Opto coupler as voltage inverter

Thread Starter

sahard22

Joined Jul 31, 2015
6
Hi,

I am trying to use an opto coupler to get a negative voltage output by controlling it with a microcontroller. Is it possible to do this? If so, would you please help me with the design.

Thank you,
 

crutschow

Joined Mar 14, 2008
34,284
Negative compared to what?
Are you talking about inverting the signal or generating an actual negative polarity with respect to ground?

What do you want to use an opto coupler? Do you need galvanic isolation of the signal?

What power supplies do you have available?
 

Thread Starter

sahard22

Joined Jul 31, 2015
6
Negative compared to ground. I am using a microcontroller, so I will have a positive input and would like to get a negative output.

I have a 3V battery available. I know it is easy to use the opto isolator as a switch to switch between high and zero. However, I would also like to have negative output so I have the option of switching between positive, negative and zero volts.

I do not need to have galvanic isolation, so if you have other ways of doing this, please let me know.
 

Thread Starter

sahard22

Joined Jul 31, 2015
6
Sorry, I forgot to mention that I need variable voltage between -1V to 1V. for 0-1V I am using a PWM to vary the voltage. Now I need a way to also have -1V to 0V.

Thank you.
 

DickCappels

Joined Aug 21, 2008
10,152
Use an optocoupler with a photo diode output, that will get you a few hundred millivolts below ground.

This trick can also be done with a coupler with a transistor output where the base and emitter connections are accessible.
 

AnalogKid

Joined Aug 1, 2013
10,987
Sorry, I forgot to mention that I need variable voltage between -1V to 1V. for 0-1V I am using a PWM to vary the voltage. Now I need a way to also have -1V to 0V.
Finally, some requirements. As DC noted, some (very few) optocouplers transfer measurable amounts of energy from the input to the output. Most do not, they use energy from the input to control energy externally applied to the output. For your application, a better approach would be a charge pump, a type of switched-capacitor DC/DC converter. What is the output current requirement?

ak
 

Thread Starter

sahard22

Joined Jul 31, 2015
6
Thank you Dick and Analog. The output current should be as low as possible as I need low current consumption.
Charge pumps do seem to be a good option for me. I looked at TI and Maxim for the their inverting charge pumps and they do not offer anything in the range of 0 to 1V. The lowest Vin they allow is 1.5V which corresponds to -1.5V out. Do you suggest any vendors?

Thank you.
 

dl324

Joined Mar 30, 2015
16,846
Charge pumps do seem to be a good option for me. I looked at TI and Maxim for the their inverting charge pumps and they do not offer anything in the range of 0 to 1V. The lowest Vin they allow is 1.5V which corresponds to -1.5V out. Do you suggest any vendors?
You could roll your own switched capacitor network...
 

crutschow

Joined Mar 14, 2008
34,284
Below is the LTspice simulation of a simple negative charge pump driven by a 0 to +3V PMW signal.
The output is shown with a 10kHz PWM frequency for duty-cycles of 90%, 75% and 57%.
Note that the negative output voltage is inversely proportional to the (positive) PWM duty-cycle.
An output of -1V is achieved with about a 57% duty-cycle (rather than the theoretical 66.6%) due to the offset from the diode.

[Edit: Changing D1 to a Schottky type diode will reduce this offset and make the output closer to the theoretical duty-cycle value as shown in the second simulation below where 65% duty-cycle gives a -1.01V output.]

The capacitor and resistor values are determined by the PWM frequency, how much ripple you can tolerate, and the required response time of the circuit.

PWM Neg Chg Pmp.PNG

PWM Neg Chg Pmp 2.PNG
 

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Thread Starter

sahard22

Joined Jul 31, 2015
6
Below is the LTspice simulation of a simple negative charge pump driven by a 0 to +3V PMW signal.
The output is shown with a 10kHz PWM frequency for duty-cycles of 90%, 75% and 57%.
Note that the negative output voltage is inversely proportional to the (positive) PWM duty-cycle.
An output of -1V is achieved with about a 57% duty-cycle (rather than the theoretical 66%) due to the offset from the diode.

[Edit: Changing D1 to a Schottky type diode will reduce this offset and make the output closer to the theoretical duty-cycle value as shown in the second simulation below where 65% duty-cycle gives a -1.01V output.]

The capacitor and resistor values are determined by the PWM frequency, how much ripple you can tolerate, and the required response time of the circuit.

View attachment 93649

View attachment 93652

Great, thank you very much. I guess another option is to just use an inverting opamp with gain of -1. Is there an advantage of using a charge pump over inverting amplifiers?
 

crutschow

Joined Mar 14, 2008
34,284
That should be easy, I can use a voltage divider to get +1.5 and -1.5 out of a 3v battery.
If you don't need the 3V for your microcontroller than, yes you can do that.
And you would need an op amp that will operate on ±1.5V which are likely somewhat rare.
 

RichardO

Joined May 4, 2013
2,270
Where did you see that info in the data sheet? :confused:
Well, I did guess a bit... :(
The open circuit voltage is 7 volts minimum and the short circuit current is 12 ua minimum. The graph at the top left of page 5 is inscrutable to me in that it does not seem to reflect the open and short circuit values in the Coupled Electrical Characteristics table on page 3.
 

crutschow

Joined Mar 14, 2008
34,284
Well, I did guess a bit... :(
The open circuit voltage is 7 volts minimum and the short circuit current is 12 ua minimum. The graph at the top left of page 5 is inscrutable to me in that it does not seem to reflect the open and short circuit values in the Coupled Electrical Characteristics table on page 3.
Okay. I see what your are referring to.
Yes, it would seem that you could get a few volts of output voltage if you have high output load resistance ( a few hundred kΩ or more).
Obviously the frequency response would be limited at those high impedances but it should work for low frequencies, depending upon the stray output capacitance, or DC.
That's an interesting observation.

The graphs on page 5 only reflect the coupling with an applied voltage at the output.
 
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