Starting from the depth of field and background blur equations, we can derive the following approximate equations for a background at infinity:
b = f^2 / (x_d * N)
DoF = 2 * x_d^2 * N * C / f^2
b is background blur disc size,
DoF is depth of field,
f is the focal length,
x_d is the subject distance,
N is the aperture F-number and
C is the circle of confusion.
We immediately notice that
b can be represented in terms of
DoF and vice versa:
b = 2 * x_d * C / DoF
DoF = 2 * x_d * C / b
Now, since we are after large background blur
b, we should consider how it varies as a function of sensor size.
x_d, the subject distance, is the same for equivalent image.
DoF is also the same for an equivalent image, provided that you can achieve the
DoF you're after on both full frame and crop body. What is different is the circle of confusion
C: it's divided by the crop factor on a crop sensor. So, background blur is also divided by the crop factor on a crop sensor. However, the sensor size is too divided by the crop factor on a crop sensor.
So, as conclusion, the amount of background blur as a percentage of the sensor size stays the same (with sensor size and background blur disc size being divided by the crop factor). Thus, on full frame and crop sensor bodies, you achieve equivalent amount of background blur, provided that you can achieve the
DoF you're looking for.
So, we must analyze whether it's possible to achieve the desired
DoF on a crop sensor body. By mounting a f/1.8 lens (such as the Canon Nifty Fifty, EF 50mm f/1.8 STM), it's possible to have a depth of field of merely centimeters as this example shows:
Clearly, we can see from the image the depth of field is too shallow (image info: 50mm focal length, 1000mm subject distance, f/1.8, crop sensor body, giving DoF of 27mm). Thus, it is usually possible to achieve a shallow enough DoF on a crop sensor body.
Now, in some cases the photographer may intentionally want an extremely shallow depth of field. Can such a shallow depth of field be achieved on crop sensor body? The answer depends on whether an equivalent lens is available, which depends on the focal length. An equivalent lens has both the aperture F-number and the focal length divided by the crop factor. Then both the numerator and denominator of the fraction formula for
b is divided by the crop factor squared, giving equivalent depth of field.
For example, for Canon telephoto primes, an equivalent lens is available:
FF 135mm f/2 -> crop 84mm f/1.25 (it's 85mm f/1.2, close enough)
FF 200mm f/2.8 -> crop 125mm f/1.75 (it's 135mm f/2, close enough)
FF 300mm f/4 -> crop 188mm f/2.5 (it's 200mm f/2.8, close enough)
I won't include the 500mm f/4 because of its high cost, and because a 300mm lens with equivalent cost is not available.
But for Canon short telephoto and standard lenses, the situation is different:
FF 85mm f/1.2 -> crop 53mm f/0.75 (it's 50mm f/1.2, not available)
FF 50mm f/1.2 -> crop 31mm f/0.75 (it's 28mm f/1.8, not available)