1

I want to reduce a transistor's Vbe variation, but I can't find anything online, except some research papers that don't really answer the question.

Here is the test circuit:

enter image description here

Where the resistors have no tolerance, and the transistor can vary Vbe and β. Applying 50 Monte Carlo simulation, we see that the transistor's Vbe is varying a lot, from 870 mV to 905 mV.

enter image description here

Using resistors with a tolerance of 1% or higher only worsens the situation, of course. So for the test circuit, for isolating the transistor, I had imposed 0 drift.

Is there a method, using resistor networks, or an active circuit with other transistors, to reduce the Vbe variation without selecting the transistor? This can improve, for example, balancing between the two pair of a differential, or increase the precision of circuits a lot.

I already tried to use Sziklai (CFP), or Darlington, or a schematic similar to a constant-current generator with feedback like this:

enter image description here

But nothing worked. Is it an impossible task?

ocrdu
  • 8,705
  • 21
  • 30
  • 42
Brando
  • 33
  • 4
  • 1
    [This might help you understand what the options are](https://electronics.stackexchange.com/a/556905/20218) – Andy aka Jan 28 '23 at 11:37
  • Since R1 is as high as 10k ohms then supplies a base current of only (11V/10k=) 1.1mA but the saturated collector current is 12V/100 ohms= 120mA then the transistor cannot saturate. The datasheet says that for the transistor to saturate, the base current MUST be about 1/10th the collector current. Then 11V/12mA= 917 ohms for R1. hFE is used for an amplifier that always has plenty of Vce, not for a saturated switch. The very small change in Vbe is almost nothing. – Audioguru Jan 28 '23 at 16:22

2 Answers2

4

You can't change or modify the exponential relationship Ic = f(Vbe).

However, you can drastically reduce the influence of the Vbe uncertainty (for example: temperature induced variations) on the current Ic using negative feedback (for example, inserting an emitter resistor Re).

In this case, unwanted "variations" of Vbe are compensated by an opposite feedback voltage. These Vbe variations can be caused by tolerances of the divider chain at the base or by temperature effects. Moreover, when the divider chain is relatively high-resistive (bad design), also B-variations can cause such Vbe changes.

Example: When the temperature increases and Vbe must be reduced to compensate the ficticious Vbe increase (Ic increase), the voltage drop across Re (Ve) increases and makes Vbe somewhat smaller.

ocrdu
  • 8,705
  • 21
  • 30
  • 42
LvW
  • 24,857
  • 2
  • 23
  • 52
1

In this situation, you can think of your vbe variations as the error signal of your amplifier (no different from the V(+)-V(-) from an op-amp with negative feedback.).

This typical common-emitter amplifier uses series-series feedback.

You're simply suffering from a lack of loop gain. The consequence is that your loop gain isn't enough to suppress the error signal, therefore, you see significant variations in your vbe.

The only solution to this is to have more loop gain. Why don't you use a two stage amplifier?

Designalog
  • 3,232
  • 1
  • 13
  • 23
  • Can you please explain how the voltage which you call "error signal" [V(+)-V(-) in case of an opamp or Vbe for a BJT] will be "suppressed" as a result of negative feedback? Or do you mean that the INFLUENCE of variations of this differential voltage will be reduced? – LvW Jan 28 '23 at 14:15
  • There, vbe variations. – Designalog Jan 28 '23 at 14:24