{"id":858,"date":"2014-11-14T01:13:40","date_gmt":"2014-11-13T23:13:40","guid":{"rendered":"http:\/\/blog.the-leviathan.ch\/?p=858"},"modified":"2015-01-04T03:26:52","modified_gmt":"2015-01-04T01:26:52","slug":"current-mirror","status":"publish","type":"post","link":"https:\/\/blog.the-leviathan.ch\/?p=858","title":{"rendered":"Current mirror"},"content":{"rendered":"<p>Given, a complicated darlington circuitry:<br \/>\n<a href=\"https:\/\/blog.the-leviathan.ch\/wp-content\/uploads\/2014\/11\/darlington.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-861 size-medium\" src=\"https:\/\/blog.the-leviathan.ch\/wp-content\/uploads\/2014\/11\/darlington-231x300.png\" alt=\"darlington\" width=\"231\" height=\"300\" srcset=\"https:\/\/blog.the-leviathan.ch\/wp-content\/uploads\/2014\/11\/darlington-231x300.png 231w, https:\/\/blog.the-leviathan.ch\/wp-content\/uploads\/2014\/11\/darlington.png 660w\" sizes=\"auto, (max-width: 231px) 100vw, 231px\" \/><\/a><br \/>\nWe now replace the darlington pairs with single NPN transistors by defining: $$I_C=I_B \\cdot a$$ with $$a=(\\beta+1)\\cdot\\beta$$ and $$I_E=I_B \\cdot b$$ with $$b=(\\beta+1)^2$$<br \/>\n<a href=\"https:\/\/blog.the-leviathan.ch\/wp-content\/uploads\/2014\/11\/darlington_replacement.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-896 size-medium\" src=\"https:\/\/blog.the-leviathan.ch\/wp-content\/uploads\/2014\/11\/darlington_replacement-225x300.png\" alt=\"darlington_replacement\" width=\"225\" height=\"300\" srcset=\"https:\/\/blog.the-leviathan.ch\/wp-content\/uploads\/2014\/11\/darlington_replacement-225x300.png 225w, https:\/\/blog.the-leviathan.ch\/wp-content\/uploads\/2014\/11\/darlington_replacement.png 454w\" sizes=\"auto, (max-width: 225px) 100vw, 225px\" \/><\/a><\/p>\n<p>Now we set:<br \/>\n$$I_{KD} = I_{B1} + I_{C3} = I_{B1} + I_{B3} \\cdot a$$ and $$I_{RL}=I_{C4}=I_{B4}\\cdot a$$ and $$I_{B2}=I_{B5}$$ and $$I_{KD} + I_{C1} + I_{RL} = I_{E5} + I_{E2} $$<br \/>\n$$I_{KD} + I_{B1} \\cdot a + I_{RL} = I_{B5} \\cdot b + I_{B2} \\cdot b = 2\\cdot I_{B5}\\cdot b = 2\\cdot I_{B2}\\cdot b$$<br \/>\n$$\\Rightarrow I_{RL}=- a \\cdot I_{KD} &#8211; I_{KD} + a^2 \\cdot I_{B3} + 2 \\cdot b \\cdot I_{B2}$$<\/p>\n<p>And resolve the current equation:<br \/>\n$$I_{B3} \\cdot b = I_{E3} = I_{C5}+I_{B2}+I_{B5}$$<br \/>\n$$=I_{C5}+2 \\cdot I_{B2}$$<br \/>\n$$=I_{C5}+2 \\cdot I_{B5}$$<br \/>\n$$=a \\cdot I_{B5}+2 \\cdot I_{B5}=(a+2)\\cdot I_{B5}=(a+2)\\cdot I_{B2}$$<\/p>\n<p>$$\\Rightarrow I_{B2} = I_{B5} = \\frac{1}{a+2} \\cdot I_{E3} =\\frac{b}{a+2} \\cdot I_{B3}$$<br \/>\n$$I_{E1} = I_{B1} \\cdot b = I_{B3}+I_{B4}$$<\/p>\n<p>$$I_{C2} = I_{B2} \\cdot a$$<br \/>\n$$= I_{E4}=I_{B4} \\cdot b$$<span id=\"MathJax-Element-14-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-301\" class=\"math\"><span id=\"MathJax-Span-302\" class=\"mrow\"><span id=\"MathJax-Span-326\" class=\"msubsup\"><span id=\"MathJax-Span-328\" class=\"texatom\"><span id=\"MathJax-Span-329\" class=\"mrow\"><span id=\"MathJax-Span-331\" class=\"mn\"><\/span><\/span><\/span><\/span><\/span><\/span><\/span><br \/>\n$$\\Rightarrow I_{B4} = \\frac{a}{b}\\cdot I_{B2}=\\frac{a}{b}\\cdot\\frac{b}{a+2} \\cdot I_{B3}$$<span id=\"MathJax-Element-14-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-301\" class=\"math\"><span id=\"MathJax-Span-302\" class=\"mrow\"><span id=\"MathJax-Span-326\" class=\"msubsup\"><span id=\"MathJax-Span-328\" class=\"texatom\"><span id=\"MathJax-Span-329\" class=\"mrow\"><span id=\"MathJax-Span-331\" class=\"mn\"><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<p>$$I_{RL}=\\frac{a^2 b}{a^2 b + 2 a b +2}$$<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Given, a complicated darlington circuitry: We now replace the darlington pairs with single NPN transistors by defining: $$I_C=I_B \\cdot a$$ with $$a=(\\beta+1)\\cdot\\beta$$ and $$I_E=I_B \\cdot b$$ with $$b=(\\beta+1)^2$$ Now we set: $$I_{KD} = I_{B1} + I_{C3} = I_{B1} + I_{B3} \\cdot a$$ and $$I_{RL}=I_{C4}=I_{B4}\\cdot a$$ and $$I_{B2}=I_{B5}$$ and $$I_{KD} + I_{C1} + I_{RL} = I_{E5} &hellip; 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