I'm talking in general terms. In a given circuit you have things that combine additively and things that combine multiplicatively. If what you want to cancel can be combined multiplicatively with something else, that is when you look for something that will have a reciprocal relationship or, if you have something that has a reciprocal relationship to what you want to cancel, you look for ways to combine it multiplicatively with what it is you want to compensate for. I am NOT saying that it is easy or even possible to do it either way for any given circuit or parameter.

Saturation doesn't mean that if Ib changes that Ic doesn't change. It merely means that Ib no longer has the strong, largely linear effect on Ic that it has in saturation.

It is a constant current source ONLY when the transistor is in the active region. Once that transistor saturates or gets cutoff then it is no longer a constant current source.

why it is always advised to keep Ve a bit large (lets say 1V) to minimize effects of Vbe in a constant current source circuit,in order to increase its performance?

 

why it is always advised to keep Ve a bit large (lets say 1V) to minimize effects of Vbe in a constant current source circuit,in order to increase its performance?

I'm not following. Where is it advised to keep the emitter voltage at 1V (or so)?

Are you sure you aren't talking about keeping Vce above a minimum voltage that is comfortably above Vcesat?

 

I'm not following. Where is it advised to keep the emitter voltage at 1V (or so)?

Are you sure you aren't talking about keeping Vce above a minimum voltage that is comfortably above Vcesat?

Pls refer the attachment..
The text of attachment is taken from the book called the 'art of electronic 'by horowitz and hill..

 

Pls refer the attachment..
The text of attachment is taken from the book called the 'art of electronic 'by horowitz and hill..

It's hard to be sure because you don't provide the circuit that is being discussed by that paragraph.

My guess is that they are talking about putting in a ballast resistor that has about 0.1V across it at the intended current level. Even that much will result in a significant improvement, though not nearly as much as if you drop more voltage across the ballast resistor. But the more you drop across the ballast resistor the less overhead you have to work with. Everything is a compromise.

 

It's hard to be sure because you don't provide the circuit that is being discussed by that paragraph.

My guess is that they are talking about putting in a ballast resistor that has about 0.1V across it at the intended current level. Even that much will result in a significant improvement, though not nearly as much as if you drop more voltage across the ballast resistor. But the more you drop across the ballast resistor the less overhead you have to work with. Everything is a compromise.

Please find the circuit in attachment...
also please try to elaborate its working ..

 

The description given in the test you cited is pretty decent. Having someone else give yet one more description is going to be of little help in terms of aiding you in understanding the circuit and its performance. You need to get your pencil dirty and start analyzing the circuit.

Presumably this transistor has a Vbe of 0.65V when Ic is 1 mA and Vce is 0.75V.

What is the load current under these conditions?

Now let's change the temperature such that Vbe is 0.55V at that same current and value of Vce. What will the load current be given that Ic changes by a factor of 10 for every 60 mV change in Vbe?

Now do the same calculation for a similar circuit but where there is no emitter resistor and the voltage divider chain is adjusted to lower both of the transistor base voltages by 1V.

Note that the top transistor is there to act as a current buffer between the current source transistor and the load. This is called a "cascode" configuration. The purposes if to present the current source transistor with a stable collector voltage so as to largely eliminate the effect of the Early effect (output resistance) of the current source transistor.

 

The description given in the test you cited is pretty decent. Having someone else give yet one more description is going to be of little help in terms of aiding you in understanding the circuit and its performance. You need to get your pencil dirty and start analyzing the circuit.

Presumably this transistor has a Vbe of 0.65V when Ic is 1 mA and Vce is 0.75V.

What is the load current under these conditions?

Now let's change the temperature such that Vbe is 0.55V at that same current and value of Vce. What will the load current be given that Ic changes by a factor of 10 for every 60 mV change in Vbe?

Now do the same calculation for a similar circuit but where there is no emitter resistor and the voltage divider chain is adjusted to lower both of the transistor base voltages by 1V.

Note that the top transistor is there to act as a current buffer between the current source transistor and the load. This is called a "cascode" configuration. The purposes if to present the current source transistor with a stable collector voltage so as to largely eliminate the effect of the Early effect (output resistance) of the current source transistor.

I ll get back to you once i get the concept....

 

The description given in the test you cited is pretty decent. Having someone else give yet one more description is going to be of little help in terms of aiding you in understanding the circuit and its performance. You need to get your pencil dirty and start analyzing the circuit.

Presumably this transistor has a Vbe of 0.65V when Ic is 1 mA and Vce is 0.75V.

What is the load current under these conditions?

Now let's change the temperature such that Vbe is 0.55V at that same current and value of Vce. What will the load current be given that Ic changes by a factor of 10 for every 60 mV change in Vbe?

Now do the same calculation for a similar circuit but where there is no emitter resistor and the voltage divider chain is adjusted to lower both of the transistor base voltages by 1V.

Note that the top transistor is there to act as a current buffer between the current source transistor and the load. This is called a "cascode" configuration. The purposes if to present the current source transistor with a stable collector voltage so as to largely eliminate the effect of the Early effect (output resistance) of the current source transistor.

Trying to make it very simple….

Vbe=.65V then Ic =1mA and Vce=0.75V ,I load=1mA.

when Vbe=0.55V then Ie1=1.1mA (Q1 emitter current),Ic1=1.1mA(Q1 collector current),and Vce now=.65V(only if we assume Vc of Q1 to 1.75) but infact this cant be possible becoz its a constant current source and thus to regulate the value of Iload Vce should change appropriately. So the collector voltage of Q1 should then be 1.65V in order to compensate change in Vbe to maintain Iload to constant value… but my inference dont agree to your quote in which you are telling to change Vbe =0.55V for same Ic and Vce…why?

also
Since as the temperature changes Vbe of Q2 should change ..and that should change Ic of Q2 (i.e I load) and in return Ie of Q2 also .Then how can we say that the collector voltage of Q1 i.e. 1.75 V in this case is held fixed by Q2 emitter ??