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Attached is my current circuit for using a 5V logic output device to control 24V going to a load with max current draw of 14mA. (Ignore the name for transistor Q1, that is just a random NPN BJT).

BJT NPN, circuit

The max current draw of the logic source is 300mA (a Velocio e-1450 PLC).

I am new to transistor design and selection, and I would like to know what you consider when choosing a transistor - what range of values do you allow for a given circuit? Should the base voltage be exactly what you expect it to be? (5V rating for 5V logic control, etc.)

Also, are the resistor values relatively ok for the desired output? For R1, I took (24V - 0.3V) / 14mA ~= 1.69 kOhm.

Thanks for your input.

EDIT

To sum it all up, is this the right design to use 5V logic to power a solenoid (J1) with 24V? Forgetting the resistors and transistor values.

1) The 5V logic output (V1) from the Velocio e-1450 can provide up to 300mA

2) The solenoid (J1) draws max 14mA.

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  • \$\begingroup\$ You said the load might take 14 mA. Do you need to still deliver something close to 24 V when the load is drawing 14 mA? \$\endgroup\$ – The Photon Oct 4 '19 at 20:02
  • \$\begingroup\$ With that schematic, R1 is supplying power to the PLC. At 14mA, the voltage drop across R1 would be over 24V -- there would be nothing left over for the PLC. \$\endgroup\$ – TimWescott Oct 4 '19 at 21:09
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You want something like this (note, it could be done with FETs -- but I like BJTs).

You need to supply current to the load, and you want to do that through a transistor. I'm assuming you want the load to be grounded (you can do this with one transistor if not).

Because you are driving your load from a voltage higher than your processor's VCC, you need voltage amplification. That requires a transistor stage (Q1). In addition to assuming that you want the load to be grounded that you don't want to consume great gobs of current when the load is off. That requires an additional stage that will conduct no current when the load's voltage is 0 (as opposed to a circuit with your topology, which basically requires that the load be shorted). This assumed requirement is satisfied by Q2, which only conducts when the load is on.

Going backwards from the load:

  1. The load draws 14mA. A 2N3906 is good for 60V and 100mA, so it's more than adequate for switching 24V and 14mA.
  2. To Q2 in saturation, you need to pull about 1/10th the collector current from the base (you can use less if you use a "super beta" transistor, but why?).
  3. In saturation, Q2's base-emitter voltage will be about 0.7V.
    1. R3 pulls about 1.55mA from the node at the base of Q2
    2. R4 pushes about 0.15mA into it. The sum is 1.4mA -- good!
  4. R4 is there to make sure that Q2 turns off positively when Q1 goes off.
  5. Q1 needs to pull 1.55mA. So it needs a base current of 0.155mA (it's the same 1/10th rule).
    1. R1 supplies 0.43mA to the node at the base of Q1
    2. R2 sucks 0.07mA from that node. That leaves 0.38mA for the base of Q1 -- that's plenty.
  6. R2 is there for the same reason as R4 -- to make sure the transistor turns of promptly.

And that's it, pretty much. If this were for production, I'd suggest you use pre-biased transistors (basically what you see here, but with the resistors built in). If it's a one-off, just grab resistors and transistors from your junk drawer.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ watch out for safe-operating-area-pulse-operation \$\endgroup\$ – analogsystemsrf Oct 5 '19 at 2:14
  • \$\begingroup\$ Thanks for the response and the math/reasoning, Tim. How does two transistors benefit (or is it necessary for the situation?) Would one transistor not do the trick? (assuming I want to use it just as a switching transistor that either is off or in saturation). Thanks \$\endgroup\$ – Anthony the Kid Oct 7 '19 at 12:31
  • \$\begingroup\$ I've edited my answer to address your additional question. \$\endgroup\$ – TimWescott Oct 7 '19 at 14:35

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