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I was reviewing/following a tutorial describing the pros/cons of various transistor base biasing configurations. Configuration #4 is shown above. Resistor Rb2 was added to what was previously Configuration #3. The circuit with the text is shown below (in quotes)....

Why does adding Rb2 increase stability with respect to variations in Beta, ( β ) by increasing the current flowing through the base biasing resistors? The unclear (to me) original text is below. Thanks

"Adding an additional resistor to the base bias network of the previous configuration improves stability even more with respect to variations in Beta, ( β ) by increasing the current flowing through the base biasing resistors.

The current flowing through RB1 is generally set at a value equal to about 10% of collector current, IC. Obviously it must also be greater than the base current required for the minimum value of Beta, β.

One of the advantages of this type of self biasing configuration is that the resistors provide both automatic biasing and Rf feedback at the same time."

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Why does adding Rb2 increase stability with respect to variations in Beta

That's not difficult to see. Adding Rb2 would "steal" some current from the base of the NPN, so to prevent that we decrease the value of Rb1 such that it provides extra current.

Now if the base current is 1 uA and we make 100 uA flow through Rb1 that leaves 99 uA for Rb2. If now for some reason beta is halved, the base current would become 2 uA. So now 98 uA flows through Rb2. Thatś not much of a difference now is it ?

Compare that to the situation where Ib = 1 uA but Rb1 provides only 2 uA so for Rb2 thereś only 1 uA left. Now if beta halves there would be zero current left for Rb2. That would not actually happen of course, it would settle somewhere in the middle.

But notice how by "wasting" current through Rb1, Rb2 I can basically ignore what happens to the base current and therefore beta as well.

For small signals adding Rb2 also has an advantage as Rb2 with Rb1 forms a voltage divider controlling how much of the output signal is fed-back.

Without Rb2 there will only be the internal small signal input resistance of the NPN, it has value beta/gm. Note how beta is in there again !

By adding Rb2 and making it much lower value than beta/gm Rb2 "takes over" and allows us to have more control and also making the influence of beta smaller.

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  • \$\begingroup\$ Thanks for this help, but Can you elaborate just a little on "for some reason beta is halved, the base current would become 2 uA. So now 98 uA flows through Rb2."? I would think that beta can be halved because of, say, temperature, or a new transistor. If this were the circumstance, the collector current should be cut in half and Vc should double. If Vc doubles, than the current through the base biasing side should also double? \$\endgroup\$ – Jeffrey Edward Messikian Sep 11 '17 at 16:07
  • \$\begingroup\$ Wouldn't the current, formerly 100ua, through R1 go up to 200ua? If everything doubles on the base biasing side, I can see how 1ua goes to 2ua through the base. I'm fairly new to this, but I don't see the connection from Beta being halved to Rb2 changing by only 1 ua. I would think that it should change similar to the current through R1, which I would think is doubled? If not, can you explain how the current changes on the base biasing side when Vc doubles? \$\endgroup\$ – Jeffrey Edward Messikian Sep 11 '17 at 16:08
  • \$\begingroup\$ No, there is no reason why the 100 uA would become 200 uA unless that 100 uA is the full base current and beta halves. But of that 100 uA only 1 uA goes into the base ! In my explanation Vc remains the same, otherwise things would get complex. Maybe we would need to change Rc to keep Vc constant between beta and half beta but that does not change the situation at Rb1, Rb2 and base which is what I was discussing. A change in Vc complicates things further. But Vc will not change much anyway due to the feedback loop involving Rb1, Rb2. \$\endgroup\$ – Bimpelrekkie Sep 11 '17 at 16:51
  • \$\begingroup\$ I'm all for keeping it simple, Vcc can remain constant. Just focusing on Rb1, Rb2 and base current, and increased stability from Rb2. If for some reason beta is halved, How does the base current become 2 uA instead of the 1ua? I see that this is double, but what causes the doubling? where does the current come from? I understand that if B is halved, the collector current would be cut in half. If you wanted to keep the same level of collector current, you would need to double the base current. I'm just wondering the actual cause of the base currents doubling? \$\endgroup\$ – Jeffrey Edward Messikian Sep 11 '17 at 18:17
  • \$\begingroup\$ beta is Ic/Ib so assuming Ic is constant halving beta means doubling Ib. In electronics there are often many interdependencies. Taking them all into account all at once (what most beginners do) leads to confusion and no solution. So what experienced EEs do is to assume other parameters to be constant (here Ic) and see what happens: Ib doubles. Then you can go back and check, if Ib doubles can Ic remain constant ? Well we saw that Vc will not change. So Ic will not change due to Rc, so Ic does remain fairly constant even if beta halves. \$\endgroup\$ – Bimpelrekkie Sep 11 '17 at 19:41

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