Home  www.playhookey.com  Thu, 11232017 

Direct Current

Alternating Current

Semiconductors

Digital

Logic Families

Digital Experiments

Computers

 Analog  Analog Experiments  Oscillators  Optics  HTML Test  


The Basics

Variations in Feedback Circuits

Mixing Analog and Digital Technologies

Generating Waveforms

 Operational Amplifiers  Current Mirrors  Differential Amplifiers  


The Basic Current Mirror

Errors in Current Mirrors

The Buffered Current Mirror

Adding Emitter Resistors

 The Cascode Current Mirror  The Wilson Current Mirror  PNP Current Mirrors  FET Current Mirrors  Active Loads  
The Buffered Current Mirror 

As we mentioned on the page discussing errors in current mirrors, the basic current mirror circuit shown to the left has, among other problems, a mismatch between I_{REF} and I_{O} caused by the unavoidable presence of base current for the two transistors flowing through R. With highgain transistors the difference is small and sometimes negligible, but sometimes it must be recognized and minimized.
A simple way to reduce the base current flowing through R is to add a third transistor, Q3, to the circuit as shown to the right. Now the base current for Q1 and Q2 is also the emitter current of Q3, and only the base current for Q3 must still flow through R. This effectively reduces the base current error by a factor of (β + 1).
Adding Q3 to the circuit does change I_{REF} slightly, since Q1 no longer has its collector and base connected together. Instead, we have an extra V_{BE} to account for. For this circuit, I_{REF} = (+V  2V_{BE})/R. This is not normally a problem, but it can be in lowvoltage circuits.
Assuming Q3 has the same parameters as Q1 and Q2, the base current for Q3 will be 2I_{B}/(β + 1). Applying this change, the mathematical expression that relates I_{O} to I_{REF} becomes:
I_{REF}  =  I_{C1}  +  2I_{B} 


β + 1  
I_{REF}  =  I_{C1}  +  2I_{C1} 


β(β + 1)  
I_{REF}  =  I_{C1}  +  2I_{C1} 


β² + β  
I_{REF}  =  (β² + β)I_{C1}  +  2I_{C1} 



β² + β  β² + β  
I_{REF}  =  β² + β + 2  I_{C1}  


β² + β  
I_{O} = I_{C2} = I_{C1}  =  β² + β  I_{REF}  


β² + β + 2  
For transistors with a gain of 100, the ratio is 10,100/10,102 = 0.9998. This is a much smaller error than the percentage variation between the transistors themselves, even within an IC. Older transistors with a gain as low as 30 will produce a ratio of just under 0.9979, which is still only about 0.2% and can usually be neglected.
If multiple mirror transistors are driven from a single reference, the only part of the equation that changes is the "+ 2" constant. As with the basic mirror circuit, that number simply reflects the number of transistors whose base current will flow through Q3. In almost all cases, this change can be neglected.
Because the output transistor is not affected in any way by the addition of Q3, its output resistance is likewise unaffected. Thus, the problems associated with this output resistance and its changes as V_{CE} changes will remain exactly as they were in the basic current mirror circuit.


 
All pages on www.playhookey.com copyright © 1996, 20002015 by
Ken Bigelow Please address queries and suggestions to: webmaster@playhookey.com 