# Darlington Emitter-Follower Circuit

#### aac044210

Joined Nov 19, 2019
178
Hi:

This darlington emitter-follower circuit attenuates the output current rather than
increasing it. Not sure what is wrong.

iRL = 9.87 * 10-8 while ivs = 2.05 * 10-7

Any suggestions?

Thanks

#### Attachments

• 2.5 KB Views: 34

#### AlbertHall

Joined Jun 4, 2014
11,311
Because in your circuit the input impedance is lower than the load impedance because of R8 and RE1 (RE1?)
Increase both of them by a factor of 100.

#### aac044210

Joined Nov 19, 2019
178
Because in your circuit the input impedance is lower than the load impedance because of R8 and RE1 (RE1?)
Increase both of them by a factor of 100.
Thanks Albert I will try that.

#### aac044210

Joined Nov 19, 2019
178
Thanks Albert I will try that.
Ai = 4.78 which I guess is good.

Last edited:

#### crutschow

Joined Mar 14, 2008
27,213
I measure an AC Vs current of 409nApp and and AC Q2 emitter current of 2.16μApp, which a significant current gain.

#### aac044210

Joined Nov 19, 2019
178
I measure an AC Vs current of 409nApp and and AC Q2 emitter current of 2.16μApp, which a significant current gain.
Hi crutschow:

ie / ivs = 1.08 * 10-6 / 2.05 * 10-7 = 5.3
Is that considered significant, I am fairly new to this stuff so I wasn't sure what to expect "Ai" wise.

aac

#### aac044210

Joined Nov 19, 2019
178
Hi crutschow:

ie / ivs = 1.08 * 10-6 / 2.05 * 10-7 = 5.3
Is that considered significant, I am fairly new to this stuff so I wasn't sure what to expect "Ai" wise.

I might as well run this buy you folks. I computed Zin ≈ 46KΩ and Zout ≈ 10Ω. Make sense?

aac

#### Jony130

Joined Feb 17, 2009
5,230
I computed Zin ≈ 46KΩ and Zout ≈ 10Ω. Make sense?
Well, in your circuit the Zin cannot by larger than RB||RE1 = 8.2kΩ||12kΩ = 4.87kΩ

#### aac044210

Joined Nov 19, 2019
178
Well, in your circuit the Zin cannot by larger than RB||RE1 = 8.2kΩ||12kΩ = 4.87kΩ
Zin = 4.6KΩ, the decimal got dropped. Is the Zout = 10Ω correct?

#### crutschow

Joined Mar 14, 2008
27,213
Is that considered significant,
It's determined by your circuit configuration and signal level.
At microamp/nanoamp current levels, the gain of a transistor drops significantly.
Use a higher signal level and a lower impedance load to see higher values of current gain.

What's the purpose of the circuit as I see no good reason to use a Darlington with those component values in that configuration?

#### aac044210

Joined Nov 19, 2019
178
It's determined by your circuit configuration and signal level.
At microamp/nanoamp current levels, the gain of a transistor drops significantly.
Use a higher signal level and a lower impedance load to see higher values of current gain.

What's the purpose of the circuit as I see no good reason to use a Darlington with those component values in that configuration?
I found this circuit online. I was just looking for a circuit that would illustrate the use of a darlington pair
in the emitter-follower configuration, with voltage divider bias.

#### Jony130

Joined Feb 17, 2009
5,230
Is the Zout = 10Ω correct?
Yes Zout ≈ 2re_2||RE = 2*5.53Ω||1kΩ ≈ 11Ω

I was just looking for a circuit that would illustrate the use of a Darlington pair
in the emitter-follower configuration, with voltage divider bias.
But for such a high load resistance there is no point in using the Darlington stage. Also, the voltage divider resistor was poorly selected. And the voltage stage gain is Av = RE/(2re2 + RE)

#### aac044210

Joined Nov 19, 2019
178
Yes Zout ≈ 2re_2||RE = 2*5.53Ω||1kΩ ≈ 11Ω

But for such a high load resistance there is no point in using the Darlington stage. Also, the voltage divider resistor was poorly selected. And the voltage stage gain is Av = RE/(2re2 + RE)
Hi Jony

I calculated Zin thus.

r'e1 = (25.86 mV / IE1) = 25.86mV / 0.05 mA = 522Ω
r'e2 = (25.86 mV / IE2) = 25.86mV / 5mA = 5Ω
 re = RE║RL = 1KΩ / 2KΩ = 667Ω ​

 βdc = (βdc1 * βdc2) + βdc1 + βdc2 = 10,200
Zin(base) = βdc * (r'e1 + r'e2 + re) = 10200 * (522Ω + 5Ω + 667Ω) = 12,178,800Ω
 Zin = RB║Zin(base) = 4,697Ω ?
I changed the load to 2KΩ.

Also, you mentioned that the designer used a poor voltage divider?

aac

#### Attachments

• 2.5 KB Views: 6
Last edited:

#### Jony130

Joined Feb 17, 2009
5,230
Also, you mentioned that the designer used a poor voltage divider?
Yes, because the designer does not take full advantage of a Darlington high current gain.

If Ie2 = 5mA than the Q1 base current will be in the range of 5mA/(100 * 100) = 0.5μA Therefore the voltage divider current can be in the range of 5μA or more.

So for Ve = 5V we can use RB1 = 300kΩ RB2 = 510kΩ or even RB1 = 680kΩ; RB2 = 1.2MΩ

And this simple procedure will increase the amplifier Zin value.

#### MrAl

Joined Jun 17, 2014
8,248
Hi:

This darlington emitter-follower circuit attenuates the output current rather than
increasing it. Not sure what is wrong.

iRL = 9.87 * 10-8 while ivs = 2.05 * 10-7

Any suggestions?

Thanks
Hi,

Your input resistance is about 5k and your output resistance is 10k, so it takes twice as much current on the input as it does produce on the output given a voltage gain of 1.
These configurations are usually used only when the output current is going to be greater than the input current, usually much greater. What this means is that you would not use a Darlington for this application, and what is more is that you would not want to use a Darlington for this application simply because you would not need the extra current
gain of the second transistor.

So the reduced current gain is by design not because the Darlington is not working.

#### crutschow

Joined Mar 14, 2008
27,213
Below is the simulation of the Darlington all by itself at a higher current:
The current-gain mean value is about 90,000 (yellow trace) with the variation due to the transistors' change in gain with collector current.

#### aac044210

Joined Nov 19, 2019
178
Below is the simulation of the Darlington all by itself at a higher current:
The current-gain mean value is about 90,000 (yellow trace) with the variation due to the transistors' change in gain with collector current.

View attachment 197651
Thanks!

#### aac044210

Joined Nov 19, 2019
178

I still have a question regarding the output impedance of the circuit.

re = RE║RL = 1KΩ ║ 2KΩ = 667Ω

Based on the sim vout / ieQ2 = 0.097529 volts / 0.000146294 amps = 667Ω

So is Zout not = 667Ω?

aac

#### crutschow

Joined Mar 14, 2008
27,213
So is Zout not = 667Ω?
No, you calculated the equivalent load resistance (1k in parallel with 2k).

The output impedance can be calculated as the difference in the output voltage for a given input voltage with a change in load current.
For example, this is shown below for a change in load current from 1mA to 1.1mA (0.1mA change)
The output voltage change is 4.95mV, giving an output impedance of 4.95mV/0.1mA = 49.5 ohms.

Last edited:

#### aac044210

Joined Nov 19, 2019
178
No, you calculated the equivalent load resistance (1k in parallel with 2k).

The output impedance can be calculated as the difference in the output voltage for a given input voltage with a change in load current.
For example, this is shown below for a change in load current from 1mA to 1.1mA (0.1mA change)
The output voltage change is 4.95mV, giving an output impedance of 4.95mV/0.1mA = 49.5 ohms.

View attachment 197700
Thanks. So is the equivalent load resistance really a "dc" quantity while output impedance is
an "ac" quantity? It makes sense that impedance is a dynamic quantity as it changes with the
output current and voltage.