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I am reading through "Op Amps for Everyone" by Ron Mancini of Texas Instruments and I can't understand why they have made an assumption:

The specifications for the amplifier are an ac voltage gain of four and a peak-to-peak signal swing of 4 volts.

This is the circuit they give:

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IC is selected as 10 mA because the transistor has a current gain (β) of 100 at that point. The collector voltage is arbitrarily set at 8 V; when the collector voltage swings positive 2 V (from 8 V to 10 V) there is still enough voltage dropped across RC to keep the transistor on. Set the collector-emitter voltage at 4 V; when the collector voltage swings negative 2 V (from 8 V to 6 V) the transistor still has 2 V across it, so it stays linear. This sets the emitter voltage (VE) at 4 V.

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I can understand this fine. Next step.

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This is also fine. However I don't understand the next bit of text:

We want the base voltage to be 4.6 V because the emitter voltage is then 4 V. Assume a voltage drop of 0.4 V across RTH, so Equation 2–35 can be written. The drop across RTH may not be exactly 0.4 V because of beta variations, but a few hundred mV does not matter is this design. Now, calculate the ratio of R1 and R2 using the voltage divider rule (the load current has been accounted for).

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Why did they assume a voltage drop of 0.4V across the Thevenin resistor? If it an assumption, then why say it will be close to this value? How did they arrive at the value of 0.4V?

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The 0.4 V across the Thevenin resistor (R1//R2) is a somewhat arbitrary design choice based on the emitter voltage. It is the voltage drop caused by the Ib of the transistor because it loads the voltage divider made by R1 and R2.

Choosing more than 0.4 V would result in higher values for R1 and R2 resulting in a larger voltage drop caused by Ib resulting in ending up with a different emitter voltage than the 4 V we wanted.

And likewise a lower value would result in less voltage drop.

I think they just used 10% of the emitter voltage (4V) as the value. That makes sense as the 0.6 V drop from \$V_{be}\$ is relatively constant. And if the emitter voltage was lower, 1 V for example, you would want the voltage drop due to \$I_b\$ to be smaller as well to prevent ending up too far away from the initial calculations.

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  • \$\begingroup\$ I am not sure I understand. If we chose a larger value than 0.4, you are right, there would be a larger drop across Rth (with the same Ib), but Vth would also be larger (because its defined by R1 and R2); so Vb would still be 4.6V and thus the emitter voltage remains unchanged.... \$\endgroup\$ – Tim Mottram Feb 22 '16 at 15:43
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    \$\begingroup\$ "so Vb would still be 4.6V" that is not true ! Just assume R1 and R2 have a value of several Mega ohms. The current through R1 would be much larger (due to the base current) than the current through R2. This would result in more voltage drop across R1, making Vbase much lower than 4.6 V. It is all about making the current through R1 and R2 significantly larger than Ib so that we can ignore Ib. \$\endgroup\$ – Bimpelrekkie Feb 22 '16 at 15:55
  • \$\begingroup\$ Now that I understand. I can see that if I want Ib to be 0.1mA I need to make the resistors R1 and R2 sufficiently small to allow enough current to pass; but that only sets an upper limit of their size. It doesn't explain why they have picked 0.4V. \$\endgroup\$ – Tim Mottram Feb 22 '16 at 16:00
  • \$\begingroup\$ Indeed is not explained in the text, so it is just my guess that they chose 0.1 * Ve = 0.4 V as a design constraint. Often this choice comes from experience. I can see that this makes sense, it is just unfortunate that it was not explained properly. \$\endgroup\$ – Bimpelrekkie Feb 22 '16 at 16:06
  • \$\begingroup\$ Yep, OK, thank you for your help, I am trying to model it now.To get a feeling of how the values of R1 and R2 will effect the result. \$\endgroup\$ – Tim Mottram Feb 22 '16 at 16:07

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