I am reading in a book about the forward bias of a diode. It mentions that in order for a diode to be in the forward mode and its current to rise exponentially:
$$V_{D}\geq + 4V_{T}=0.1V$$
where \$V_{D}\$ is the diode's voltage and \$V_{T}\$ the thermal voltage (approximately 0.025V as a typical value).
My problem is that from what I know and from what I read in another section of the same book and also in other books, then for a diode to be in forward bias:
$$0.5V\leq V_{D}\leq 0.8V$$
If we use the equation of a diode with \$V_{D}=0.1V\$ then:
$$I=e^{0.1/0.025}\approx 53\cdot I_{S}$$
where \$I_{S}\$ is the reverse saturation current.
But with \$V_{D}=0.5V\$ then:
$$I=e^{0.5/0.025}\approx 485165194\cdot I_{S}$$
Clearly with \$V_{D}=0.1V\$ the current through the diode is very small compared to the current that will pass if the voltage is \$V_{D}\geq 0.5\$, as most sources mention as the necessary threshold for a diode to be in forward bias.
Am I missing something?