Newbie: Simple amplifier and voltage to current converter

Part of the answer depends on how much current you'll be switching on the 10V side. Another part on how quickly (and how often) you'll need to operate it.

A simple, reliable approach is to have the microcontroller I/O pin drive a transistor switch that in turn controls a relay. If the current's you're switching are low enough (and you can accommodate a small voltage drop) you could switch the load directly with a transistor or pair of transistors.

Also, on the "make 10 volt" question, do you also need to create a 10V source from 5 VDC?

If you've got a 0-5V output and a 12VDC supply, it's easy with an op amp (view in fixed font or cut&paste to M$ Notepad):

| | .----------. | | | | | | | | uC | 1/2 LM35 | | | | | |Analog output +12V | | | 0-5V |\\| | | o----------------|+\\ 0-10V Out | | | | >---o-------o | | | .---|-/ | | | | | |/| .-. | | | | === | |10K | | | | GND | | | | | | '-' | | | | | | | | '----------o | | | | | | | .-. | | | | |10K | '----------' | | | '-' | | | === | GND (created by AACircuit v1.28.6 beta 04/19/05

The simplest way to get a 4-20mA output with a microcontroller is to make a 0-20mA output:

| | .----------. +12V | | | | | | | '---------o | | uC | 1/2 LM35 | | | .---------o | | |Analog out +12V | | | | 0-5V ___ |\\| | | | o------|___|-o-------|+\\ |/ | | | 47K | | >-------| 2N3904 | | | .-. .---|-/ |>

| | | 12K| | | |/| | | | | | | | === | | | | '-' | GND | | | | | | | | | | === | | | | | GND '----------------o | | | | | | | | | | | .-. | '----------' 47| | | | | | '-' | | | === GND (created by AACircuit v1.28.6 beta 04/19/05

This circuit works in a very similar manner to the one above. The 47K and 12K resistors cut the op amp input to about 0-1VDC. The op amp raises the output voltage as necessary to achieve that 0-1V across the

Also, the current source is compliant to about 10V. If that isn't enough for the instrument or device you're driving, raise the op amp/ transistor power supply voltage accordingly. But in that case, you'll need a transistor that can handle more power.

Oh, and since the LM358 has two op amps iin one package, you can do both at the same time.

Cheers Chris

Sorry -- a small tweak. If you're deriving your +5V supply from the

| | | .----------. | | | | | | | | uC | 1/2 LM35 | | | | | |Analog out +12V | | | 0-5V |\\| | | o------o---------|+\\ 0-10V Out | | | | | >---o-------o | | | .-. .---|-/ | | | | 22K| | | |/| .-. | | | | | | === | |10K | | | '-' | GND | | | | | | | '-' | | | === | | | | | GND '----------o | | | | | | | .-. | | | | |10K | '----------' | | | '-' | | | === | GND | (created by AACircuit v1.28.6 beta 04/19/05

Cheers Chris

Ok, Am I the only one having a problem with the above component being used which is a temperature sensor if I remember ? LM35 ? Maybe the OP meant a LM358 ?

lobotomy"

Hi, Jamie. The temperature sensors are the 1N4148 diodes, in another post. The LM358 is a dual single supply op amp, a "jellybean" IC mentioned in both posts. And yes, the LM35 is a temperature sensor.

Although somewhat dated (it's been around for over 25 years), the LM358 and LM324 (a quad) are the most common (and probably just about the cheapest) op amps out there. You can easily get one of either for $0.12 each or so if you buy by the reel. And the onesie price for the LM358 in an 8-pin DIP package is $0.21 at Mouser. Two for the price of a postage stamp. C'mon -- can't beat that.

Cheers Chris