Do Fixed Bias BJT 's have a cut in voltage?

I am doing a DC analysis of a Fixed bias BJT and am trying to calculate its collector current. So far I have these following values.

I am using a MMUN2211LT1 bias resistor transistor network.

My Emitter is shorted to GND (so VE = 0V). My Voltage supply on my Collector is VCC=2.5V with an RC of 22kOhms. The Voltage supplied to my base is VCB =.785V with an RB of 7.6kOhms.

So usually when I do DC analysis, I usually calculate my IB using Beta and the cut in voltage (VBEON), and then I am good to go calculating my other values. However, the datasheets I have found for this part do no not mention a cut in voltage or Beta. Should I be using a different set of values to analyze my circuit, or should I be assuming a constant current is going through my BJT. Any advice to deal with these lack of parameters would be greatly appreciated.

The MMUN2211LT1 already has a base resistor of 7~13kΩ with an hFE of 35 ~60.

If you apply 2.5V to base resistor it will saturate to 0.2Vmax with 5mA collector current.

added- - I read the above in the product spec. Have you read it? My analysis of this switch is as follows; Output 5mA for input 2.5V/10kΩ nom = 0.25mA which translates to current gain of 20x which is typical for a saturated switch. 5x is common for high current switch.

So apply more base voltage to this device and remove the external base resistor. The 0.785 V you supply is too low to let the transistor conduct. The datasheet doesn't mention base-emitter voltage since the base isn't available to the user anyway. But let's assume it's 0.65 V. The internal resistors are both 10 kΩ. Then the base to ground resistor will see 0.65 V/ 10 kΩ = 65 µA. If the input current is anything lower than that everything will go to the resistor, and none to the base.

The total input resistance is your 7.6 kΩ in series with the internal 10 kΩ = 17.6 kΩ. The voltage difference is 0.785 V - 0.65 V = 0.135 V. Then the input current is 0.135 V/ 17.6 kΩ = 7.67 µA. Which is much lower than the 65 µA we needed so the transistor will do nothing. Removing the external 7.6 kΩ doesn't help, the current will still be only 13.5 µA. Your input voltage is simply too low.

Let's use the 2.5 V in, and put the 7.6 kΩ back in. Then the input current is (2.5 V - 0.65 V)/ 17.6 kΩ = 105 µA. Now we're getting somewhere. Subtract the 65 µA through the ground resistor and you have 40 µA base current. At a worst case HFE of 35 this will give you 1.4 mA collector current, which will drop 30.8 V across the load resistor. You only have 2.5 V, then the 22 kΩ will limit the current to (2.5 V - 0.2 V)/ 22 kΩ = 105 µA.

You can also calculate the minimum input voltage to get the transistor to saturate. That will be at the 105 µA collector current. At an HFE of 35 that's 3 µA base current. We add our 65 µA ground resistor current to that and we have a total input current of 68 µA. Through 17.6 kΩ gives 1.20 V, which the input has to be higher than the base voltage of 0.65 V, so that's 1.85 V for saturation.