Electronics > Beginners

Current mirror fail?

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Zero999:

--- Quote from: magic on December 08, 2019, 12:12:22 pm ---
--- Quote from: Zero999 on September 20, 2019, 08:52:34 pm ---The VBE must be matched without the transistors in the circuit.

Set the multimeter to diode test. Connect the base and collector to positive lead and the emitter to the negative lead. Repeat the process with several different transistors until a matched pair is found.

--- End quote ---
Folk wisdom has it that transistors from one production batch tend to match to 3mV. I have actually tested some small (5~20) lots of a few different transistors using exactly your technique and to my moderate surprise I found it to be true, except for super cheap no-name BC547/BC557s where maybe 33% were outliers but still within 10mV between the extremes.

--- End quote ---
You're probably right. I haven't tried this with transistors but have found LM334s from  the same bag are matched very closely. I made the circuit on page 8 of the data sheet and trimmed it to 50µA. I noticed the regulators from the same bag matched. The diodes will all from the same batch.
http://www.ti.com/lit/ds/symlink/lm334.pdf


--- Quote from: Lucky-Luka on December 08, 2019, 08:26:32 am ---Finally I've found time to improve my circuit as you suggested.
I didn't use emitter resistors but nonetheless the circuit performance improved.
Cheers

--- End quote ---
Good, I'm glad you got it working.

Yes, emitter resistors aren't completely necessary and they do help but do have their disadvantages, apart from more parts, the volt drop across the resistors reduces the maximum voltage across the load, before it stops working. In hindsight, 470R is overkill for 10mA, as it would cause a volt drop of 4.7V. I'd probably opt for 68R to 120R for a 680mV to 1.2V drop.

Yansi:
Or use a transistor pair dedicated for current mirror applications, such as BCV61, BCV62.

magic:
Note that the datasheet is silent on the question of internal thermal tracking and even the Early effect. They only guarantee 30% current match at modest 5V and 2.5mW power dissipation.

Lucky-Luka:

--- Quote from: w2aew on December 09, 2019, 11:58:24 pm ---Also keep in mind that you're not accounting for the burden voltage of the ammeter.  When you have the ammeter on the left side (where the current is set), the series resistance of the ammeter is added to the 1K resistor, which lowers the current.  When you move the ammeter to the LED side, the reference current on the left is higher because the ammeter's resistance isn't in series with the 1K resistor anymore.

--- End quote ---

It doesn't have anything to do with the post but I just wanted to say that whenever I've got spare time I watch your videos and I love them.
Cheers

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