Consider the following Common Collector & Darlington emitter follower circuits,

CASE 1:
Let V(in) = 5V
Vcc(supply) = 10V
Emitter(load) resistance = 100 ohm
thus, V(out) should be 5V (following input)
Corresponding Load current will be 5/100 = 50 mA

CASE 2:
Let V(in) = 5V
Vcc(supply) = 10V
Emitter(load) resistance = 33 ohm
thus, V(out) should be 5V (following input)
Corresponding Load current will be 5/33 = 152 mA

Which source will provide this additional current requirement [Vcc or V(in)] ?

 

Hello,

The main current will be delivered by the Vcc.
Keep in mind that the emitter follower will have a lower output voltage as Vin,
due to the voltage drop of the base - emitter of the transistors.
In the case of the darlington circuit, you will even have 2 X as much voltage drop.

Bertus

 

Hello,

The main current will be delivered by the Vcc.
Keep in mind that the emitter follower will have a lower output voltage as Vin,
due to the voltage drop of the base - emitter of the transistors.
In the case of the darlington circuit, you will even have 2 X as much voltage drop.

Bertus

That is V(out)= 5V - V(be) for CC
V(out)= 5V - 2*V(be) for Darlington

And I have to set the current limit of supply to above 152mA and don't bother about base current even if Load current= beta * Base current for CC and Load current= 2*beta * Base current for Darlington.

 

Hello,

Have a look at the following page of the eBook:
http://www.allaboutcircuits.com/vol_3/chpt_4/6.html

Bertus

 

I want the output should exactly be 5V not (5-junction drops). So that I have to add extra junction drops to source volts. But my source is fixed at 5V.

Shall I design one unity gain Op Amp summer with one 5V and other fixed .7 or 1.4V inputs to feed the emitter follower ?

 

Let an opamp drive an emitter-follower (a darlington is not needed for your low output current).
The opamp should be non-inverting with a gain of one.
Take the negative feedback from the emitter of the emitter-follower then the output voltage will be the same as the input voltage.

 

What is the purpose of C3? To bypass supply spikes ?

 

C3 helps prevent oscillation of the opamp.
EVERY electronic circuit should have a supply bypass capacitor on the pcb.

 

Let an opamp drive an emitter-follower (a darlington is not needed for your low output current).
The opamp should be non-inverting with a gain of one.
Take the negative feedback from the emitter of the emitter-follower then the output voltage will be the same as the input voltage.

What would be the output if I apply 0.5V which is less than junction drop 0.7V to Op Amp input ? Can I get same 0.5V at emitter ?

 

What would be the output if I apply 0.5V which is less than junction drop 0.7V to Op Amp input ? Can I get same 0.5V at emitter ?

The output voltage of the opamp is higher than the emitter of the transistor by about 0.7V.

If you use an opamp that has inputs that work at a very low voltage and an output that goes down close to ground in that circuit like a modern rail-to-rail opamp or an opamp in an LM358 dual or LM324 quad then the output will be +0.5V when the input is +0.5V.

But most ordinary opamps will need a dual-polarity supply (positive and negative) for their inputs to work close to 0V and an output that goes close to 0V.

 

If you use an opamp that has inputs that work at a very low voltage and an output that goes down close to ground in that circuit like a modern rail-to-rail opamp or an opamp in an LM358 dual or LM3234 quad then the output will be +0.5V when the input is +0.5V.

But most ordinary opamps will need a dual-polarity supply (positive and negative) for their inputs to work close to 0V and an output that goes close to 0V.

LM324,OPA 341, OPA 2341, TLCV2772, TLC2774, LM358, LM3234--- Which one is better?

 

YOU must decide what is the most important specification for an opamp then YOU must look through all the datasheets for the opamp that has it.

 

Below is a complementary emitter follower circuit that cancels most of the offset (determined by the relative base-emitter voltage of the two transistors). The simulation shows an output offset of about 120mV at 5V input.



Edit: Note that if you want the output voltage to go near zero with the saturation point at the top of the graph instead of the bottom, just reverse the position of the NPN and PNP transistors as shown below.

 

Let an opamp drive an emitter-follower
Take the negative feedback from the emitter of the emitter-follower then the output voltage will be the same as the input voltage.

I have setup the Op Amp-Current booster circuit as shown below using LM324 & 2N2222 for loads 33/68/100 Ohms one at a time.

But the output voltage is not following the input as shown below:

INPUT(40mA max)// EMITTER VOLTAGE// DEVIATION //SUPPLY(10V/200mA max)

0 // 0 // 0 // 10,< 100mA

0.5 // 0.4 // 0.1 // 10,< 100mA

1 // 0.9 // 0.1 // 10,< 100mA

1.5 // 1.3 // 0.2// 10, 100mA

2 // 1.8 // 0.2 // 10, 100mA

2.5 // 2.3 // 0.2// 10, 100mA

3 // 2.7 // 0.3// 10, 100mA

3.5// 3.2 // 0.3// 10, 100mA

4// 3.6 // 0.4 // 10, 200mA

4.5 // 4 // 0.5 // 10, 200mA

5 // 4.3 // 0.7 // 9.8, 200mA

5.5 // 4.3 // 0.7 // 9.8, 200mA

6 // 4.3 // 0.7 // 9.8, 200mA

6.5 // 4.3 // 0.7 // 9.8, 200mA

7 // 4.3 // 0.7 // 9.8, 200mA

7.5 // 4.3 // 0.7 // 9.8, 200mA

8 // 4.3 // 0.7 // 9.8, 200mA

8.5 // 4.3 // 0.7 // 9.8, 200mA

9 // 4.3 // 0.7 // 9.8, 200mA

9.5 // 4.3 // 0.7 // 9.8, 200mA

10 // 4.3 // 0.7 // 9.8, 200mA

Anomalies:
1. Output is not following the input
2. After 5V input, the output is struck at 4.3V
3. Supply voltage dips to 9.8V when input is >= 5V
4. Deviation incraeses as the input voltage and set at 0.7 level
5. The measurement is repeated for inputs set at 10mA max & 100mA max. But the issue continues.
6. Can I use complementary emitter follower as suggested by Crutschow for my above application by leaving the Op Amp?

 

Your circuit is not working properly probably because you swapped the emitter and collector pins of the transistor. Then it has a very low current gain and the opamp output is limiting the current by reducing the output voltage.

Normally the transistor has plenty of current gain then the output current of the opamp is low and it can reduce the error to be very very small.

 

................................
6. Can I use complementary emitter follower as suggested by Crutschow for my above application by leaving the Op Amp?

The complementary follower will have higher offset (typically in the neighborhood of 100mV or less) as compared to the op amp circuit, where the offset is basically equal to the op amp input voltage offset value (typically a few mV but low offset op amps have less than 1mV). So if having an offset much less than 100mV is important, then use the op amp circuit.

 

Your circuit is not working properly probably because you swapped the emitter and collector pins of the transistor.

Normally the transistor has plenty of current gain then the output current of the opamp is low and it can reduce the error to be very very small.

1. OK, I will check it and report on this forum
2. Can I use the same Op Amp-2N2222 settings for all three loads 33/68/100 ohms only by setting the supply to its max requirement of above three cases i.e, 10V, 160mA current limit for all the cases.
Thus I can simply swap loads without any settings change while getting output which follows input.

 

1. OK, I will check it and report on this forum
2. Can I use the same Op Amp-2N2222 settings for all three loads 33/68/100 ohms only by setting the supply to its max requirement of above three cases i.e, 10V, 160mA current limit for all the cases.
Thus I can simply swap loads without any settings change while getting output which follows input.

You don't have to set a current limit on the supply unless you are trying to protect the load or the supply from overcurrent.

 

Something is wrong with your 10V power supply since its output drops to 9.8V when it has a small load current. But it will not affect the Opamp-2N2222 circuit.

Why do you say the input is 10mA, 40mA or 100mA? The input to what?
The input bias current of the LM324 opamp is only 0.25uA maximum.

If you disconnect the output load resistor while changing its value then the output voltage might float higher until the next load resistor is connected.

 

Something is wrong with your 10V power supply since its output drops to 9.8V when it has a small load current. But it will not affect the Opamp-2N2222 circuit.

Why do you say the input is 10mA, 40mA or 100mA? The input to what?
The input bias current of the LM324 opamp is only 0.25uA maximum.

If you disconnect the output load resistor while changing its value then the output voltage might float higher until the next load resistor is connected.

1.I am using linear power supply to power the circuit

2. I am using 1ms single pulses from monoshot(with 0.5v stepped out) as the input to the op-amp and monitoring voltage at emitter on CRO.
To simulate the monoshot, I gave inputs from an exponential power supply manually with current limits set at 10mA, 40mA or 100mA.

3. I am changing the loads only after turning off the circuit not online. So I think I can use the same settings for any loads without any adjustment.

Can OP-AMP & transistor can track the 1ms pulse change?