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From: Rocky Stevens on 26 May 2010 19:19
From what I have read, the gate of a PUT is what determines its
threshold voltage, so you would control that with two resistors, one
between positive and the gate, and the other between the gate and
ground. I put together a circuit using 1K for both resistors, and got
a certain behavior (pulsing an LED). I then tried using 2 470K
resistors instead, and got a different behavior (basically, nothing).
My question is, shouldn't the voltage of the gate be solely determined
by the ratio of the two resistors, and not depend on the actual values
of the resistors? Or does the amount of *current* through the gate
effect something as well?

thanks
From: John Larkin on 26 May 2010 20:26
On Wed, 26 May 2010 16:19:12 -0700 (PDT), Rocky Stevens
<rocky.stevens(a)gmail.com> wrote:

>From what I have read, the gate of a PUT is what determines its
>threshold voltage, so you would control that with two resistors, one
>between positive and the gate, and the other between the gate and
>ground. I put together a circuit using 1K for both resistors, and got
>a certain behavior (pulsing an LED). I then tried using 2 470K
>resistors instead, and got a different behavior (basically, nothing).
>My question is, shouldn't the voltage of the gate be solely determined
>by the ratio of the two resistors, and not depend on the actual values
>of the resistors? Or does the amount of *current* through the gate
>effect something as well?
>
>thanks

The gate must see a low enough equivalent resistance for the internal
SCR positive feedback mechanism to snap. So yes, you do need gate
current, and you can't get enough if those resistors are too big.

John

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