# I am trying to understand amplifier gain in transistor.

#### ashokraj

Joined Feb 1, 2018
109
Here in this circuit LED is placed at the collector. i am with an impression that, IE=IB+IC. so it is better to keep at the emitter, right ? all the gain of the transistors are not utilized.

What is the purpose of LED, placing at collector? Please educate me

#### MrChips

Joined Oct 2, 2009
23,495
You have no current limiting resistors. You are going to blow the LED and transistors.

#### Alec_t

Joined Sep 17, 2013
12,060
If the LED were in the emitter path the voltage available to drive it would be 3V- 3 x Vbe = 3V - 3 x 0.6V = 1.2V, which would not be enough to light it. (Vbe is the base-emitter junction voltage drop). As indicated above, there should be at least one current-limiting resistor in the circuit.

#### ashokraj

Joined Feb 1, 2018
109
You have no current limiting resistors. You are going to blow the LED and transistors.
how much resistance do i need to keep how can i calculate ?

#### ericgibbs

Joined Jan 29, 2010
12,883
hi A,
Using a 3V battery and say a RED LED with a 2v forward drop at say 10mA, thats 1v/0.01A =100R.

E

#### Bordodynov

Joined May 20, 2015
2,865
If the LED were in the emitter path the voltage available to drive it would be 3V- 3 x Vbe = 3V - 3 x 0.6V = 1.2V, which would not be enough to light it. (Vbe is the base-emitter junction voltage drop). As indicated above, there should be at least one current-limiting resistor in the circuit.
You're making this too easy. The Base Emitter voltage depends on the current. This voltage for the second transistor will be less than the third one by 120 mV. That's the voltage for the first transistor, which will be even lower. Also, the voltage on the first transistor may be greater than the power supply.

#### Bordodynov

Joined May 20, 2015
2,865
See