How can I shift analog voltages ranging from -5V...+5V to 0V...5V without using an op amp? I was thinking about some kind of current mirror?
\$\begingroup\$
\$\endgroup\$
2
-
2\$\begingroup\$ What sort of signals? What is the application? \$\endgroup\$Kevin White– Kevin White2024-04-04 19:04:50 +00:00Commented Apr 4, 2024 at 19:04
-
\$\begingroup\$ What is the motivation for not using an op amp? \$\endgroup\$le3th4x0rbot– le3th4x0rbot2024-04-05 04:08:53 +00:00Commented Apr 5, 2024 at 4:08
Add a comment
|
2 Answers
\$\begingroup\$
\$\endgroup\$
1
-
\$\begingroup\$ For high-speed signals it helps to add capacitors in parallel with the resistors, creating a frequency-compensated divider. \$\endgroup\$jpa– jpa2024-04-05 05:38:14 +00:00Commented Apr 5, 2024 at 5:38
\$\begingroup\$
\$\endgroup\$
Provided your source can provide far more current than what you need at the output (without disturbing your source), you have a very simple solution, with just 2 resistors : a voltage divider towards +5V :
simulate this circuit – Schematic created using CircuitLab
You get Vout = R1/(R1+R2) * Vin + R2/(R1+R2) * 5V Which simplifies to Vout = (Vin + 5V)/2 = 2.5V + Vin/2 if R1=R2 So Vin=-5V -> Vout=0V ; Vin=0V -> 2.5V ; Vin=5V -> Vout=5V
Note that the current flowing between Vin and 5V needs to be far bigger than the current flowing to Vout if you want accurate results. This might be blocking depending of your source and load (if your source is weak, you need big values of resistors, which means you can draw only very little current on Vout).
