# Darlington transistor with LED ..#2

#### pinkyponky

Joined Nov 28, 2019
351
You don't need a voltage divider to feed voltage to the base of an emitter-follower. The base of an emitter-follower also does not need a current-limiting resistor.

A transistor has a base-emitter diode that has a voltage drop of about 0.6V to 0.7V. If 4.8V is applied to its base then 4.1V to 4.2V will appear at its emitter.
But a darlington transistor has two transistors with their base-emitter diodes in series. Each base-emitter diode also has a voltage drop of 0.6V to 0.7V.
If 4.8V is applied to the base then about only 3.6V will appear at the emitter.
Thank you for the explanation.

I have a question here:

I'm planning to use the darlington circuit for current increment, but I have also seen the voltage drop in my simulation. I would like to increase the current and maintain the voltage without drop. So, in this case, which configuration is better to use and to meet my requirements (maintain constant base voltage 5V at the emitter of darlington and also the current needs to be increased from few micro AMPs to 300mA).

Thank you!

#### dl324

Joined Mar 30, 2015
16,898
I'm planning to use the darlington circuit for current increment, but I have also seen the voltage drop in my simulation. I would like to increase the current and maintain the voltage without drop. So, in this case, which configuration is better to use and to meet my requirements (maintain constant base voltage 5V at the emitter of darlington and also the current needs to be increased from few micro AMPs to 300mA).
You must have posted to an old thread and your question was split from it.

Post a schematic of the circuit you simulated and specify your tolerance for voltage drop. There will be a voltage drop because you have to use real (vs ideal) components.

#### pinkyponky

Joined Nov 28, 2019
351
Post a schematic of the circuit you simulated and specify your tolerance for voltage drop. There will be a voltage drop because you have to use real (vs ideal) components.

#### dl324

Joined Mar 30, 2015
16,898
What is the output voltage that you want and its tolerance?

#### pinkyponky

Joined Nov 28, 2019
351
What is the output voltage that you want and its tolerance?
The output should be 5V

#### crutschow

Joined Mar 14, 2008
34,392
Sorry, you can't get 5V out of a Darlington with 5V at the base and a 5V supply.
There is always a couple base-emitter drops between input and output.
Try again as to what you can live with, not the idealized circuit you have requested.

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#### BobTPH

Joined Jun 5, 2013
8,932
Why is the LED not on the collector side? With that configuration you only lose less than half a volt. And no LED I know of requires 5V, in fact they would burn up if you gave them that.

#### dl324

Joined Mar 30, 2015
16,898
The output should be 5V
You can't get 5V out with V1=V2=5V.

What is the load current range and your output voltage tolerance? You won't be able to control output voltage without a feedback circuit.

#### pinkyponky

Joined Nov 28, 2019
351
You can't get 5V out with V1=V2=5V.

What is the load current range and your output voltage tolerance? You won't be able to control output voltage without a feedback circuit.
My output voltage should be same as input voltage, in this case if my input is 5V then output should also be 5V with above 300mA output current.

#### dl324

Joined Mar 30, 2015
16,898
My output voltage should be same as input voltage, in this case if my input is 5V then output should also be 5V with above 300mA output current.
If your input is 5V, why do you need a 5V regulator?

#### crutschow

Joined Mar 14, 2008
34,392
My output voltage should be same as input voltage, in this case if my input is 5V then output should also be 5V with above 300mA output current.
Then you will need a more complex circuit, such as a rail-rail op amp driving a P-MOSFET buffer (LTspice sim example below):
The P-MOSFET must be a logic-level type with the max Vgs(th) ≤ 2V.
The capacitors are needed to keep the loop stable due to the added loop gain from the MOSFET.

#### pinkyponky

Joined Nov 28, 2019
351
Then you will need a more complex circuit, such as a rail-rail op amp driving a P-MOSFET buffer (LTspice sim example below):
The P-MOSFET must be a logic-level type with the max Vgs(th) ≤ 2V.
The capacitors are needed to keep the loop stable due to the added loop gain from the MOSFET.

View attachment 315999
Instead of MOSFET, we can also use the darlington pair circuit to reach 300mA output current, Right?.

#### crutschow

Joined Mar 14, 2008
34,392
Instead of MOSFET, we can also use the darlington pair circuit to reach 300mA output current, Right?.
No. Not if you want to go all the way to 5V.
BJT's have an inherent small voltage drop when fully on that is unavoidable.
A Darlington has a minimum drop of about 1V.
(You need to read up on how Darlingtons work).

A MOSFET has only its on-resistance when fully on, so the drop can be made very small at low currents.

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#### BobTPH

Joined Jun 5, 2013
8,932
My output voltage should be same as input voltage, in this case if my input is 5V then output should also be 5V with above 300mA output current.
Why? Does the LED require 5V across it? If not, you will need a resistor in series to reduce the voltage anyway. If it is 4V, that just requires you to use a different resistor.

And why are you switching the high side, when NPN transistors operate as low side switches, not high side?

#### khizer kashif

Joined Mar 27, 2024
1
Thank you for the explanation.

I have a question here:

I'm planning to use the darlington circuit for current increment, but I have also seen the voltage drop in my simulation. I would like to increase the current and maintain the voltage without drop. So, in this case, which configuration is better to use and to meet my requirements (maintain constant base voltage 5V at the emitter of darlington and also the current needs to be increased from few micro AMPs to 300mA).

Thank you!
If you want to maintain a constant base voltage at the emitter of the Darlington pair while increasing the current from a few microamps to 300mA without experiencing a significant voltage drop, you might want to consider using a Darlington pair in a configuration known as an "Emitter Follower" or "Common Collector" configuration.

In this configuration, the input is connected to the base of the first transistor, and the output is taken from the emitter of the second transistor. This setup provides a high input impedance and a low output impedance, making it suitable for buffering purposes while maintaining relatively constant voltage.

Here's why this configuration might be suitable for your requirements:

1. **Low Voltage Drop**: The emitter follower configuration has a voltage gain very close to 1. This means that the voltage at the emitter (output) will closely follow the voltage at the base (input), resulting in minimal voltage drop.

2. **Current Gain**: Darlington pairs inherently provide high current gain, which means that small changes in base current result in large changes in collector current. This property can help you achieve the desired increase in current while maintaining a relatively constant voltage.

3. **Base Voltage Stability**: Since the emitter follower configuration has a very high input impedance, it can help stabilize the base voltage, especially if driven by a stable voltage source.

Here's a basic schematic of how you might configure a Darlington pair in an emitter follower configuration:


Vcc

└─────── R1 ───── Base of Q1 (Darlington)

└─────── R2 ───── Emitter of Q1 (Darlington) ───── Collector of Q2 (Darlington)

└─────── GND


In this configuration:
- Q1 and Q2 form the Darlington pair.
- R1 provides the base current to Q1.
- R2 provides negative feedback to stabilize the output voltage.
- The load is connected between the emitter of Q1 (collector of Q2) and ground.

By appropriately selecting the resistors R1 and R2, and ensuring that the Darlington pair transistors are capable of handling the desired current, you should be able to achieve your requirements of maintaining a constant base voltage while increasing the current from microamps to 300mA.