Dynamically create blazor component and call its methods

Question

I am using the MudBalzor table in my component but also want to use Syncfusion Grid as it contains some useful methods like ExportToExcel and ExportToPdf.

I want to create component of Syncfusion Grid but do not want to render it to view.

First, I try to simply create an object of that component but it didn't work, it throws a null reference when I call its methods.

SfGrid<GetEquipmentsModel> grid = new SfGrid<GetEquipmentsModel>();
grid.DataSource = data;
await grid.ExportToPdfAsync(); // inside this method it throws exception.

then I search for how to create components dynamically and found this.

var componentRef = new RenderFragment(builder => {
  builder.OpenComponent(0, typeof(SfGrid<GetEquipmentsModel>));
  builder.AddComponentReferenceCapture(1, inst => { grid = (SfGrid<GetEquipmentsModel>)inst;  });
  builder.CloseComponent();
  });

I am assuming this below line will capture the instance of a created component.

builder.AddComponentReferenceCapture(1, inst => { grid = (SfGrid<GetEquipmentsModel>)inst

await grid.ExportToPdfAsync(); // it is still trhrowing null reference exception.

but RenderFragment isn't invoked. also, i tried calling

componentRef.Invoke() //but it needs RenderTreeBuilder object which I don't know where to get from.

What I want to achieve is to create a SfGrid<GetEquipmentsModel> component in code create a column in that grid (SfGrid has a property Columns where I can add columns) Export data and that's it.

If I use that component in view and use that reference it works but that way I have to hide the component, I am thinking if I can achieve this in code without using it in view?

<SfGrid @ref="grid">
<SfGrid >

now if I use grid it will work.

I try to use a few open-source javascript libraries to export table data and they work as well but Syncfusion exporting data the same as it is in grid means using the same widths and alignments.