voltage level shifter

Thread Starter

anhnha

Joined Apr 19, 2012
880
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?


Specs.png
 

EM.

Joined Nov 13, 2015
13
If you can show the circuit it who'd be easier to try to figure out

What is going on with the circuit?

But I believe that the 1V is actually less than 0.7V and wen the current

coms in, it goes throw a diode, and the diode doesn't let throw current

at all as long the voltage is at least more than 0.7V

you might see on the tester 1V if the tester is not 100% accrete.
 

Kermit2

Joined Feb 5, 2010
4,163
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.

You need to prevent your base voltage from turning on the transistor when Vdd is 1 volt but let enough voltage through when Vdd is 2 volts to forward bias the base.

Do you know how much voltage 2 diodes in series drop when forward biased?
 

Thread Starter

anhnha

Joined Apr 19, 2012
880
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.
Is this what you meant?
What you describe is common collector or emitter follower not common emitter.
Common Collector.PNG
Do you know how much voltage 2 diodes in series drop when forward biased?
The answer depends on the type of diodes.
For silicon diodes, the typical forward voltage is 0.7 volts, so two diodes in series will drop about 1.4V.
For germanium diodes, the forward voltage is only 0.3 volts, so two diodes in series will drop about 0.6V.
 

Lestraveled

Joined May 19, 2014
1,946
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??
 

hp1729

Joined Nov 23, 2015
2,304
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
A simple voltage comparator? LM393?
At what specific point do you want to recognize the 2 V level?
 

WBahn

Joined Mar 31, 2012
24,854
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?

Instead of a plot of Vdd and Vout as a function of time, you need to plot Vout as a function of Vdd.
 

Thread Starter

anhnha

Joined Apr 19, 2012
880
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??
Not sure...
Level shifter.PNG


A simple voltage comparator? LM393?
At what specific point do you want to recognize the 2 V level?
I just want to try a simple solution.
Also, I am not sure if the comparator will work at 1V VDD.
I will check it now.
 

Thread Starter

anhnha

Joined Apr 19, 2012
880
I am designing a digital 2 to 1 multiplexer.
Vout is will be the voltage for select pin for the multiplexer.
VDD will have only two values 1V and 2V with some tolerance for example 1%.
With VDD = 1V, I want the select pin will be 0V and when VDD = 2V, the select pin is 2V.

See post #9
Yes, I will check if it works.
 
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