Level shifting from 3.3V to 5V not working: voltage drops when attaching load

Question

I am trying to trigger a laser using an Arduino. I used a power supply to determine that the laser needs a 3.5V (or higher) signal, at which point it draws around 2.2mA from the power supply. When supplying the laser with 5V, it draws 3.7mA from the power supply. The signal is delivered through a BNC cable. My Arduino can only supply 3.3V on its digital pins. I've tried using a level shifter and also tried using two cascaded inverters to increase the voltage level.

The problem is that while the voltage increases as expected when the laser is disconnected, the moment I connect the laser the voltage drops.

1. Using the level shifter, the voltage drops from ~5V to 1.2V. The power supply says it provides 0.5mA, but using a multimeter I measure 0.58mA going into the laser.

2. Using two inverters, the voltage drops from ~4.3V to 3.4V. The power supply says it provides 0.2mA, but using a multimeter I measure 2.02mA going into the laser.

The level shifter is an AN97055, just very standard as far as I can tell. The inverter I'm using is SN74LS14N.

What could be causing this? Can the level shifter and inverter not maintain enough current? The Arduino (GIGA) can supply 8mA on its digital pins, so that should be fine. If I were to use something like BOB-12009, would that help? It seems it allows for higher current.

Answer 1

The output stage of an LS14, like almost all bipolar TTL circuits, does not swing all the way up to its Vcc (+5 V) because the pull-up device is an emitter follower. A simple change is to change the part to the CMOS 74ACT14. AC and ACT series gates have a much more robust output stage that can source and sink 24 mA with a relatively small voltage drop. When sourcing 4 mA it will swing much closer to its Vdd (+5 V).

Answer 2

That kind of level shifter uses the MCU output to sink current, so it is limited by how much current the MCU can sink. That's not your issue here, but it could be relevant later (see below).

The MCU output is not involved in sourcing current, that comes from the pullup resistor, whatever value you have used, when the MOSFET turns off. So if you have a 10kΩ pullup and load the output with 3K it won't be able to get above 1.15V.

If you need to source more current you can use a proper level shifter chip, use a 74HCTxx gate as a level shifter or buffer the output of your present level shifter with a 5V HCMOS gate etc. etc. You could, of course, try to reduce the value of the pullup but whether that will work depends on the details.

Answer 3

You could probably make this work with only the AN97055, but you'll need to wire it up like this:

^{simulate this circuit – Schematic created using CircuitLab}

I imagine you've tried to place the laser diode between HV1 and ground, but here I've put it between HV1 and +5V. That's because R4 limits current flow from HV (+5V) out of output HV1. Consequently HV1 is unable to source much current towards ground. However, it is able to sink whatever current you like directly from +5V, via load D1.

As long as the GPIO is able to hold its output at 0V, MOSFET Q1 is fully on, and there will be the full 5V across load D1. This means that all diode current must be sunk by the GPIO output. You say that it can handle 8mA, so it should do just fine.

It also means that a GPIO low output will light the laser.

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All 74LS TTL ICs have the same problem; their outputs are able to sink a lot more current than they can source. So again, you'll have better luck connecting the laser diode between +5V and the 74LS14 inverter output:

^{simulate this circuit}

Alternatively, you can parallel TTL outputs to boost their current sinking/sourcing ability:

^{simulate this circuit}

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The best solution (in my opinion), and the easiest, is to use literally any 74HCT gate to power the laser diode. 74HCT (The "T" is important) devices have 3.3V compatible inputs, so your GPIO will operate their inputs perfectly well. Power comes from the +5V supply, meaning their outputs are either 0V or +5V. The gate could be an inverter from a 74HCT14, or an AND gate from a 74HCT08, a NOR from a 74HCT02 etc.

In addition, they have CMOS outputs, meaning they can source and sink current equally well, and their output potentials will be much closer to the supplies than regular TTL could ever be. Here's how I'd do it, using a 74HCT1G08:

^{simulate this circuit}

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Don't forget you can use a plain old BJT or MOSFET:

^{simulate this circuit}

Answer 4

I am not sure why you are "level shifting". You could just drive the laser diode like this.

^{simulate this circuit – Schematic created using CircuitLab}

The BSS138 has a threshold voltage of 0.6 to 1.4 V, so it should work fine with 3.3V logic. You could substitute another similar MOSFET for this one. I chose the BSS138 only because it is available on CircuitLab.

R2 is used to ensure the gate current is not too large that it could damage the GPIO. R1 may or may not be necessary, depending upon the laser diode. It should be sized to limit the current to within the diode's specs.