galvanometervoltage
I am trying to control a galvanometer that requires +/-5V input voltage, my source is a 14Bit DAC (LTC2612) that I control via an Arduino. I use a +/-15v powersupply to power the galvanometer.
What would be the best way to convert the output of the DAC to suite the input needs of the galvanometer? I have a LM358 dual op-amp if that is a way to go?
Edit - I realised you might want from 0V to +5V for -24V to 0V, so here is another NPN based option that might do the trick, and also does not need the -24V rail. I left the FET based circuit there anyway in case it is of interest. As noted below you might have to fiddle with values a bit, this was just intended to put the idea across.
I wasn't entirely sure whether the signal is analog or digital, as you mention digital in the title, but also an acquisition device (so I thought it may be possibly heading for an ADC) so I assumed either was possible, hence the attempt at linearity.
NPN circuit
NPN circuit simulation
Alternative Option
An N-channel FET source follower might work okay with a voltage divider on the input. Say something like a 10:1 divider for a range of -2.4V-0V at the gate. The source of the FET to 0V through a resistor and drain to +5V.
This won't be the most linear but shouldn't be too bad if you fiddle around with values a bit to adjust the desired ratios. You could maybe add a constant current source (e.g another N-channel FET with source to -24V through e.g a 5K resistor, gate tied to source and drain to source of upper FET) on the source to make it more linear if necessary.
Here is a circuit and simulation of the (more linear) constant current idea (the green line is +Vout as Vin is swept from -24V to 0V):
FET circuit
FET circuit simulation
One approach is to use a non-inverting amplifier at the end of the DAC output, supplied by atleast 10 volts.
The DAC output in this case will be \$V_{in}\$
\$ V_o = V_{in}( 1 + R_1/R_2)\$
Lets say we want \$V_o = 10V\$ , when \$V_{in}= 3.3V\$
\$10 = 3.3(1+R_1/R_2)\$
\$10/3.3-1 = 2.0303 = R_1/R_2\$
So one way to go at it , is make a non-inverting amplifier with a gain of roughly 2, by using 2Kohm for feedback resistor \$R_1\$ and 1Kohm for the other resistor \$R_2\$
8-bit resolution means the range is 255 discrete voltages. So 10V/255 bits = \$ 39.2mV\$/ bit
Best Answer
Assuming the galvo amplifier has a relatively high impedance input (doesn't draw more than a few mA) you can use this circuit:-
simulate this circuit – Schematic created using CircuitLab
The amplifier has a gain of -2 from the DAC output, and D1/R2 biases the voltage at +5V out with 0V in, so the output will change from +5V with 0V in to -5V at 5V in. R4 provides some vias current for D1, which is the -5V reference to offset the amplifier.
If you want to avoid D1 and have a 5V reference already, you can use the remaining amplifier to get -5V from the +5V reference as shown in the dashed box.