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Question: What is the purpose of the current mirror? Answer: It allows to combine the current changes in BOTH transistors (T1, T2). Explanation In a classical differential amplifier (with separat...
Answer
#3: Post edited
by
LvW
·
2021-06-28T09:57:43Z (almost 3 years ago)
- **Question**: What is the purpose of the current mirror?
- **Answer**: It allows to combine the current changes in BOTH transistors (T1, T2).
**Explanation** In a classical differential amplifier (with separate collector resistances for T1 and T2) we must use the difference between the collector voltages or currents for further prodessing - when we want to use the full dynamic capabilities of the amplifier.This can be accomplished using the shown current mirror with simultaneous double-ended into single-ended conversion.- Resulting from a differential input signal the collector currents of T1 and T2 change their values in the opposite direction. However, the current mirror does not "allow" such a change because it tries to keep the currents equal.
Therefore, the difference is available as an **output current iout** which drives the next stage.- So we have:
- ic1=ic3+2ib and ic2=-ic1=ic4+iout;
- With ic4=ic3 this gives:
- **iout=-2(ic1+ib)**
- **Question**: What is the purpose of the current mirror?
- **Answer**: It allows to combine the current changes in BOTH transistors (T1, T2).
- **Explanation** In a classical differential amplifier (with separate collector resistances for T1 and T2) we must use the difference between the collector voltages or currents for further processing - when we want to use the full dynamic capabilities of the amplifier.
- This can also be accomplished using the shown current mirror - with simultaneous double-ended into single-ended conversion.
- Resulting from a differential input signal the collector currents of T1 and T2 change their values in the opposite direction. However, the current mirror does not "allow" such a change because it tries to keep the currents equal.
- Therefore, the difference between the currents is available as an **output current iout** which drives the next stage, which makes the current-to-voltage conversion.
- So we have:
- ic1=ic3+2ib and ic2=-ic1=ic4+iout;
- With ic4=ic3 this gives:
- **iout=-2(ic1+ib)**
#2: Post edited
by
LvW
·
2021-06-28T08:43:07Z (almost 3 years ago)
- **Question**: What is the purpose of the current mirror?
Answer: It allows to combine the current changes in BOTH transistors (T1, T2).**Answer:** In a classical differential amplifier (with separate collector resistances for T1 and T2) we must use the difference between the collector voltages or currents for further prodessing - when we want to use the full dynamic capabilities of the amplifier.- This can be accomplished using the shown current mirror with simultaneous double-ended into single-ended conversion.
**Explanation**: Resulting from a differential input signal the collector currents of T1 and T2 change their values in the opposite direction. However, the current mirror does not "allow" such a change because it tries to keep the currents equal.- Therefore, the difference is available as an **output current iout** which drives the next stage.
- So we have:
- ic1=ic3+2ib and ic2=-ic1=ic4+iout;
- With ic4=ic3 this gives:
- **iout=-2(ic1+ib)**
- **Question**: What is the purpose of the current mirror?
- **Answer**: It allows to combine the current changes in BOTH transistors (T1, T2).
- **Explanation** In a classical differential amplifier (with separate collector resistances for T1 and T2) we must use the difference between the collector voltages or currents for further prodessing - when we want to use the full dynamic capabilities of the amplifier.
- This can be accomplished using the shown current mirror with simultaneous double-ended into single-ended conversion.
- Resulting from a differential input signal the collector currents of T1 and T2 change their values in the opposite direction. However, the current mirror does not "allow" such a change because it tries to keep the currents equal.
- Therefore, the difference is available as an **output current iout** which drives the next stage.
- So we have:
- ic1=ic3+2ib and ic2=-ic1=ic4+iout;
- With ic4=ic3 this gives:
- **iout=-2(ic1+ib)**
#1: Initial revision
by
LvW
·
2021-06-28T08:40:55Z (almost 3 years ago)
**Question**: What is the purpose of the current mirror? Answer: It allows to combine the current changes in BOTH transistors (T1, T2). **Answer:** In a classical differential amplifier (with separate collector resistances for T1 and T2) we must use the difference between the collector voltages or currents for further prodessing - when we want to use the full dynamic capabilities of the amplifier. This can be accomplished using the shown current mirror with simultaneous double-ended into single-ended conversion. **Explanation**: Resulting from a differential input signal the collector currents of T1 and T2 change their values in the opposite direction. However, the current mirror does not "allow" such a change because it tries to keep the currents equal. Therefore, the difference is available as an **output current iout** which drives the next stage. So we have: ic1=ic3+2ib and ic2=-ic1=ic4+iout; With ic4=ic3 this gives: **iout=-2(ic1+ib)**