Current To Voltage Converter Circuit Using Lm358

I didn't say it won't work, I said it might not work.The circuit as configured is running from 3.3V. The differential amp has a gain of about 30. At 1A, that is 0.1V x 30 = 3V. Right there, the output of IC1a cannot go above about 1.8V.The output of IC1b is meant to drive the sense line of the buck regulator high, so the regulator will lower the output thinking it is too high.

But with a max of 1.8V output and a 0.6V drop across D1, it is only about 1.2V maximum. Possibly worse, as fast recovery diodes sometimes drop a bit more voltage than the garden variety silicon rectifier diodes. Many regulators use 1.25V as the reference voltage and so the Sense line is at 1.25V already.If you raise the Op Amp supply voltage to 5V, it might work. And replace D1 with a shottky diode.But it isn't as simple as just looking at the specs and saying it will work. You need to consider operation of the circuit.

Voltage To Current Signal Converter The battery voltage = 12 volts. In the first circuit The op amp works as a current to voltage converter and its. . Input Common-Mode Voltage Range Includes electronics without requiring the additional ±15V Ground power supplies. Differential Input Voltage Range Equal to the The LM358 and LM2904 are available in a chip sized Power Supply Voltage package (8-Bump DSBGA) using TI's DSBGA. Large Output Voltage Swing package technology.

It certainly could work with changes to the circuit, both have about the same offset voltage and Vcc requirements. It is the output range where the LM358 has an issue with this circuit.The MCP602 isn't really that expensive. The eBay link the author gives is only $4 for 5 with shipping, about 80 cents each.Correction: I see a problem with this circuit as designed, for either IC.

The common mode input range for the MCP602 is V+ - 1.2V, and for the LM358 is V+ - 1.5V. So the Op Amp really needs a supply voltage higher than 12V, or the circuit redesigned. Right now the common mode voltage with 12V on the current shunt is 12 x 100k/(100k + 3.3k) = 11.6V. I have a similar design in my linear power supply. I use a 0R01 instead of 0R1, a 1K instead of 3K3 and run the LM358 from +5. Works for me. Then again, I am not using this buck/boost module.I always use 3A schottky's everywhere reverse leakage is not a problem.

I have about a hundred left!He has a schottky symbol in his schematic, but uses a fast diode part.Its not the cost, LM358 is available everywhere. Microchip parts are hard to get in developing countries and shipping costs are prohibitive.

I think that would be a problem for some. I have to of these exact buck-boost modules. I have used one to keep the LED spotlights on my workbench at optimal brightness when the solar drops below 11.5v after a long day, and it works fabulous.But I believe I believe there is a simpler add-on circuit that should do the job.

Here is a link to the schematic and the article. Duralast jump starter instructions. These are used to set a constant current source on a vacuum tube's plate supply.

Hi haven't tried them with the Drok, but see no reason they shouldn't work.http://www.diyaudioprojects.com/Technical/Current-Regulator/.

Hi friends, in this post we will see how to convert voltage into current using simple circuitry. In most of the cases we get the output of measuring devices in the form of voltage. It is not good to transmit this output voltage to the destination directly. Due to addition of noise and wire impedance the output voltage may get corrupted. So in such cases we have convert that voltage into current form. So let us see voltage to current converter.

Voltage to Current Converter using Op Amp

Following circuit shows the voltage to current converter using operational amplifier. It consist of simple resistance connected to the inverting and non inverting terminals of op amp.

In this circuit the load is grounded and the current through the load can be calculated as follows.

The current through the load is given by,

The gain of the amplifier is

So

Substituting this value in above equation we get,

Thus the current is directly proportional to the applied voltage and the resistance used in the circuit. it should be noted that all the resistances used in the circuit are equal to R.

See alos: Current to Voltage Converter

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