In these cases, it can be considered as 1V.. But if VDD=1.1V or VDD=0.9V that then?
Is this what you meant?Draw a simple transistor circuit - common emitter configuration.
Put a suitable resistor in the emitter leg to ground that keeps current to a safe level for the transistor you use.
Oh, Use NPN. Connect Vdd straight to collector.
The answer depends on the type of diodes.Do you know how much voltage 2 diodes in series drop when forward biased?
A simple voltage comparator? LM393?I am stuck at design a circuit with the specs as below.
Input voltage: VDD
Output voltage: Vout
VDD is also the ONLY power supply for the circuit.
When VDD = 1V, Vout = 0V
When VDD = 2V, Vout = 2V
Any idea how this can be done without using IC?
View attachment 99097
You haven't given enough information. You've said what you want Vout to be at just two particular values of Vdd. What about all of the other values? What if Vdd is 0.5 V, or 1.5 V, or 1.9 V, or 2.2 V?I am stuck at design a circuit with the specs as below.
Input voltage: VDD
Output voltage: Vout
VDD is also the ONLY power supply for the circuit.
When VDD = 1V, Vout = 0V
When VDD = 2V, Vout = 2V
Any idea how this can be done without using IC?
View attachment 99097
Not sure...It will take one PNP transistor, one diode and two resistors (or one resistor depending on the load.) I'll give you a hint; the emitter is connected to the input. Can you figure it out??
I just want to try a simple solution.A simple voltage comparator? LM393?
At what specific point do you want to recognize the 2 V level?
Yes, I will check if it works.See post #9
Thanks. I think that is already very good result for my purpose.You can probably adjust the values of R1 and R2 downwards and get some improvement.
A simple voltage comparator? LM393?
OP asked for a solution that didn't use an IC...Any idea how this can be done without using IC?
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