# Art Of Electronics 2.10B

New to electronics and looking to deepen my understanding of things by going through the Art Of Electronics. I've hit figure 2.10B and it's amusing statement:

"... only ~0.6mA of the 4.4mA collector current comes from R3 - make sure you understand why"

Here I am, asking for a clarification/guidance on whether or not I've understood why.

1. My first thought was. 0.6V is needed on the base of Q3 to turn it in, so if that voltage exists, a current must also exist but couldn't think how to reach 0.6mA from this. Could this thought lead to an answer or am I off course?

2. My second attempt to analyse it was prompted from reading this similar Q&A: PNP Circuit-The Art of Electronics

I decided to view it as a simpler circuit with two voltage sources, where the second is Vcc after deducting the 0.6 for Veb:

Which can be sperated into separate circuits and then the currents solved for each resistor:

This seemed to make sense, with the current over R2 becoming (0.015 - 0.014 = 0.0006 = 0.6mA). This meant that the absense of V2 had a short over R1 so there was no contribution from V1, but this didn't align with the text saying that 0.6mA is contributed by V1.

Am I along the right tracks - and/or have I been clear enough to get some guidance?

You kind of have the right and wrong idea. Simulation wise you can't do what you're doing because while placing a voltage supply across a component does force the voltage across that component, it also adds POWER to the circuit which is not what happens in reality.

Figure 2.10. B, The Art of Electronics, Third Edition, page 77

Now look at this:

simulate this circuit – Schematic created using CircuitLab

What is the current through $$\R_3\$$ here?

If $$\V_{be} = 0.7V\$$ then $$\I_{R_3} = \frac{V_{R_3}} {R_3} = 0.7mA\$$ because $$\R_3\$$ must have 0.7V across it.

What's the current running through $$\R_2\$$?

If we neglect Q2's saturation voltage (I'm just going to assume 0V here) then the voltage across R2 must be whatever is left over from the 15V supply after the above is $$\V_{R_2} = 15 - 0.7V = 14.3V\$$. That means the $$\I_{R_2} = 4.3mA\$$

Which makes sense because $$\V_{be}\$$ is also conducting current. How much is it conducting? The portion of 4.3mA that is not going through $$\R_3\$$ which is a majority of it.

I'm not sure why the book is using 0.6V for $$\V_{be}\$$ instead of 0.7V.

EDIT: It was proposed that 0.6V was used because the authors had a power PNP in mind rather than a small signal PNP.

• @DKNguyen I think the authors are considering a PNP power switch, instead of a small signal PNP. For those, the base-emitter is often closer to 600 mV than 700 mV because their saturation currents are so much higher. The authors pretty much know their stuff. Chances are that's what was in their head. And writing about various cases, they figured, would just confuse more than help.
– jonk
Commented Apr 20, 2020 at 20:06
• @jonk Ah I see. Commented Apr 20, 2020 at 20:06