Darlington won't turn off, pull down resistor too low of value

Thread Starter

Hamlet

Joined Jun 10, 2015
339
A couple of questions, no project, just fooling around on a breadboard:

I have found that Darlington transistors seem to perform well with little
base current, even 1uA can switch a quarter amp. Seems really low for
base current, even though I know that's what they do best.

Also, I'm having trouble keeping Darlington off, and I need 100 ohms
or less pull down to keep it off, yet my base resistor drive resister is much
higher. I am suspecting somehow I am coupling
with my florescent overhead lights... a head scratcher.
 

ci139

Joined Jul 11, 2016
1,696
the darlingtons have current transfer ratios in multiple thousands say 195x84 = 16380 e.g. for 2A output is required 122µA , for 250mA 15µA
+ there might be some leakage currents at high temperatures
+ there might be some grease/dirt/oil/dust/condensate that conducts enough to allow the µA range
+ if there are HF or voltage spikes the base or emitter can pick them up and keep the darlington "open"
_____________
http://www.paulvdiyblogs.net/2017/07/my-new-power-supply.html
https://www.onsemi.com/pub/Collateral/TIP140-D.PDF (a newer d/s) https://www.st.com/resource/en/datasheet/tip142.pdf
 

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Thread Starter

Hamlet

Joined Jun 10, 2015
339
1580407549814.png1580407549814.pngNot sure what is going on. I've tried several Darlingtons, even discrete, and all perform the same. Perhaps I have contamination, or some kind of interferance. I'm going to try another multi-meter next.
 

Thread Starter

Hamlet

Joined Jun 10, 2015
339
What is the voltage at the base with only the 330 ohm connected?
Zero, but wait: I disassembled my breadboard before going to bed. This morning, I reassembled the circuit,
using the same componets, and now it appears to be correctly operating as expected. Full brightness with 130uA, off when disconnected, no pull-down resistor of any value required. I must have miss-wired something, or maybe a UFO was hoving nearby
last evening...
 

ci139

Joined Jul 11, 2016
1,696
the ammeter connected immediately to the base is not good to measure darlington -- rather you should have two paired ammeters or a resistor having the exact value of the (µA-m) equivalent resistance so you use voltage divider (a Pot-m) and measure sourced current from Vcc and subtract the sinked current at Vee so the base must be sinking the result
as http://tinyurl.com/wfu5tyu
 
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Thread Starter

Hamlet

Joined Jun 10, 2015
339
the ammeter connected immediately to the base is not good to measure darlington -- rather you should have two paired ammeters or a resistor having the exact value of the (µA-m) equivalent resistance so you use voltage divider (a Pot-m) and measure sourced current from Vcc and subtract the sinked current at Vee so the base must be sinking the result
i don't understand, but my guess is the lead on the ammeter/base can act as an ant., collecting spurious signals. I should measure voltage drop across a small value base resister, and use that to calculate current. Yes?
 

ebeowulf17

Joined Aug 12, 2014
3,274
Zero, but wait: I disassembled my breadboard before going to bed. This morning, I reassembled the circuit,
using the same componets, and now it appears to be correctly operating as expected. Full brightness with 130uA, off when disconnected, no pull-down resistor of any value required. I must have miss-wired something, or maybe a UFO was hoving nearby
last evening...
Those darned UFOs get me sometimes too! :D
 

ci139

Joined Jul 11, 2016
1,696
almost (see the simulation) . . . either
you should have 2 identical Ampere meters (( not good option you should adjust the "0" of one DMM - but if there are linearity errors it matches only 0 and the maximum of the selected range ))
or - Ampere meter and a resistor (of the same resistance than Ammeter) (( bad option → the resistance of the DMM may have current dependence))
or - you should know the difference of the two ammeter indications along the entire selected (say 200µA) range (( good option because it's doable . . . ))
. . . as - connect 2 DMM-s in series for 9V chose series resistance 39kΩ+5kΩ + 2MΩ potentiometer make the table of the DMM readings . . . etc.
 

ci139

Joined Jul 11, 2016
1,696
if you want to take base I-V characteristic then it's almost the only way . . .
the Vx below is given by the currents of the Base voltage divider as ::
\[ V_X=\frac{I_3}{I_2}·\left({I_1·R-V_1}\right)=\frac{I_1·R-V_1}{\frac{I_1}{I_3}-1}\ ,\ R=R_1+R_3=Const.\ ,\ I_1=I_2+I_3 \]
by any means . . . IT IS NOT A MUST
.
Darlington - IV.png
 
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BobTPH

Joined Jun 5, 2013
2,579
The circuit you have drawn is a darlington configured as a voltage follower. This is not the circuit you would use as a switch.

Move the load to be between the collector and V+. Remove the pot. Change the base resistor to 10 or 100K.

Now you can experiment with a
darlington as a switch.

Bob
 
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