[SOLVED] Voltage Level Range Converter Shifter

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

What is the tolerance on the 83.33mV steps?

crutschow said:

What is the tolerance on the 83.33mV steps?

Click to expand...

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

Cheers
M

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.

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

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.

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

erikl said:

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.

Click to expand...

apprenticemart2 said:

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

Click to expand...

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

erikl said:

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).

Click to expand...

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

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

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

Thanks erikl,

That sounds like the best way to do it.

Thanks again
Martin

erikl said:

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

Click to expand...

apprenticemart2 said:

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

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Click to expand...

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 - - -

apprenticemart2 said:

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

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)

Click to expand...

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.

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

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

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.

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

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.

I will try something like this then?


Thanks for your help.

Martin

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.

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