I came across the ICL7673 IC used for a battery backup system but I could not understand exactly what is the role of the R2 and R4 resistors in the circuit below (marked with red): enter image description here

So my question is: what role does the R2 and R4 resistors have? When Vp > Vs , P- is high and S- is low and when Vp < Vs, P- is low and S- is high. Since we have the R1 and R3 resistors to bias the base of the PNP transistors, why do we need the R2 and R4 resistors? If somebody could explain this to me (in detail) I would greatly appreciate it.


They're to prevent the transistors to conduct because of leakage current into the \$P-\$ and \$S-\$ outputs when they're supposed to be off. Any leakage current will go through the resistors and not the emitter-base junction as long as the voltage drop it causes is less than 0.7V.

If it were just the open-drain outputs and their pull-ups to define an output level the leakage would most likely be negligible. But now this current will be amplified by the transistor's \$H_{FE}\$, and that may be too high to be ignored.

At high temperatures the output leakage current can go to almost a \$\mu\$A. The transistor may amplify that to more than 100\$\mu\$A from your supply, which you don't want. If we pick a 10k\$\Omega\$ for R2 the leakage current will cause a voltage drop across it (and thus across the transistor's base-emitter junction) of maximum 10mV, which is way too low to get the transistor conducting.

Calculation of R1: If the main supply is 3V and your equipment needs 100mA you'll need at least 1mA of base current (many small signal transistors have an \$H_{FE}\$ of at least 100). Let's play safe and go for 5mA base current. Don't forget to check the current through R2: 0.7V/10k\$\Omega\$ = 70\$\mu\$A. OK, that's only 1% so we'll ignore that. Then R1 = (3V - 0.7V)/5mA = 460\$\Omega\$. (Yes, I know, I used other values in my comments, but with the given assumptions these should be better values).

  • \$\begingroup\$ On of my friends argued that the R2 and R4 resistors act as pull-ups since P- ans S- are open-collector outputs (shown in the functional diagram of the IC) and that without those pull-ups the PNP transistors won't turn on. Is this true? Here is the link with the functinal diagram: imageshack.us/photo/my-images/850/icl7673functionaldiagra.jpg \$\endgroup\$ – Buzai Andras May 11 '12 at 16:20
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    \$\begingroup\$ @Buzai - non quite. In a perfect world R2 and R4 wouldn't be required, while pull-ups are needed to define an output level. \$\endgroup\$ – stevenvh May 11 '12 at 16:25
  • \$\begingroup\$ Buzai It might work without the R pullup to shut it off, but with current gain in PNP and leakage in the IC Drain, dont take a chance that some stray electric field might trigger it. So it adds safety margin to turning it off.. Thats all. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 11 '12 at 16:27
  • \$\begingroup\$ Thank you both for your responses. I am really new to this wonderful world of electronics and I am trying to understand things as correctly as possible. So if I understand this correctly R2 and R4 are acting as pull-ups for the open-drain outputs but in the same time R2 and R4 are preventing the PNP transistors to turn on because of the amplified leakage current. Is this correct? Or the pull-ups for the open-drain outputs are in fact R1 and R3? Thank you for your patience with me :). \$\endgroup\$ – Buzai Andras May 11 '12 at 16:46
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    \$\begingroup\$ @Buzai - R2 and R4 are the pull-ups. They pull the base of the transistors to V+ if the open-drain FETs are off. Leakage in the MOSFETs may pull the base voltage down a little, but if R2 and R4 are small enough (say 1k) the voltage will never be so high that the transistor will start to conduct. R1 and R3 are there to limit the base current when the MOSFETs are on. \$\endgroup\$ – stevenvh May 11 '12 at 16:53

THe IC7673 is an Open Drain ( like Open Collector) driver. All loads which require pull-up need R to pullup. In this case pullup from the base is more direct and effective than the drain. It also affect the transition time ON to OFF or low to high my making it faster and more reliable.

this in the off state = open drain.

(steven's comments are also true)


To supplement the comments above, I redrew the hysteresis circuit with external PNP transistor for better understanding

enter image description here


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