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Please try this.

1. Let R2 be 1 kilo ohms
2. Place a zener of reverse voltage of say 2.7 V (for discussion sake) between base and the ground and make sure it is biased properly when MCU is driving it.
3. Change emitter resistor(R1) to 14 ohms
4. Now we have a constant 2.7 V applied across the base whenever the zener is driven (reverse voltage)
5. Assuming base drop of 0.7 V, we will have 2 V across this emitter resistor all the time.
6. It corresponds to a current of 2 V / 14 ohms which is roughly 142 mA.
7. Note that this is independent of positive voltage (10 V to 20 V)
8. Please find a darlington transistor for high ß
9. Depending of MCU voltage you can choose zener voltage and do similar math again to choose new emitter resistor value
10. No. Doesn't work with V+ floating.
11. R3 can be your load (LEDs for example) All above points refers to reference designatiors in OP's question

Below text uses references in my circuit

Zener and R2(in my circuit) together act as negative feedback for this system and hence when current tries to increase, the emitter voltage will increase which tends to decrease the applied base emitter voltage for the transistor there by reducing the gain a little and the nice the collector current accordingly.

Will upload the circuit soon

Example only Schematics:

Below is the sweep done for 12 V supply in 2 V steps until 20 V. you can run the simulation and tweak values.

Please try this.

1. Let R2 be 1 kilo ohms
2. Place a zener of reverse voltage of say 2.7 V (for discussion sake) between base and the ground and make sure it is biased properly when MCU is driving it.
3. Change emitter resistor(R1) to 14 ohms
4. Now we have a constant 2.7 V applied across the base whenever the zener is driven (reverse voltage)
5. Assuming base drop of 0.7 V, we will have 2 V across this emitter resistor all the time.
6. It corresponds to a current of 2 V / 14 ohms which is roughly 142 mA.
7. Note that this is independent of positive voltage (10 V to 20 V)
8. Please find a darlington transistor for high ß
9. Depending of MCU voltage you can choose zener voltage and do similar math again to choose new emitter resistor value
10. No. Doesn't work with V+ floating.
11. R3 can be your load (LEDs for example)

All above points refers to reference designators in OP's question

Will upload the circuit soon

Example only Schematics:

Below is the sweep done for 12 V supply in 2 V steps until 20 V. you can run the simulation and tweak values.

Below text uses references in my circuit

Zener and R2(in my circuit) together act as negative feedback for this system and hence when current tries to increase, the emitter voltage will increase which tends to decrease the applied base emitter voltage for the transistor there by reducing the gain a little and the nice the collector current accordingly.

Please try this.

1. Let R2 be 1 kilo ohms
2. Place a zener of reverse voltage of say 2.7 V (for discussion sake) between base and the ground and make sure it is biased properly when MCU is driving it.
3. Change emitter resistor(R1) to 14 ohms
4. Now we have a constant 2.7 V applied across the base whenever the zener is driven (reverse voltage)
5. Assuming base drop of 0.7 V, we will have 2 V across this emitter resistor all the time.
6. It corresponds to a current of 2 V / 14 ohms which is roughly 142 mA.
7. Note that this is independent of positive voltage (10 V to 20 V)
8. Please find a darlington transistor for high ß
9. Depending of MCU voltage you can choose zener voltage and do similar math again to choose new emitter resistor value
10. No. Doesn't work with V+ floating.
11. R3 can be your load (LEDs for example)

Zener and R1 together act as negative feedback for this system and hence when current tries to increase, the emitter voltage will increase which tends to decrease the applied base emitter voltage for the transistor there by reducing the gain a little and the nice the collector current accordingly.

Will upload the circuit soon

Example only Schematics:

Below is the sweep done for 12 V supply in 2 V steps until 20 V. you can run the simulation and tweak values.

Please try this.

1. Let R2 be 1 kilo ohms
2. Place a zener of reverse voltage of say 2.7 V (for discussion sake) between base and the ground and make sure it is biased properly when MCU is driving it.
3. Change emitter resistor(R1) to 14 ohms
4. Now we have a constant 2.7 V applied across the base whenever the zener is driven (reverse voltage)
5. Assuming base drop of 0.7 V, we will have 2 V across this emitter resistor all the time.
6. It corresponds to a current of 2 V / 14 ohms which is roughly 142 mA.
7. Note that this is independent of positive voltage (10 V to 20 V)
8. Please find a darlington transistor for high ß
9. Depending of MCU voltage you can choose zener voltage and do similar math again to choose new emitter resistor value
10. No. Doesn't work with V+ floating.
11. R3 can be your load (LEDs for example) All above points refers to reference designatiors in OP's question

Below text uses references in my circuit

Zener and R2(in my circuit) together act as negative feedback for this system and hence when current tries to increase, the emitter voltage will increase which tends to decrease the applied base emitter voltage for the transistor there by reducing the gain a little and the nice the collector current accordingly.

Will upload the circuit soon

Example only Schematics:

Below is the sweep done for 12 V supply in 2 V steps until 20 V. you can run the simulation and tweak values.

Please try this.

1. Let R2 be 1 kilo ohms
2. Place a zener of reverse voltage of say 2.7 V between base and the ground and make sure it is biased properly when MCU is driving it.
3. Change emitter resistor(R1) to 14 ohms
4. Now we have a constant 2.7 V applied across the base whenever the zener is driven (reverse voltage)
5. Assuming base drop of 0.7 V, we will have 2 V across this emitter resistor all the time.
6. It corresponds to a current of 2 V / 14 ohms which is roughly 142 mA.
7. Note that this is independent of positive voltage (10 V to 20 V)
8. Please find a darlington transistor for high ß
9. Depending of MCU voltage you can choose zener voltage and do similar math again to choose new emitter resistor value
10. No. Doesn't work with V+ floating.
11. R3 can be your load (LEDs for example)

Zener and R1 together act as negative feedback for this system and hence when current tries to increase, the emitter voltage will increase which tends to decrease the applied base emitter voltage for the transistor there by reducing the gain a little and the nice the collector current accordingly.

Will upload the circuit soon

Please try this.

1. Let R2 be 1 kilo ohms
2. Place a zener of reverse voltage of say 2.7 V (for discussion sake) between base and the ground and make sure it is biased properly when MCU is driving it.
3. Change emitter resistor(R1) to 14 ohms
4. Now we have a constant 2.7 V applied across the base whenever the zener is driven (reverse voltage)
5. Assuming base drop of 0.7 V, we will have 2 V across this emitter resistor all the time.
6. It corresponds to a current of 2 V / 14 ohms which is roughly 142 mA.
7. Note that this is independent of positive voltage (10 V to 20 V)
8. Please find a darlington transistor for high ß
9. Depending of MCU voltage you can choose zener voltage and do similar math again to choose new emitter resistor value
10. No. Doesn't work with V+ floating.
11. R3 can be your load (LEDs for example)

Zener and R1 together act as negative feedback for this system and hence when current tries to increase, the emitter voltage will increase which tends to decrease the applied base emitter voltage for the transistor there by reducing the gain a little and the nice the collector current accordingly.

Will upload the circuit soon

Example only Schematics:

Below is the sweep done for 12 V supply in 2 V steps until 20 V. you can run the simulation and tweak values.