In the figure, you could connect the transistor Q1 as a diode by shorting its collector with its base, but instead it is done by keeping both base and collector at ground.

The text says that if it were connected the other way (by shorting collector and base) then the base current would have caused an error (base current error.)

  • What type of error is the author talking about?
  • Why and how does the error manifest?

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1 Answer 1


If you short collector to base and use it as a diode between the opamp output and the inverting input then any base current has to flow from the input signal causing an error, especially at low currents where the HFE drops off.

In the arrangement shown the base current flows to ground and so the current flowing into the input is slightly less. This could cause an error of a few % at low currents.

The configuration shown can provide a logarithmic function over about 9 decades, a diode-connected version can work for 5-6 decades.

Here is some good information that has a brief mention of both types Log Amplifiers

  • \$\begingroup\$ What kind of changes can this error brings into the output voltage...? \$\endgroup\$
    – partykid
    Commented Oct 2, 2015 at 4:30
  • \$\begingroup\$ A few % of the input current would go into the base and cause that percentage of error. \$\endgroup\$ Commented Oct 2, 2015 at 23:26
  • \$\begingroup\$ For temparature compensation of the circuit Q2 is involved because involving will make the circuit output independent of the reverse leakage currents which are temparature dependent for temparature compensation both the transistor Vbe should be equal and opposite and the net difference of both voltages should be equal to output voltage When diodes are involved in compensation the equations are Vo=[25mVln(Iref/Is2)] - [25mVln(I1/Is1)] =25mV*ln(Iref/I1) Is diode leakage current I1 diode current Io Iref What are the equations when transistor are involved ?? I m not sure of the equations... \$\endgroup\$
    – partykid
    Commented Oct 3, 2015 at 13:52
  • \$\begingroup\$ why does the transistor-as-diode version of the log amp provide a wider range (9 decades) than the diode version (5-6 decades)? [edit: your link to the document at the end of the post is not working] \$\endgroup\$
    – wcc
    Commented Jul 16, 2019 at 1:46

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