Hello, currently I am using a this very long circuit to convert a 0-5V pulse into -2 to -6V volt pulse.
Is there a a good circuit which I could learn how to effectivly shift the pulse levels?and coltrol the high and low levels, while keeping the pulses rise and fall time?

Thanks.

 

hi yef,
The usual method is to use an OPA.
Dual rail supplies +/-Vsupply
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Hello Eric,I am reading currently this arcle ,can you reccomend also some good articles I could simulate?
Thanks.
https://www.analog.com/en/resources/design-notes/level-shifting-digital-signals.html

 

hi yef,
This is a draft example.
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Hello Eric, I tried to analize your circuit to get mathematical formula Vout=F(Vin)
I used the virtual ground assumption of V-=V+.
but its not working so much
I got in the end Vout=6V and Vin cancels.
Where did I go wrong in the mathematical logic?
Thanks.

 

hi yef,
What are you assuming for the Gain of the Non Inverting OPA input?

The Vin = 5v attenuated 10k/(10k+15k) == 0.4*5V = 2V
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Hello Eric there is such a thing a virtual ground in opamp when the gain is very large so V+=V-

Yes you are correct V+=2V because of voltage divider of R1 and R4.

how do I know the next logic that defines Vout?
Thanks.

 

Hello Eric there is such a thing a virtual ground in opamp when the gain is very large so V+=V-

hi yef
What virtual grounds are you relating too??
Where is: the gain is very large

What are you assuming for the Gain of the Non Inverting OPA input?
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In analog design class I was told when finding the formula for Vout as a function of Vin that V+=V- when opamp gain is very high.

In your circuit how the high value and low value of the output are defined?
Thanks.

 

In your circuit how the high value and low value of the output are defined?

Hi yef,
Forgive me for being blunt.
I am getting the impression that you have no idea on the operation of OPA's in a simple amplifier circuit.

You should be able to answer the question.

What is the Gain of the Non Inverting OPA input for the circuit I have posted??
Post the equation you are using.

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the gain is very very large thus we have virtual short. by the photo shown below taken from the link.
What assumption am I missing?
Thanks.

https://toshiba.semicon-storage.com...p/what-is-the-virtual-short-of-an-op-amp.html

 

the gain is very very large thus we have virtual short. by the photo shown below taken from the link.

hi yef,
The Gain is not very large.

For that inverting OPA image the Gain is: Gain = -Vin(R2/R1), the voltage sense of the Output is inverted with respect to OPA Inv Input sense.

For a Non Inverting OPA the Gain is Vin(1+R2/R1), the Vout is the same polarity sense as the Non Inv input voltage.

Read these two PDFs'
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I suppose the most important question is: what is supposed to happen if the input is a 0 to 4V pulse? Should the output pulse be scaled to 80% of the amplitude. Should the output pulse therefore be from -2V to -5.2V, or should it be from -2.8V to -6V.n Or should the output be -2V to -6V regardless of the amplitude of the input pulse? And should the output be -2V when the input is 0V, or -6V when the input us 0V?

 

we have virtual short.

Those two voltages may be essentially the same, but they are not a short.

The high open-loop gain of the op amp means, with negative feedback, the amp will do whatever it can to keep the two inputs at the same voltage.
From this it is easy to calculate the gain, since i1 must equals i2 with the (-) junction essentially at ground potential.

 

hi,
All the TS has asked for is:

I am using a this very long circuit to convert a 0-5V pulse into -2 to -6V volt pulse.

No mention of other than a 0-5V pulse input.

 

hi,
All the TS has asked for is:

I am using a this very long circuit to convert a 0-5V pulse into -2 to -6V volt pulse.

No mention of other than a 0-5V pulse input.

And does no-one in product design every consider what happens if the input pulse is slightly out of spec?
Does no-one designing a product for a 230V supply ever consider what happens if the supply is 220V or 240V?

 

 

Hi Ian,
Engineering should create a design to meet the specification stated by the client.

In this case, the 'client' has stated a basic specification of a 0v/+5v pulse he needs to convert to a -2V > -6V , nothing more.

Unfortunately, 'Yef' has a problem in understanding the very basics of OPA's.

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Ok so this is not a linear amplifier but the opamp acts as a comparator outputing the power rails?
How can I know that its outputs are saturated in this situation?

 

Hi Ian,
Engineering should create a design to meet the specification stated by the client.

In this case, the 'client' has stated a basic specification of a 0v/+5v pulse he needs to convert to a -2V > -6V , nothing more.

Unfortunately, 'Yef' has a problem in understanding the very basics of OPA's.

E

if that’s what engineering did when faced with a vague and incomplete specification, then the result would be an ever decreasing number of clients.
Finding out what the customer actually wants is good engineering (especially when the customer cannot express it precisely in engineering terms)