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I need a simple exponential converter that with 9V and allows me to control a VCO. I've tried the one the schematics below but it needs bipolar power supply and just doesn't work. I have plenty of transistors and op amps, a simple circuit would be appreciated!

VCO

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enter image description here Hello, I developed this exponential converter for my filter, but it's usable for other filters or oscillators, too. The resistors R3 and R6 limit the maximum output current. Feel free to use this expo-converter for your own device. It's a kind of poor men's converter.

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Worth reading this paper: http://electronotes.netfirms.com/s019.pdf

The circuit in the picture works well in simulation.

enter image description here

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There are a series of questions from JackDamery which relate to what you want - it's probably worth reading through these to get the jist of what worked for him. The problem is not the VCO of course but the log convertor.

At one point I suggested using a lower power dc-to-dc convertor to produce split rails so maybe this can be considered. Other than that it boils down to using digital methods i.e. reading a pot voltage into an ADC, converting to digital, applying a log function digitally and reproducing an analogue voltage output that is the log of the input voltage.

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  • \$\begingroup\$ Thank you. Digital is the last thing I want. All I need is an antilog amplifier. \$\endgroup\$
    – alkopop79
    Commented Jul 18, 2013 at 12:44
  • \$\begingroup\$ @alkopop79 I gave you the option - either use a dc-to-dc convertor to get split rails or use digital. \$\endgroup\$
    – Andy aka
    Commented Jul 18, 2013 at 14:50
  • \$\begingroup\$ How about JFETs? It seems to me that they have a more linear response as opposed to transistors. \$\endgroup\$
    – alkopop79
    Commented Jul 18, 2013 at 15:17
  • \$\begingroup\$ I've tried split rail and it didn't work. I've built the exact same circuit and it didn't work though I used 9V (-4.5V to 4.5V). \$\endgroup\$
    – alkopop79
    Commented Jul 18, 2013 at 15:20
  • \$\begingroup\$ @alkopop79 not enough voltage I think \$\endgroup\$
    – Andy aka
    Commented Jul 18, 2013 at 17:17

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