After modification

If you are trying to run the transistor in saturation, then you generally work with a lower beta. For a Darlington a saturation beta of 100 isn't too far off, but the data sheet characterized the device for saturation betas of 65 and 10. I would just pick 65 since you can pull value from datasheet curves without having to interpolate them as much.

You are ALMOST tracking your units through your work. Put in the effort to track them completely through your work -- it will pay huge dividends.

Always as if your answers (including intermediate ones) make sense.

Knowing how metric scaling prefixes work, does it make sense that when you divide 5 A by 100 you get 5 mA?

The value you are using for Vbesat is the maximum voltage it will be under a specific set of conditions, namely 6.5 A of collector current and running at a beta of 65. The diagram in the datasheet is also for this test condition and shows that the typical Vbesat will likely be somewhat under 2 V. Whether the distinction is worth considering depends on what is important in your application.

Unless you heat sink the transistor, you are going to have to be very careful about your duty cycle. The thermal resistance from the junction to the ambient air is about 42 °C/W. If you have it well saturated, then the Vce is about 1 V and so you have 5 W at 5A. That means that the junction would tend to be about 210 °C above the ambient air temperature, taking it to somewhere in the 235 °C range. The max junction temperature is only 125 °C, or about 100 °C above ambient, meaning that you need to keep the power dissipation below about 2.3 W. So your duty cycle has to be well below 50% and the pulse width needs to be short enough that the temperature can't rise too much before you turn it off and it starts cooling down again.

Note that at high junction temperatures, the nominal beta at an Ic of 5 A falls down below 100. If you assume a higher beta than you will actually have, you don't get as much collector current which means that transistor may be in the active region, which results in a higher Vce which may result in increased power dissipation and hence even higher junction temperatures. When you want to drive a transistor into saturation to use it as a switch, drive it hard into saturation unless there is some other issue that makes that impractical.

Redo your calculations and I think you will find your power in your base resistor is going to be much higher, even if you use a beta of 100. You are probably looking at somewhere in the 300 mW to 600 mW range. But if you run the numbers for operating at a beta of 10, you'll really see them climb.

 

If you are trying to run the transistor in saturation, then you generally work with a lower beta. For a Darlington a saturation beta of 100 isn't too far off, but the data sheet characterized the device for saturation betas of 65 and 10. I would just pick 65 since you can pull value from datasheet curves without having to interpolate them as much.

You are ALMOST tracking your units through your work. Put in the effort to track them completely through your work -- it will pay huge dividends.

Always as if your answers (including intermediate ones) make sense.

Knowing how metric scaling prefixes work, does it make sense that when you divide 5 A by 100 you get 5 mA?

The value you are using for Vbesat is the maximum voltage it will be under a specific set of conditions, namely 6.5 A of collector current and running at a beta of 65. The diagram in the datasheet is also for this test condition and shows that the typical Vbesat will likely be somewhat under 2 V. Whether the distinction is worth considering depends on what is important in your application.

Unless you heat sink the transistor, you are going to have to be very careful about your duty cycle. The thermal resistance from the junction to the ambient air is about 42 °C/W. If you have it well saturated, then the Vce is about 1 V and so you have 5 W at 5A. That means that the junction would tend to be about 210 °C above the ambient air temperature, taking it to somewhere in the 235 °C range. The max junction temperature is only 125 °C, or about 100 °C above ambient, meaning that you need to keep the power dissipation below about 2.3 W. So your duty cycle has to be well below 50% and the pulse width needs to be short enough that the temperature can't rise too much before you turn it off and it starts cooling down again.

Note that at high junction temperatures, the nominal beta at an Ic of 5 A falls down below 100. If you assume a higher beta than you will actually have, you don't get as much collector current which means that transistor may be in the active region, which results in a higher Vce which may result in increased power dissipation and hence even higher junction temperatures. When you want to drive a transistor into saturation to use it as a switch, drive it hard into saturation unless there is some other issue that makes that impractical.

Redo your calculations and I think you will find your power in your base resistor is going to be much higher, even if you use a beta of 100. You are probably looking at somewhere in the 300 mW to 600 mW range. But if you run the numbers for operating at a beta of 10, you'll really see them climb.

At the beginning I thank you for your thorough analysis of the circuit.
Second, I apologize for my poor English
I think It is a bad idea to control that type of lamp as switch by that way, especially this type of bulbs that use the filament. As the lamp heat rise up resistance of the filament changes with the temperature(not ohms resistor) and therefore varies collector current and thus will change many of the parameters.
I think the right way is to use a low-power transistor such as 2N2222 with a relay to run the load as a switch , is not it?

 

I think the right way is to use a low-power transistor such as 2N2222 with a relay to run the load as a switch , is not it?

I see no reason for adding an expensive relay as long as the transistor is adequately sized to switch the load voltage and current.

 

At the beginning I thank you for your thorough analysis of the circuit.
Second, I apologize for my poor English
I think It is a bad idea to control that type of lamp as switch by that way, especially this type of bulbs that use the filament. As the lamp heat rise up resistance of the filament changes with the temperature(not ohms resistor) and therefore varies collector current and thus will change many of the parameters.
I think the right way is to use a low-power transistor such as 2N2222 with a relay to run the load as a switch , is not it?

While you could use a relay, there probably isn't much of a reason to. But switching to a power MOSFET is something to consider. Find one with a low Rds_on resistance so that you dissipate significantly less power in the transistor. You can also find bipolar transistors that have much lower Vcesat voltages and then drive them with a second transistor to reduce you base current requirement from your control circuit. The Darlington configuration results in a significantly higher Vcesat (~1 V instead of ~0.3 V), which is an issue.