Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Trending News

Promoted
treadnotonme
Lv 5
treadnotonme asked in Science & MathematicsEngineering · 8 years ago

Why does a Darlington circuit need 1.4v instead of 0.7?

It is my understanding that each Bipolar Junction Transistor (BJT) in a Darlington configuration needs 0.7 volts input at it's base in order to switch on.

Everything I read says that for a Darlington pair, the input of the first transistor needs to be 1.4v (0.7 times two, one for each transistor).

I'm sure it's because I just don't understand the circuit/process, but it seems to me that 0.7 to the front transistor should turn that transistor on. Once it's on, the current passing through it would provide the necessary 0.7v to turn on the second transistor. So it seems to me that 0.7 at the first base should turn on the entire Darlington pair, but like I say, the docs I've found says you have to have 1.4v.

Can someone explain what I'm missing in my understanding?

Thanks!

Update:

Ahh.... the first transistor has no path to ground until the second transistor turns on. Therefore, it's like the emitter of the first is just dangling loose in the air, disconnected, and whether I put 0.7 volts, or 1.4, or 20, or 110 volts to the base of the first, without a path to ground, the first transistor will never turn on without that emitter-to-ground path.

So, all the voltage to turn on both the first transistor and the second must come through the base of the first. I had been thinking all that was needed was the required 0.7 volts to the base of the first, and then whatever else that's needed for the base of the second (another 0.7) could come through the collector-emitter path of the first. But until that second base is turned on, the collector-emitter path of the first doesn't exist. So ALL the base voltage for BOTH of the first and second transistor has to come through the base of the first transistor. Thus, 1.4 instead of 0.7.

Now I get it. Thanks Nathan!

1 Answer

Relevance
  • Anonymous
    8 years ago
    Favorite Answer

    The diodes are in series. 0.7 volts will not turn either diode on. If you have n diodes in series it takes n*0.7 Volts to turn them 'on'.

Still have questions? Get your answers by asking now.