0
\$\begingroup\$

Is calling open collector output as high impedance due to the value of pull up resistor or due to the fact that the IC itself do not source any current?

And what determines the upper limit value of the pull-up resistor used for an open collector output?

\$\endgroup\$
  • \$\begingroup\$ Do you know how it "looks" on the schematic? What is the impedance of a transistor "switched off" ? \$\endgroup\$ – Eugene Sh. May 3 '19 at 16:10
  • \$\begingroup\$ Yes I know how it works basically sjc1.discourse-cdn.com/digikey/uploads/default/… I just wonder why called high z. \$\endgroup\$ – floppy380 May 3 '19 at 16:12
  • \$\begingroup\$ Impedance of MOSFET measured between which terminals? \$\endgroup\$ – floppy380 May 3 '19 at 16:14
  • \$\begingroup\$ Sorry, MOSFET is about open-drain, this one is open-collector, hence BJT. When the transistor is cut-off, the output can be seen as disconnected from whatever it is connected to. \$\endgroup\$ – Eugene Sh. May 3 '19 at 16:15
  • \$\begingroup\$ Yes disconnected but what is the reasoning between being disconnected and being called high impedance. I guess Im stuck at that point. \$\endgroup\$ – floppy380 May 3 '19 at 16:18
1
\$\begingroup\$

An open-collector output can be thought of as a switch connected to ground. When the switch is open, it has a high impedance. When the switch is closed, it has a low impedance.

So, when the transistor is conducting, the output is low, and it can sink current from whatever the load is, limited by the ratings of the transistor. But when the transistor is cut off, the output is high, and any current supplied comes through the pullup resistor.

The pullup resistor must be selected to meet the needs of the load. Current through the pullup also adds to the current through the transistor when it is on, so you mustn't overload it — but this is rarely an issue.

\$\endgroup\$
  • \$\begingroup\$ “When the switch is open, it has a high impedance”. What is meant by impedance in that context? What is high? Is it the output impedance of IC or is it the pull-up resistor value?? That’s the point Im stuck. More the the term’s reference. \$\endgroup\$ – floppy380 May 3 '19 at 16:41
  • \$\begingroup\$ The impedance between the pin and anything else inside the IC. It is not related to the pull-up. \$\endgroup\$ – Eugene Sh. May 3 '19 at 16:48
0
\$\begingroup\$

I think it is best to show by example from calculations using a datasheet;

MMBT3904 Specs

Collector Cut-Off (x) Current \$ I_{CEX} = 50nA @ V_{CE} = 30V, V_{EB(OFF)}= 3.0V\$

Therefore \$Rcex= 30~ V ~/~ 50~ nA= 0.6~ GΩ \$

Thus turn-off or rise time \$τ=R\cdot C\$ with stray + load capacitance C

\$τ=0.6 ~GΩ \cdot C\$

But turn-on or fall time with same load and time constant formula is ;

from datasheet :
Collector-Emitter Saturation Voltage
\$V_{CE(SAT)}=0.20V ~ @ I_C = 10mA, I_B= 1.0mA \$
\$V_{CE(SAT)}=0.30V~ @ I_C= 50mA, I_B= 5.0mA\$

thus @ Vce(sat) \$ R_{ce(sat)}= 0.2V/10mA= 20 ~Ω\$

\$τ=20Ω⋅C\$

Therefore the choice of R depends on rise/fall time required, PD during Rce sat, Ic/Ib=10 unless it is a super beta which ma be slower.

Looking further at the datasheet for switching characteristics using a standard test config. , if you want, you can estimate the intrinsic junction capacitances, but this is the limiting case.

For slow turn off times, reducing Vce may also increase Rcex.

For practical values of R pullup, often you "may" find 100 Ohm , 1k 10k, 100k , 1M depending on the application, speed, power and load C requirements. But when used as a current limiting switch then using Ohm's Law with Rce*Ic+Vce(sat) you can choose your Rc value according to left-over voltage drop across Rc.

Since the value of \$r_{ce}~ or ~R{cex}\$ or Rce(sat) is small, it is often neglected, this term is rarely used, but becomes important when you exceed the test Ic levels towards Ic max for Vce(sat).

Diodes Inc use the value Rce in their super-beta switch transistors, but for all OEMS, you can estimate it as I have shown.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.