in the coil). \$\endgroup\$
    – Circuit fantasist
    Commented Jun 1, 2020 at 6:06
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    \$\begingroup\$ Thanks, now I completely got it. When connecting load across emitter and the ground: The VBE required to maintain the forward bias will go away as voltage will drop across the load. If VBE is set at 0.7 V, and then as soon as the voltage across load becomes 0.7V transistor will no longer act as a switch. As their will not be potential difference across VBE. Thanks Circuit fantasist (Cyril Mechkov) for giving correct reply. \$\endgroup\$
    – TapasX
    Commented Jan 2, 2021 at 18:01

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