0

In the transistor biasing using diode below assume that at time t1 the circuit is stable and apply KCL at the junction of the three devices you have: 

IR1 = IB1 + ID1

(the index 1 just to indicate it's at the time t1.)

enter image description here

Now let's assume that after some time for example at t2 for some reason VBE suddently decreases and this causes IB and ID decrease as a result. However, IR = (VCC - VBE)/R increases.

Thus,

IB2 < IB1
ID2 < ID1
IR2 > IR1

However, KCL at the node at t2 should also be right meaning that:

IR2 = IB2 + IR2

This doesn't seem to be possible as both IB2 and IR2 decrease while IR2 increases compared to their values at t1.

So what is wrong with this reasoning?
How can KCL be satisfied here?

emnha
  • 1,547
  • 4
  • 34
  • 84
  • Your question can be reduced, without losing its effect, by just having a resistor in series with a diode. Not so? – jonk May 01 '22 at 06:05
  • @jonk yes, right. How would you explain it? – emnha May 01 '22 at 06:13
  • 1
    So the answer is pretty simple. The only way to reduce the voltage is to have an effect that causes a reduced voltage where the increased current in the resistor also makes sense. A temperature change (higher, for example, with a diode) lowers the difference across it for the same current. But what really happens then is that the current increases and therefore also the voltage across the diode increases, too. They meet "in the middle" where the equilibrium state once again makes sense. And any attempt to force the situation means you've added something else to the node.) – jonk May 01 '22 at 06:17
  • @jonk ` But what really happens then is that the current increases and therefore also the voltage across the diode increases, too.` I did a graphical method and see that the voltage across diode actually decreases but now the I-V of the diode is different so even VD decreases but ID increases. – emnha May 01 '22 at 06:37
  • You *must* have a mechanism to reduce the voltage. There are a number of them and all of them will work out in the end. But temperature is easy. For example, if you assume for a moment that the temperature increases by 10 K and that the diode's voltage difference is diminished (if held at the same current) by 40 mV, then you might say that the current increase by 40 mV divided by the resistor. But in increasing the current, the diode voltage difference increases because of it and counters, reducing the current change to something less than 40 mV/R, etc. It all works out fine. – jonk May 01 '22 at 06:44
  • I can write all this up pretty easily using the Lambert-W function for an R and diode. If you want to see a sample of how that proceeds, [look here](https://electronics.stackexchange.com/a/592785/38098). But it's also solvable, iteratively. The results are always similar. – jonk May 01 '22 at 06:47
  • @jonk thanks, I see it now from graphical method by plotting I-V characteristic of diode and resistor at tempterature t1 and when temperature increases to t2. – emnha May 01 '22 at 06:55

1 Answers1

1

Think about reasons why Vbe would decrease. It could be caused by lower base current as you suggest. The requirement for that is either RB increasing, VCC decreasing or Id increasing.

Vbe could also decrease with constant or increasing Ib for example when temperature of the transistor increases. This would lead to a lower diode current Id.

enter image description here

Source: Researchgate

Elliot Alderson
  • 31,192
  • 5
  • 29
  • 67
Lars Hankeln
  • 2,513
  • 1
  • 8
  • 15
  • How would you explain it if Vbe decreases due to increasing temperature? In that case, Vcc and RB does not change. – emnha Apr 30 '22 at 11:25
  • In that case IR and IB increase and ID decreases. – Lars Hankeln Apr 30 '22 at 13:00
  • Can you explain how ID decreases? If Vbe decreases due to increasing temperature then VD or ID decreases but how IB increases? BE is like a diode so if VBE decreases should IB also decreases? – emnha Apr 30 '22 at 19:32