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I have a voltage divider configuration for biasing my transistor. The gain of transistor is lets say 150 which is typical value for BC547.

What I want is limit the current flowing through the base to be 10 micro amps.

Doing the Thevenin's analysis gives me the Thevenin Voltage(Vth) of 6V and the Thevenin resistance (Rth) of 500 ohms. So I redraw the circuit with Rth in series with the transistor and the input voltage being Vth=6V. The B-E on voltage is 0.7V. So doing the calculations gave me the current of 10.6 milli amps.

Instead, I choose to go with the design in figure 2. Since I know that the voltage divider produces the voltage of 6V, keeping a resistor of 530 kohms does the job of limiting the current through the base to 10 micro amps.

schematic

simulate this circuit – Schematic created using CircuitLab

So my questions are:

  1. Why do we calculate the Thevenin's resistance and keep it in series with the transistor? We can create a voltage divider to create a voltage of 0.7V and feed it directly into the base. Why don't we do that?

  2. Can't we just calculate the voltage from voltage divider(since the current going into the base is very less) and place the resistor of our choice in series? This would enable me to change the current going into the base just by applying Ohm's law and changing the resistor accordingly. Why bother with Thevenin when Ohm's law is simple enough?

Looking for conceptual explanation rather than mathematical one.

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  • \$\begingroup\$ Side note: having the emitter tied directly to ground like this will make your amplifier SUPER sensitive to things like temperature variation and component-to-component variation. You'd be wise to add in an emitter resistor to make things more stable and consistent. \$\endgroup\$ – Mr. Snrub Sep 6 at 19:48
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If you were going to do what you proposed in Figure 2, then why bother with the divider at all? Just use a resistor connected to +V with the appropriate value to produce the desired bias base current.

But this is BIASING, not just sending a lone current through the base-emitter. Biasing implies you might end up having another signal ride on top of (superimposed onto) the bias current something (i.e. using a DC-blocking capacitor to couple an AC signal onto your bias current). A series resistor like Rb will muck that up. This is why you don't want anything resembling a series resistor like Rb there.

To directly answer question #1: The voltage across base-emitter won't a diode junction of exactly 0.7V (and neither will your resistors) so that won't work in real life. You're probably more interested in the current flowing through the base-emitter junction anyways than the voltage across it. With a diode in the base-emitter junction, little changes in voltage make huge changes in current so it's not really a good way to try and control the voltage across the junction when you're interested in the current.

Using a divider with Thevnin to analyze it lets you come at it from the perspective of current through the base-emitter while not actually having a series base resistor which would interfere with any AC coupling you might attempt.

My analog knowledge is not well-developed and very rusty, but I think that's pretty much the gist of it.

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  • \$\begingroup\$ Rth is the input impedance of the transistor and I am calculating it by Thevenin's method of opening the load terminal. But when I don't have to impose any analog signals, why can't I ignore the Rth and simply say that the voltage at transistor's base is 6V? I know 6V is way too much from the diode's point of view. But still why can't I say its 6V and do not attach the Rth? \$\endgroup\$ – JuneStar_2918 Sep 7 at 3:28
  • \$\begingroup\$ @JuneStar_2918 I'm not sure what it is you're asking. You can't ignore it because it is there and has a significant effect. You are applying 6V to Rb with the base on the other side so the voltage at the base is less than 6V. There isn't 6V at the base. It's not the same thing. This is completely thindependent of superimposing anything. Having a divider instead of R lets you superimpose. Having Rth doesn't let you superimpose. That's all. It doesn't mean you can just ignore. \$\endgroup\$ – DKNguyen Sep 7 at 4:06
  • \$\begingroup\$ I meant to ask why can't I just take Vth in reduced circuit. Why should I take Rth in account? \$\endgroup\$ – JuneStar_2918 Sep 7 at 4:10
  • \$\begingroup\$ @JuneStar_2918 Because you can't. What makes you think you can ignore Rth to begin with? They are completely different circuits. 6V through Rth is not the same as 6V at the base. There is less than 6V at the base when Rth is there. \$\endgroup\$ – DKNguyen Sep 7 at 4:11
  • \$\begingroup\$ SInce the base is directly connected between those two resistors, I need to get that voltage at my base directly. Shouldn't I? \$\endgroup\$ – JuneStar_2918 Sep 7 at 4:16

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