Transistor in pull-down "switch" application

Question

What is the transistor circuit which acts as a SPST switch given: 1) +24vdc, 2) ground, 3) a "floating" pin that the switch takes to ground when triggered (there is no "load")? I can use either +24vdc, ground, or the closing of a dry contact as a trigger -- whichever is simplest to implement the transistor switch described.

Answer 1

Example for NPN with emitter connected to GND and the load in the collector path:

1. Define the expected load current and voltage, here e.g. 100 mA and 24V
2. Find a transistor, that can operate with a continous collector current of at least a bit more, and that can sustain the supply voltage + margin, here we take 2N3904TA, limits are 200 mA and 40 V
3. Now we take the minimum DC current gain (hFE) of that transistor, the datasheet says 30 at 100mA. Because we want to be on the safe side and the value may be lower, we calculate with 25 here.
4. To calculate the needed base current, we divide the collector current by the gain, here 100 mA / 25 = 4 mA
5. To calculate the base resistor we must know, what voltage is used as source for the base current, here we take 5 V. The basis-emitter junction will subtract about 0.6 V from this.
6. Rb = (5V - 0.6V) / 4 mA = 1.1 kohm will turn on this transistor
7. Sometimes there is a second resistor between basis and emitter, either to force the transistor off if the other resistor is unconnected, or to speed up the switching frequency. But this is worth another question.
8. We calculate the heat in the transistor and decide if cooling is necessary. Vce (sat), the remaining voltage between collector and emitter in conducting state is tpypically 0.5V, but for safety we take 1V. P = 1V * 0.1A = 100 mW. The thermal resistance of this case is 200°C/W, so we expect a temperature rise of 20°C. This is no problem for a TO 92 case, no cooling needed here, but it is not a cool design. A MOSFET is much better in this context.

We have calculated 1.1 kohm here, 1 kohm is a good final choice. There is some estimation in here, but I didn't want to overload this.

Answer 2

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Answer 3

If we call the potential you use to control your switch "IN", and the node to be pulled down "OUT", any of these will do:

^{simulate this circuit – Schematic created using CircuitLab}

Q1 will connect OUT to ground if IN exceeds about +0.8V. That will be the case regardless of which bipolar transistor you use. Q2 would pull OUT low when IN exceeds about 2V. The MOSFET you choose will decide the actual switching thresholds.

Q2 might not like a control signal "IN" higher than +20V or so (or lower than -20V), depending on your choice of MOSFET. You can "scale" IN down with a resistor divider, if there's a chance that the potential at IN might lie outside of the MOSFET's comfort zone. The bottom version protects Q3's gate using resistors to divide IN by 2. For this reason, Q3 will require IN to be 4V or more to switch on.

Note that there's no need for +24V in any case.

The bipolar junction transistor (BJT) Q1 will not work to "pull up" a negative potential at OUT (OUT must be floating or positive), but the MOSFETs can do that. The MOSFET behaves more like a physical switch than a BJT.

Q1, being bipolar, cannot pull OUT all the way to ground. It can get within 0.2V or so of ground (0.1V if you're lucky), but the MOSFETs will perform much better in this respect.

The top-right implementation is attractively simple, but be aware you may never leave a MOSFET's gate (IN) floating. There must always be a potential explicitly applied there. That means a simple switch from IN to +10V (for example) will not work, because the gate will float when the switch is open. The other two (top-left, bottom) do not have this problem.

If OUT is unloaded, as you suggest is the case, all of these transistors may leak enough current to pull down OUT even when they are "off". They are not perfect on/off switches. Without knowing more about your application, it's not possible to say what measures need to be taken to solve this issue.