How to generate opposing outputs at 24V from 3.3V control signal

Question

We have a set of 4 control signals at 3.3V. For each of these control signals, we want to generate a pair of output signals. For example, if \$control_1\$ is high, then \$out_1\$ is 24V and \$out_1inv\$ is 0V. We do have a 24V rail to work with. So far we've accomplished this with a set of 4 DPDT relays. I was looking to see if there was a "better" way. When I say "better" I mean smaller, more power efficient, more reliable, and less expensive. I've been reading about push/pull and totem pole drivers, but I'm wondering if there is something simpler to generate the complementary outputs in the same small sub-circuit, rather than 2 separate push/pull circuits per control input.

EDIT: I should note that each of the 24V outputs (inverting and non-inverting) need to drive an optocoupler when high that has an input impedance of 4.7kOhms

EDIT 2: Here is the diagram showing the optocoupler load.

EDIT 3: There are 8 individual optocouplers to drive. 4 should be driven at any given time. All of the 0V connections to the optocouplers are connected to the system ground inside of the device. So I don't think low side switching is going to work for this application.

Answer 1

The IRF9530 is overkill but you can substitute for whatever you might have on hand. This assumes you're just driving an optocoupler with the non-inverted output and only drawing a few mA (unsure what "24V output" you were referring to).

^{simulate this circuit – Schematic created using CircuitLab}

Answer 2

I interpreted your question to mean that you wish to control the opto-coupler LED (input) side from a 3.3V signal source.

The circuit below will take a single 3.3V signal, and switch one opto-coupler on and the other off, or vice versa, depending on the input signal level:

^{simulate this circuit – Schematic created using CircuitLab}

If the input signal changes like this:

then current in the opto-coupler LEDs will look like this:

This relies on the fact that Q1's drain is either 0V (left opto-coupler is "on", passing LED current) or 24V (left opto-coupler is "off"), and Q2 can invert that voltage to provide the converse state in the right opto-coupler.

Take care to use a MOSFET that can handle 24V long-term at its gate. I chose the VN0606 for precisely that reason.

High-side switching version

In response to the new information regarding the need for high-side switching, here's a version using P-channel FETs instead:

^{simulate this circuit}

The only other change is the level translator formed by Q3 & R1. It's needed because Q1's gate must be near 24V for it to switch off, which cannot be achieved with a 3.3V input signal.

Using jelly-bean MOSFETs

If the MOSFETs have a maximum gate-source voltage of less than 24V, we need to protect them. For Q1's gate, this is easy, but Q2 requires some extra care:

^{simulate this circuit}

We use a simple potential divider (R1 and R2) to raise the lowest voltage ever applied to Q1's gate from 0V to +12V.

We can't use this trick directly for Q2's gate, because a similar potential divider connected between Q1's drain and +24V would create a permanent current through the left opto-coupler.

Therefore Q4 is employed as a buffer for the drain voltage of Q1. It's a source follower, presenting no load to Q1, and its source voltage is always slightly above Q1's drain. We are free to load that voltage source with another potential divider (R3 and R4) to constrain the voltage seen by Q2's gate.

Answer 3

The way I read the question, there should be two optocouplers to provide complimentary outputs. I would adjust the vir schematic thus:

24 V minus 3.5 V (1 IR and 1 regular LED in series) = 20.5 V

This times 10 mA = 0.205 W, the total power dissipated in each resistor string when that string is on.

One (opto + indicator) in series with R3 and the other in series with R2.

Adjust R2 = 2K; R1, R3 = 1K each. If your optocoupler Vf is greater than 1.5 V @ 10 mA, adjust the resistors accordingly

Separate from that, there is another way to do this that eliminates M2, R2, and R3 (but adds a small signal diode). The parts are from my design library; substitute as necessary.

UPDATE: Added visible LED indicators to the two optos.

Answer 4

Using comparators

Since the circuit in my previous answer has become quite complex, I had the idea of using comparators to the same end:

^{simulate this circuit – Schematic created using CircuitLab}

The most difficult part of this approach is keeping the comparator inputs within their acceptable common mode range. I achieve this by biasing the input signal to become either 1V or 3V, with R5, R1 and R2. This signal is compared with the fixed 2.2V derived from R3 and R4.

Note how the two comparators have their inverting and non-inverting inputs swapped (with respect to each other), so that one has a high output while the other is low.