[SOLVED] Voltage Level Range Converter Shifter

[apprenticemart2]

Hi All,

I need a system were I can input a 0V to +5V DC range and get a -2V to +3V DC range out

It needs to be stable/accurate enough to handle voltage steps of 83.33mV. There are 12 steps for every Volt so,

If I input +83.33mV, I get -1916.67mV volts out, if I input +166.66mV volts, I get -1833.34mV volts out and so on.

I have access to +/- 15 Volts, +/-5 Volts and an accurate +10 Volts supply. The application is for 4 voice CV/Gate control of my Teisco SX-400 synthesiser driven by a Roland CMU-800 sequencer.

I am aware of another thread here https://www.edaboard.com/threads/189649/
but I don't understand from the schematic where I inject my signal, or what the circle with the centre offset plus sign means(Vref or source?). There may also be a better part for the job than the MAX4250 suggested currently available or it may not suit my application, and finally, the forum warned me about hijacking old threads and I decide not to take the risk

There are other threads on this too that I am reading now but I still don't know if I can achieve a repeatable accuracy of 83.33mV steps with any of those circuits.

I am trying to learn as much as I can by myself but any suggestions will help speed up my understanding and will be very welcome.

Thanks in advance
Martin

 

[crutschow]

What is the tolerance on the 83.33mV steps?

 

[apprenticemart2]

What is the tolerance on the 83.33mV steps?

I would hope to get no more than about +/- 2mV deviation at each step.

Cheers
M

 

[crutschow]

Below is a circuit that should do what you want.

Just about any low input-offset op amp should work.

V4 represents the input voltage source.

The accuracy is mainly determined by the resistor tolerance and the accuracy of the 10V Vref voltage.

 

[apprenticemart2]

Brilliant. Thank you very much. It looks a much simpler layout to most I have seen. I'll post back with more praise if I can get it working.

Is slew rate going to be an issue here?

I could try an LF411 maybe or AD711?

Thanks again
Martin

 

[erikl]

3 small notes in addition to crutschow's suggestion:

1. Instead of the ±5V power supply I'd suggest to use a ±15V (or: +15/-5V) power supply, if you can't get a true rail-to-rail input amplifier (the max. 5V input voltage must fit into the admissible input range of the opAmp).

2. The load resistance for your input voltage source (V4) is R2+R1. If you want to get a larger load resistance, you may of course scale all 4 resistors by the same factor, e.g. by 10.

3. 2mV/83mV ≈ 0.024 , i.e. use resistors with a max. (in-)accuracy of 2% or better.

 

[apprenticemart2]

I will look into running the opamp at =/-15V.

I will need a quad opamp as I have 4 channels and these are quite expensive, so I am considering the Ti LT1014CN.It seems the best compromise.

50K ohm resistors aren't very common it seems. The only ones I can find are Vishay ones, 0.125w and 0.1% tolerance so that works out OK. I will get the 10K version of the same type.(unless scaling the resistor values gets me cheaper prices)

Thanks
Martin

3 small notes in addition to crutschow's suggestion:

1. Instead of the ±5V power supply I'd suggest to use a ±15V (or: +15/-5V) power supply, if you can't get a true rail-to-rail input amplifier (the max. 5V input voltage must fit into the admissible input range of the opAmp).

2. The load resistance for your input voltage source (V4) is R2+R1. If you want to get a larger load resistance, you may of course scale all 4 resistors by the same factor, e.g. by 10.

3. 2mV/83mV ≈ 0.024 , i.e. use resistors with a max. (in-)accuracy of 2% or better.

 

[erikl]

I will look into running the opamp at =/-15V.

Hi Martin,
I was wrong with my 1^st statement above, sorry:

1. Instead of the ±5V power supply I'd suggest to use a ±15V (or: +15/-5V) power supply, if you can't get a true rail-to-rail input amplifier (the max. 5V input voltage must fit into the admissible input range of the opAmp).

Because of the R2-R1 voltage division, the voltage from V4=5V at the non-inverting input doesn't rise higher than 4.17V, so a ±5V power supply should be sufficient, as crutschow already had suggested. Just make sure that the acceptable input voltage range is high enough.

erikl

 

[apprenticemart2]

Hi erikl,
I will re-evaluate that. I am not decided on the opamp to use yet as they are quite expensive on their own. Duals are a little better priced but they are still a lot.

I am going to socket in an MC33079 when my other parts arrive and see how that behaves and get some measurements while I read a few more datasheets.
Thanks for your help.
Martin

 

[erikl]

Hi Martin,

the MC33079 you'd actually better supply with ±15V (or +15V/-5V), because its upper input common mode range (pnp input with current source) is limited to +VDD - 1V (typ.) or +VDD - 2V (w.c.). Already the typ. positive input common mode range isn't high enough for your application with a ±5V supply.

Cheers, erikl

 

[apprenticemart2]

Thanks erikl,

That sounds like the best way to do it.

Thanks again
Martin

Hi Martin,

the MC33079 you'd actually better supply with ±15V (or +15V/-5V), because its upper input common mode range (pnp input with current source) is limited to +VDD - 1V (typ.) or +VDD - 2V (w.c.). Already the typ. positive input common mode range isn't high enough for your application with a ±5V supply.

Cheers, erikl

 

[crutschow]

.............................
Is slew rate going to be an issue here?

......................

Depends upon the speed of your ramp.

Edit: As noted, the op amp I used in the simulations does not have a sufficient guaranteed input CM range to be used with 5V supplies. You either need to use a rail-rail type op amp or go to 15V supplies.

- - - Updated - - -

...................................

50K ohm resistors aren't very common it seems. The only ones I can find are Vishay ones, 0.125w and 0.1% tolerance so that works out OK. I will get the 10K version of the same type.(unless scaling the resistor values gets me cheaper prices)

Yes, the values I picked where arbitrary and not common standard values. Below is a list of standard 1% value resistor pairs that have an exact 1 to 5 ratio, as required by the circuit. You can use any pair combination that you like. The 11.8kΩ and 59.0kΩ pair are the closest to the values I used.

 

[apprenticemart2]

Thanks for these updates. Very much appreciated.

Martin

- - - Updated - - -

It's much cheaper buying 11.8K and 59K resistors and I've just remembered that I've got some OPA4277 op amps somewhere. They tick almost every box. The LT1058 seem good but I would need to find an 'A' spec and they are costly.
Cheers All
Martin

 

[apprenticemart2]

The reason for me needing a voltage shifter is because I want to be able to control my Synthesiser with CONTROL VOLTAGES (CV) generated externally by a CV sequencer.
CV/Gate as it's known operates between 0 and +5 in 83.33mV steps representing each note.

Aside from some other problems, the keyboard I have (Teisco SX-400) uses a system that operates in the range -2 to 3 volts and this is the reason I needed a shifter.

I've just made some changes to my original plan based on the new information that I have been provided with( Thanks for that )

Originally I was trying to inject the CV's in to a 4051 I piggybacked on top of the one already in the circuit, but I was having trouble getting it to play notes in tune, so it seems there was something in the circuit sucking the life out of the voltages meaning that I have had to take a different approach.

I have experimented over the last couple of days with inputting the CV voltages as they are ( not yet level shifted) into each channel after the 4051 rather than before. It seems to play the notes in tune now and each voice is independent ( still experimenting )

All I need to do now is to add the voltage shifter, but I have a slight question.

Do I need to add an analogue switch ( or similar) so that I can disconnect the newly added voltage shifter from the circuit, would this effect the original configuration in any way when I play from the keyboard as normal?

I'm wondering if the new additions will be passive or will they influence the original circuit in any way when I am not using them?

Thanks
Martin

 

[erikl]

Sure: if you want to use CV channel 1 independently from the level-shifted CV Out voltage, you'll have to disable CV Out by one more analog switch.

You cannot bind together two different active outputs: they will fight against each other, and the stronger one will win - by that possibly destroying the weaker one.

 

[apprenticemart2]

Great. Thanks. I can power the 4066 with ±5V because the signals are being shifted to -2 to 3V.

Would a pull down resistor go to GND or -5V?
10K,100K?

Cheers
Martin

 

[erikl]

If either CV Channel 1 or CV Out will be activated, you don't need a pull down resistor. If both could be deactivated, you should use a pull down resistor to that voltage you like to be output by U4a. If it doesn't matter in that case, I'd suggest to pull down to GND.

 

[apprenticemart2]

I will try something like this then?


Thanks for your help.

Martin

 

[erikl]

Yes, that's fine I think. Just make sure that U2's Y1 output and U5a's output aren't activated at the same time.

You could replace the Latching Switch by an appropriate logic combination of Q5-Q6-Q7 (A-B-C inputs of U2) creating not-Y1 (i.e. the logical combination which does not activate the Y1 output of U2) to activate U5a. Or simpler: use the inverted logic combination of Q5-Q6-Q7 which activates the Y1 output to disable the U5a output.

 

[apprenticemart2]

I hadn't thought about doing it that way. I would have to learn a few more things before I could get my head around that. It's an interesting concept that might give better control and more options.

I had thought about using the inhibit on the 4051 and tying that to the same switch that I am using for the 4066, so when I switch the 4066 for external CV in I am disabling the 4051.

Thanks again for your help

Martin