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The datasheet says "Css is charged via a PNP transistor that is connected to the error amplifier." Does this mean that current flows from the emitter through the base?

I have drawn on this diagram to show what I think must be happening:

enter image description here

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    \$\begingroup\$ It appears to me that VREF also supplies current to CSS via RSS. That seems to be where much of the charging current arrives. The base current from the BJT adds to this. But due to the limited compliance current from the error amp, it is negligible. The purpose of the BJT is to "mess with" the error amplifier output before it reaches the PWM comparator. The error amplifier's maximum output current is just dumped to ground via the collector. Eventually, the voltage on CSS gets charged up and the BJT is effectively removed from the circuit and it stops screwing with the PWM comparator. \$\endgroup\$
    – jonk
    Dec 18, 2020 at 22:46
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    \$\begingroup\$ But I'm not sure (since you didn't provide a link) if VREF is shut down during certain phases of the operation. If so, then the base current could be the significant source while the error amplifier's compliance current is shorted to ground. Not sure. \$\endgroup\$
    – jonk
    Dec 18, 2020 at 22:50
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    \$\begingroup\$ @jonk What you said made everything click for me! The PNP conducts until Css is charged enough to drive the base high. Basically, this shorts the error amplifier output to ground, which slows the start-up of the whole converter. Thank you so much! \$\endgroup\$ Dec 18, 2020 at 23:03
  • \$\begingroup\$ Glad to hear it helped. I only wish I could have looked at the actual datasheet. I might have been able to be clearer than I was. I was just "guessing" a lot, here. \$\endgroup\$
    – jonk
    Dec 18, 2020 at 23:06
  • \$\begingroup\$ Here is a link to the datasheet: ti.com/lit/ds/symlink/ucc28c44-q1.pdf?ts=1607698658196 Page 12 has the functional block diagram, and page 18 has the soft-start diagram I took a screenshot of. \$\endgroup\$ Dec 18, 2020 at 23:11

2 Answers 2

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Does this mean that current flows from the emitter through the base?

It does, but it is negligible compared to the current supplied by Vref. Not only is the base current typically at least 100 times smaller than the collector current, the error amplifier can only source 1mA anyway. So the base current will be something less than 10µA.

At least based on the snippet you've provided, the datasheet is not claiming Css is charged via the PNP, only that the PNP is connected to Css. That connection actually affects the PNP more than it does the capacitor.

Specifically, while the capacitor is not charged, the PNP is hard on and sinks all the current from the error amplifier (dumping it out the collector to ground), effectively starving the feedback network and inhibiting the IC. As the capacitor charges, the PNP is pinched off and no longer sinks the current from the error amplifier. The error amplifier is then free to supply the feedback network and the IC begins normal operation.

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  • \$\begingroup\$ Thank you so much Heath; this really helped a lot! \$\endgroup\$ Dec 18, 2020 at 23:13
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Does this mean that Current flows from the Emitter through the Base?

No because this PNP will operate in active mode. That means that the current amplification:

\$\beta = \frac{I_C}{I_B}\$

still holds.

So if for example \$\beta = 100\$ then around 99% of the current will flow through the Collector and only around 1% will flow though the base.

Because I felt like it, I give you another example:

Here the NPN is used as a diode.

schematic

simulate this circuit – Schematic created using CircuitLab

Note how the current nicely divides so that \$\beta = \frac{I_C}{I_B}\$ remains valid.

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  • \$\begingroup\$ Ok gotcha. So a very small amount of current does flow through the base to charge the capacitor. Does the datasheet's explanation of what is going on make sense to you then? \$\endgroup\$ Dec 18, 2020 at 21:56
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    \$\begingroup\$ It looks to me like they use the current division of the amplifier (\$I_B << I_C\$) to charge the capacitor with a much lower current so that it takes more time and you get a slow startup without having to use a very large value capacitor. \$\endgroup\$ Dec 18, 2020 at 22:06

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