As you said in my previous thread—->“But the more you drop across the ballast resistor the less overhead you have to work with. Everything is a compromise.”



my inference of above :The addition of emitter resistor can be advantageous as if we increase voltage drop at emitter resistor the lesser Vbe is(due to negative feedback)..and the LESSER we have to worry about Vbe since the MAGNITUDE of Vbe is now quite manageable and also the effects of Vbe mismatch are also less since the value of Vbe have decreased..

Is this what you meant by term “overhead”..

The magnitude of Vbe is going to be the same. But without it the voltage that sets the current is JUST Vbe and a small change in Vbe results in a very large change in current. With the ballast resistors, the voltage that sets the current is the sum of the voltage across the ballast resistor AND Vbe, so a small change Vbe is absorbed by the voltage change across the ballast resistor.

Let's take things to ridiculous extremes and make the ballast resistors 100kΩ and say that we want a load current of 10mA. That means that we will have a voltage drop of 1000V across the ballast resistors (let's ignore the 10W of power that they are dissipating and the resulting tempco issues that this would lead to in real life). Now let's assume that our Vcc is 1020V and that we have a 1kΩ load resistor in the collector path. The voltage applied to the base will be about 1000.7 V. What is the Vbe that this transistor needs to conduct 10mA changes from 0.7V to 0.8V, which is a significant change? Well, instead of having 1000V across the ballast resistor we will have 999.9V across it and our current will go down from 10 mA to 9.999 mA for a change of 0.01%.

I could swap out the transistor with a transistor that only needed a Vbe of 0.35V to produce 10 mA of current and 0.7V might correspond to 100 A of current, but that is irrelevant because the ballast resistor will now have about 1000.35 V across it at the collector current will increase to 10.004 mA, a change of less than 0.05%.

 

Do you agree with post #19

No, I do not; the addition of those resistors does NOT reduce Vbe.

Pay close attention to what WBahn is saying here:

The magnitude of Vbe is going to be the same. But without it the voltage that sets the current is JUST Vbe and a small change in Vbe results in a very large change in current. With the ballast resistors, the voltage that sets the current is the sum of the voltage across the ballast resistor AND Vbe, so a small change Vbe is absorbed by the voltage change across the ballast resistor.

There, he explains the effect of those emitter resistors about as succinctly as it can be explained.

 

To stop thermal runaway.

 

In discrete circuits thermal runaway usually is not a problem (notice I'm not saying that it is never a problem) except when you are paralleling transistors. This is because leadwire and trace resistance usually provides adequate degeneration resistance to prevent complete runaway.

But the point is a good one to make.

 

It's more typically called a compensation cap and is used to modify frequency response. It's not uncommon to split the emitter resistance into multiple resistors so you can tweak stage gain at multiple frequencies.

My teachers always referred to it as a decoupling capacitor, going back to the same reason when used in a tube amplifier.
Essentially preventing the reduction of the AC gain.
Max.

 

The magnitude of Vbe is going to be the same. But without it the voltage that sets the current is JUST Vbe and a small change in Vbe results in a very large change in current. With the ballast resistors, the voltage that sets the current is the sum of the voltage across the ballast resistor AND Vbe, so a small change Vbe is absorbed by the voltage change across the ballast resistor.

Let's take things to ridiculous extremes and make the ballast resistors 100kΩ and say that we want a load current of 10mA. That means that we will have a voltage drop of 1000V across the ballast resistors (let's ignore the 10W of power that they are dissipating and the resulting tempco issues that this would lead to in real life). Now let's assume that our Vcc is 1020V and that we have a 1kΩ load resistor in the collector path. The voltage applied to the base will be about 1000.7 V. What is the Vbe that this transistor needs to conduct 10mA changes from 0.7V to 0.8V, which is a significant change? Well, instead of having 1000V across the ballast resistor we will have 999.9V across it and our current will go down from 10 mA to 9.999 mA for a change of 0.01%.

I could swap out the transistor with a transistor that only needed a Vbe of 0.35V to produce 10 mA of current and 0.7V might correspond to 100 A of current, but that is irrelevant because the ballast resistor will now have about 1000.35 V across it at the collector current will increase to 10.004 mA, a change of less than 0.05%.

thanks for making it very lucid...

 

No, I do not; the addition of those resistors does NOT reduce Vbe.

Pay close attention to what WBahn is saying here:



There, he explains the effect of those emitter resistors about as succinctly as it can be explained.

thanks ..i ve got it

 

My teachers always referred to it as a decoupling capacitor, going back to the same reason when used in a tube amplifier.
Essentially preventing the reduction of the AC gain.

I've seen references to decoupling and bypass being used where I was taught that they were compensation caps. It seems that a lot of terms were being used "interchangeably" by instructors.

Do opamp datasheets refer to compensation as "decoupling" in your area?