Voltage translation with change in polarity

Thread Starter

Sitara

Joined May 2, 2014
57
Hi,

I'm stumped with the following problem: I have an arbitrary 0-5V waveform which ranges between 2 to 30 Hz and which I would like to translate to -60 to +60 V. The translated voltage feeds a constant resistive load which will draw 10mA at the peaks (ie at +/- 60V).

I've simulated (in LTSpice) amplitude modulating a 30 KHz square wave with this arbitrary waveform, then using a step-up transformer with a split secondary to amplify the voltage, then rectifying the secondary outputs to demodulate back to the lo-frequency signal but that cannot give me the required bipolarity.

Some help please!
 

Thread Starter

Sitara

Joined May 2, 2014
57
Thank you #12! Option 2 will do nicely. Option 1's 1uF (or other value) capacitor will likely distort the waveform by preferentially amplifying the higher frequency components. I wish I could have given you an extra star to bring you up to a nice round 9500, but I will wait till the next occasion.
 

#12

Joined Nov 30, 2010
18,190
Oh bleep! I drew that wrong.:mad:
(I was in a hurry to go get my driver's license renewed.)
Just move one resistor from parallel to series connection and the grounding resistor (theoretically) becomes 874K for minimum thermal drift.
That should work, but the chip has way too much high frequency response and might oscillate. Feel free to add a capacitor across the 22M resistor or an R and a C in series, and that across the 22M resistor. You could cut the high frequency response down to 300 Hz and be within a few % of the proper high frequency gain...not that 30 Hz is a high frequency.:D The gain is also a wee bit high, so you could redesign in 1% resistors or place a small resistor in series with your 6K load.

The range of resistor values is high enough that you can buy a very low leakage film cap at 1 uf and not have to do all kinds of guard rings on the board to eliminate leakage paths. It's all in the proportionality. The op-amp can do a lot higher impedance and frequencies, but I only chose it as the cheapest chip that can survive the 120 volt power supply.

I was observing that your thought process was sophisticated enough that I only had to show you a shape and a part number and you could roll from there.;)
 

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Thread Starter

Sitara

Joined May 2, 2014
57
I was observing that your thought process was sophisticated enough that I only had to show you a shape and a part number and you could roll from there.;)

I wish! I cannot even figure out which op amp you meant, at least its part name isn't marked on the schematic. So, would you please name the chip? Thanks!
 

#12

Joined Nov 30, 2010
18,190
I did name it...in post #2
LTC6090
I found it at www.mouser.com but that's not the only source.

ps, you need to learn that @#12 is the way to summon me.
The @ symbol will summon any other player when you follow it with their moniker.
 

Thread Starter

Sitara

Joined May 2, 2014
57
@#12
Oh, I see. I was under the impression that "There is more than one way to do this." and the two uploads in post #12 (now, isn't that an eerie coincidence!), meant that the two uploads involved the more than one ways. But if both referred to the LTC6090, then a direct connection between the 0 to +5V signal source and the non-inverting input of the opamp would be enough, per the datasheet (see uploaded snippet from pg1 of the datasheet). The absence of an intermediate capacitor would mean no distortion to the waveform. Anyway, your solution is neat, and very practical. I've ordered a couple of LTC6090s and will give the idea a try.
 

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#12

Joined Nov 30, 2010
18,190
One purpose of the capacitor is to let the input signal be centered around the zero volt position. That's how you get a monopolar signal to split into a bipolar output centered at zero. I think if you leave out the capacitor you will get an output of zero to +volts only.

You keep talking about distortion caused by a capacitor. I calculated that capacitor to cut the gain by 3db at 0.175 Hz.
I think that is low enough to pass 2Hz without much distortion. However, you can place (2) of the 1uf capacitors in parallel to get the -3db point down to 0.09Hz. The problem is that you want to center the output signal around zero volts AND not use a capacitor. You can do that by attaching the resistor on the non-inverting input to +2.5VDC. Then you can leave out the capacitor and still get the output centered around zero.

I see your diagram in post #7 illustrates that pretty well except that diagram will use 1.25V for the reference and center a signal that ranges from 0V to 2.5V. You just need to change the reference voltage to 2.5V to center a 0V to 5V signal.
 

Thread Starter

Sitara

Joined May 2, 2014
57
@#12
You keep talking about distortion caused by a capacitor. I calculated that capacitor to cut the gain by 3db at 0.175 Hz.
I think that is low enough to pass 2Hz without much distortion.

You are right. I just worked it out. The increase in gain at 20Hz relative to 2Hz is just 7.5%. Small potatoes! Thanks again!
 
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