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.

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}

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}

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.

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}

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}

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

^{simulate this circuit}

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.

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}

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.

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}

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}