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I'm a super beginner, but I've developed a circuit that appears to work. My question is if there is a better way to do it.

I have two power supplies. The primary supply will power a load. When the primary supply is cut, the secondary should power the load.

There are two possible solutions:

  1. Use a transistor in a not circuit. When the power fails, it allows power to the relay to switch it.

  2. Use the relay, have it always on, and when the power fails, the relay falls to its natural state connecting the secondary supply.

I know we're talking about very little power, but I'm curious which one uses more? If the relay requires 20ma to stay activated for the primary power, then it's always using that. If I use the transistor/NOT circuit, then the relay is using 0ma to connect the primary power, but the transistor looks like it is using 20ma instead to just short the relay circuit.

So my question is, is one of those preferable over the other? Or is there a way to build a NOT circuit that uses less power than the load it needs to power?

enter image description here

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  • \$\begingroup\$ Is R1 really 5 ohms? Did you build this circuit or only simulate it? \$\endgroup\$
    – The Photon
    Oct 3 '20 at 15:08
  • \$\begingroup\$ The base resistor is 5k I guess, not 5ohm. What load will be connected to these 12V? \$\endgroup\$
    – Peter MP
    Oct 3 '20 at 15:12
  • \$\begingroup\$ I'm a super beginner, but I've developed a circuit that appears to work So you're (trying) to design a circuit with no circuit design experience? In my opinion that's the road to broken components and lots of dissapointments. You're not the first to think "I understand what a transistor does, now lets build a circuit" without first having a look at how transistors are commonly used. What circuits are common I mean. The result of that is often transistors being blown up even though "it works in the simulator". So do yourself a favor and have a look at common LED switching circtuits. \$\endgroup\$ Oct 3 '20 at 16:05
  • \$\begingroup\$ Example: R1 being 5 ohms with nearly 12 V across it, unless it is a 30 W resistor, it will go up in flames immediately possibly taking Q1 with it. The fact that you didn't see: "hey, 2.2 A is flowing there, maybe that's a bit much, I should check that out" tells me enough. Sure, in the simulator components don't blow up but if you actually build this they will. \$\endgroup\$ Oct 3 '20 at 16:08
  • \$\begingroup\$ The transistor should be between the lower terminal of the relay coil and Ground, not in parallel with the relay coil. \$\endgroup\$ Oct 3 '20 at 16:12
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Yes, the direct answer to the question is that for sure it is possible to build a transistor circuit that would trigger relay and use much less power than the relay for its own internal needs (and in total when relay is off).

A common emitter with the load in the collector circuit is not actually an "inverter" with respect to the collector current, as the current through the load increases when the base voltage / base current rises. I think you need two transistors to build the needed circuit, for instance:

schematic

simulate this circuit – Schematic created using CircuitLab

In this solution, the first common emitter amplifier acts as true inverter (resembling the "not" element from very old times). It is open (I use the literature word "open" in the sense it is conducting) when the second voltage, V2, is present. Most of the power consumed is due current on resistor R1 and should be about 0.5 mA. It keeps the base potential of Q1 low enough so it does not open.

Another transistor acts as current amplifier to drive the relay. As soon as V2 "disappears" one or another way, Q2 closes (also in the sense stops conducting), the base potential of Q1 raises and the current through R1 starts flowing into the base of Q1. This should be about 0.5 mA and that much is enough for a 50 mA low voltage relay to trigger. Maybe 20 mA if it is a really bad transistor.

Hence we have a circuit that uses 0.5 mA on its own but can drive a 50 mA relay that draws two orders of magnitude more current. I have verified the behavior with the simulator:

enter image description here

Here in the middle of the chart the second voltage source V2 "wakes up", powering off the coil of the 50 mA relay. The total power consumption drops from about 50 mA (mostly for relay) to something that looks almost zero in the chart.

This explanation contains some simplifications: I do not talk about base-emitter voltage drop for transistors and the like, but for this particular circuit they should not have much impact.

How much it is "worth building", depends on the power situation. If you have 1 A power consumers, probably a 50 mA relay would not make a difference. But if it is something more towards 100 mA, this "always on" 50 mA relay starts looking funny. Likely even better efficiency is possible if to spend more time on the design.

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Instead of BJT use Mosfet. Your current consumption will reduce significantly.

schematic

simulate this circuit – Schematic created using CircuitLab

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