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!
3 Answers
Worth reading this paper: http://electronotes.netfirms.com/s019.pdf
The circuit in the picture works well in simulation.
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\$ Commented Jul 18, 2013 at 12:44
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\$\begingroup\$ @alkopop79 I gave you the option - either use a dc-to-dc convertor to get split rails or use digital. \$\endgroup\$– Andy akaCommented Jul 18, 2013 at 14:50
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\$\begingroup\$ How about JFETs? It seems to me that they have a more linear response as opposed to transistors. \$\endgroup\$ Commented Jul 18, 2013 at 15:17
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\$\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\$ Commented Jul 18, 2013 at 15:20
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\$\begingroup\$ @alkopop79 not enough voltage I think \$\endgroup\$– Andy akaCommented Jul 18, 2013 at 17:17