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Old 10th March 2015, 10:57 PM   #1
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Default Complementary transistor matcher?

Does this circuit actually work? I tried testing a BC560 with Hfe of 585 against a BC550 with Hfe of 509 and didn't get either of the LEDs to light.
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Old 10th March 2015, 11:12 PM   #2
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Which Vs did you try? R1 and R2 look like LED pull-down values from +6V.
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Old 11th March 2015, 01:33 AM   #3
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Seems that the leds would only light if there were a rather large mismatch since the voltage mismatch has to be 1.6-1.7 before they light. A variation of 9-10 mA would be necessary across 180 ohms. Probably a ballpark tester for output transistors at best. The one with the anode of the led attached to it's collector would be the low gain one of the pair.

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Old 11th March 2015, 01:55 AM   #4
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Quote:
Originally Posted by Willi Studer View Post
Which Vs did you try? R1 and R2 look like LED pull-down values from +6V.
The suggested Vs was 10v but I tried 5 and 12. Neither did anything.
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Old 11th March 2015, 02:47 AM   #5
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A test jig to do small signal transistors. Scale would be 1uA per 1mV. Having a pair of transistors that are with in 10mV on all 3 settings would be a fairly good match. R4/R5 and R6/R7 should be closely matched to each other. Intended power supply is a 9 volt battery.
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Old 11th March 2015, 04:57 AM   #6
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RNMarsh had posted this on some thread.

Gajanan Phadte
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Old 11th March 2015, 10:41 AM   #7
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I thought about putting an instrumentation opamp in place of the LEDs. With enough gain I could measure small differences.

What do you think about that idea?
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Old 11th March 2015, 11:59 AM   #8
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A DMM will do the job, better.

Gajanan Phadte
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Old 11th March 2015, 04:01 PM   #9
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Another question: I don't see how any current is going to flow in both bases in any case but I measured 2.5 volts across the 30k base resistor. What is causing current to flow between the two bases?
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Old 11th March 2015, 05:44 PM   #10
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Look at Q1 base and notice that Q1 is a PNP transistor. And to open PNP transistor we need a closed path for the Q1 base current from base to ground. As you can see Q1 emitter is tied to Vs via R1. So the only path from Q1 base to ground is via R3 and Q2 base-emitter junction.
And also for NPN transistor to open we need a closed path from Vs. So eventually the current will flow in this circuit:

Vs ---> R1 ---> Q1 base-emitter junction ---> R4 ---> Q2 base-emitter ---> GND. And this current will turn ON both transistors.
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