Possible methods to shift a voltage level?

Hello,

I am now trying to shift a voltage by adding an offset. For example, a voltage relative to ground is 5V and now I wish to offset this voltage by, say -30V. Therefore the shifted voltage now is -25V relative to ground. The key to this design is that the original voltage level is fixed = 5V while the voltage offset can vary.

Are there any possible ways that can let me achieve this? I don't think I can use an op-amp since normally the op-amp output range is limited to 15V.

Please help.

Thanks.
Jingzhe

I am now trying to shift a voltage by adding an offset. For example, a voltage relative to ground is 5V and now I wish to offset this voltage by, say -30V. Therefore the shifted voltage now is -25V relative to ground. The key to this design is that the original voltage level is fixed = 5V while the voltage offset can vary.

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Hi Jingzhe
What kind of voltage ? DC or AC ? how much power ? how much is the BW if it is AC ?

Are there any possible ways that can let me achieve this? I don't think I can use an op-amp since normally the op-amp output range is limited to 15V.

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Who told op amps are limited to 15 ?? they can be used till 30 in single supply mode . and also there are some op amps available with pretty more ranges till 900 volts too . in addition , who told just op amps can be used for this aim ? there are too many ways for this purpose .

Best Wishes
Goldsmith

Hi,

Thanks. It is DC that Im talking about. The power requirement is very low since the current is very low. Actually I am designing a triggering circuit for IGBTs and according to the design I will have to shift the voltage first.

What I thought is that since most op-amps are supplied by +-15V power suppler, like op-amp 741. I am glad to know there are much more ways to achieve this. Is it possible for you to suggest some simple methods that can achieve this? The voltage offset can be varied from 30V to probably 100V.

Thank you.

Thanks. It is DC that Im talking about. The power requirement is very low since the current is very low. Actually I am designing a triggering circuit for IGBTs and according to the design I will have to shift the voltage first.

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Well if you're running an IGBT then if it is square wave which has no negative side it means it will require some inrush current to be turned on .
In addition if you mentioned it at first then i could guide you in better ways . IGBTs will be turned on with 15 volts ( usually ) so if you're dealing with logic level of TTL ( 5 volt ) you can use an ICL7667 which is designed to suit this aim . and it will work as well as enough .

m 30V to probably 100V.

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are you suggesting that you are going to drive GE of an IGBT via 100 volts ? is that some sort of kidding ? it will be destructed of course ! or perhaps you're going to design a float driver ? if so there are much better ideas . you do ?
Best Wishes + Good Luck

Hi Goldsmith,

It's a bit hard to explain my design. I am a PhD student in engineering, so I am pretty clear what Im doing.

Probably it is better not to discuss triggering IGBT in this thread. What I wish to do is just to try to find a method to shift a voltage down by 30-100V. So if I have a dc voltage of 5V with reference to earth and I would now like to shift the voltage down to be -25 to -95V. How could I implement such a design?

Many thanks.
Jingzhe

Probably it is better not to discuss triggering IGBT in this thread. What I wish to do is just to try to find a method to shift a voltage down by 30-100V. So if I have a dc voltage of 5V with reference to earth and I would now like to shift the voltage down to be -25 to -95V. How could I implement such a design?

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Ok i just suppose that you want shift voltage and i prefer to don't think about what is the purpose of it .
you can simply do what i say and see the result :
use a potentiometer which it's common head is is the out put . tie one of it's right or left side head to the main signal and tie the other side to the -100 volt . measure the out put and then turn the handle and then you'll see the out put is going to be shifted simply ! that's all you need to do .

Good Luck

Hi Goldsmith,

Thank you very much! I've thought of using external electrical devices, but I would prefer some electronic solutions. Is it possible to do this by usual electronic components? In my mind, it's a bit like an op amp subtracting circuit. The problem is the normal lab op amps are supplied by low voltage therefore I can't get the desired shift range.

Thank you very much! I've thought of using external electrical devices, but I would prefer some electronic solutions. Is it possible to do this by usual electronic components? In my mind, it's a bit like an op amp subtracting circuit. The problem is the normal lab op amps are supplied by low voltage therefore I can't get the desired shift range.

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Hi
If potentiometer isn't an electronic component so what is it ??!! i have used this way many times in accurate projects and it worked as well . but if you want loss money you can use another ways .
like going though the appex op amps which are dealing with high powers or high voltages or high currents and of course high prices !
why you just want use op amp ? is there something that i don't know ? you mentioned you know exactly want you want d to do . the problem is just this ! you want ask someone and use help of the others . so the others should know what you want to do exactly so they can help you in your way . ok ?
you told you want use op amp . i should ask why when it can be obtained by a potentiometer ? you told you want use electronic components i need to ask do you know that potentiometer is an electronic component too ? perhaps you meant semiconductor components ?

This is the classic method of using an op amp to sum two incoming volt levels (as long as they do not exceed the bounds of the supply rails).



However if you do not have an op amp which can allow the supply rails to be 100 V apart...

Then an alternative might be to make your own op amp, starting with a 'long-tailed pair', using transistors which are made for high volt levels.

However if you do not have an op amp which can allow the supply rails to be 100 V apart...

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Hi dear BradtheRad

PA94 op amp which made by Apex can work with 900 V supply voltage .

Best Regards
Goldsmith

Hi Goldsmith,

I see what you mean! I took for granted that the potentiometer you mentioned is similar to something very big in my lab.

I think this method should work. I should have come up with this idea earlier.

Thank you very much.

Jingzhe

I see what you mean! I took for granted that the potentiometer you mentioned is similar to something very big in my lab.

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Hi jeffwjz
Wow ! i meant small potentiometers which are just some mili meters ! :wink: i've never thought that you may even think about large potentiometers !

Ok , if you had any other question let me know .

Good Luck
Goldsmith

Hi Goldsmith,

I thought about your method but I found its still not what I wanted. By using a potentiometer I can get a desired output when the original is 5V. But if the 5v input becomes 0, the output is no longer -30V. Is it posdible to output a -25V when input is 5V and to output a -30V when input is 0V?

Thanks.

I thought about your method but I found its still not what I wanted. By using a potentiometer I can get a desired output when the original is 5V. But if the 5v input becomes 0, the output is no longer -30V. Is it posdible to output a -25V when input is 5V and to output a -30V when input is 0V?

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Hi jeffwjz

what you mean ? can you show me result of your experiment ? because the summer with a potentiometer which i've introduced you , will act exactly as a summer which is based on op amp but with this difference that it can not be active and i guess you're not going the advantage of gain og the summer which is based on op amps . isn't it ? so please show me the result of your experiment and what you've did and i'll help you to improve it to get your desired result .

Best Wishes
Goldsmith