No, higher voltage supplies can also be used. But the ideal case is that we only single supply, like 6V directly from the input source without any added power supply.Are you constrained to 6V and -12V supplies or can they be higher voltages?
I tried, but the simulation still cannot run.Change V2 to -6V.
Is this school work?Can anyone explain to me the function of R11? Why do we use a 1M ohm resistor here?
YesIs this school work?
That's not ideal, that's impossible.But the ideal case is that we only single supply, like 6V directly from the input source without any added power supply.
I think I understand how it works. The Op amp can work as a comparator by comparing one analog voltage level with another analog voltage level. Something work as follows:That's not ideal, that's impossible.
How do you expect to get -12V from a +6V supply.
An opamp can only output no more that its supply voltages.
If this is school work, why are you using someone else's design?
Shouldn't you be doing your own?
I have a design that does what you want using only 4 resistors and the opamp, but you need to demonstrate some understanding of op amps circuits before I will post that.
Yes, it can if you just want two output levels.The Op amp can work as a comparator
You can accomplish the goal also with 2 BJT transistors and a few resistors. Because this is homework, can't provide solution but something to think about.Hello, guys,
I have a circuit design to output the +6V and -12V square wave with the input 0V and 6V. Vout is connected to the oscliscope.
View attachment 249231
One of my friends designed this circuit. But it does not work in the simulation. Can anyone help me out with this?
Thank you~
That will give a +6V output for a 0V input and a -12V output for a +6V input?You can accomplish the goal also with 2 BJT transistors and a few resistors.
Assuming we have a +6V supply and a -12V supply. I was thinking the other way around: 0V goes to -12V and +6V goes to +6V.That will give a +6V output for a 0V input and a -12V output for a +6V input?
Yes, I see that his original circuit uses the non-inverting input for the signal, so that would be the polarity you are thinking.I was thinking the other way around: 0V goes to -12V and +6V goes to +6V.
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