amplifieranalogbjtoperational-amplifiersignal
I am using LED as a light sensor, the range of output is from 0-700mV DC, I want to amplify/scale this signal to 0-3.3V DC.
My question is, what is the best way to scale the output signal to the desired range using only BJTs?
The constraint of using only BJTs is because of lock-down due to Corona Virus pandemic, I cannot go and buy something like an opamp.
You can use a differential amplifier to subtract the 30 mV offset.
When R1 = R2 and R3 = R4 the transfer function is
\$ V_{OUT} = \dfrac{R3}{R1}(V_2 - V_1) \$
So set V1 to 30 mV and choose R3 = 250 \$\times \$ R1.
A problem with differential amplifiers is that R1 will load the resistor divider to get the 30 mV offset, so that you have to recalculate the resistors, and also V2 will have an input impedance which may distort the measurement.
An instrumentation amplifier is the solution.
Most instrumentation amplifiers are differential amplifiers with a buffering input stage. The input stage sets the gain, while the differential stage is usually a \$\times\$1 amplifier. The amplification is then
\$ V_{OUT} = \dfrac{2 R2}{R1}\cdot \dfrac{R4}{R3} (V_2 - V_1) \$
The Microchip MCP6N11 is a suitable device.
Other answers have already made the main point: TL084 is not a good op-amp for your application. I'll answer one of your specific questions:
And how does the input offset voltage appear at the output (suppose that the input offset voltage s 3mV, the input signal is x(t) and the opamp output is y(t))?
The input offset appears as if it were a voltage source in front of one of the input pins (doesn't really matter which one).
This typically means it appears as an error in your input signal, and sees the same gain as the input signal.
a) 1000 times amplified along with the input signal. (i.e.; y(t) = 3V + 1000x(t))
Yes, though it might be more clear to say \$ y(t) = 100 \left( x(t) + V_{os}(t) \right)\$.
If you look you will be able to find an op-amp with Vos in the 10 uV range, which might be acceptable for your application.
You might also consider that 1000x is a very high gain to achieve with a single op-amp stage. You might want to consider doing this with two stages instead of one.
Best Answer
The trick that isn't mentioned yet is that your LED may show you 700mV with your voltmeter, but that's because your voltmeter is a crazy high input impedance (like 10Meg probably). The LED can't source much current - Maybe nano-to-microamps at best. In other words, it has a very high output impedance. If attached to a load much smaller (say, the 10k base resistor of a BJT amplifier) is not going to have the current drive needed. Your 0-700mV signal will disappear. BJT's are current-devices, not voltage. This would be way easier if you had MOSFETs handy, because those ARE voltage controlled (nearly infinite input impedance). But you don't - You have BJT's - so you gotta deal with currents instead of voltages.
TO simulate the LED, a 0-700mV-0 variable voltage source and a 10MEG resistor are connected to the base of the first transistor. This resistor is NOT part of the circuit you'd build!!!! It represents the output impedance of the LED. (i.e. your LED is simulated by the voltage source AND the 10Meg resistor) I don't know what the impedance actually is, but it's gotta be in the 1-10M range at least. Note you DO need the 100k resistor here.
Since all you have are BJTs, the first stage should be a voltage follower. LIke your voltmeter, and like a good JFET OPAMP, this circuit has a high input impedance, so it'll buffer the LED current. You don't get voltage gain here - You get current gain. That'll let us drive the next stage.
The rest of the circuit is simple, just a couple typical common emitter amplifiers. The diode in stage 1 lifts the output high enough to forward bias the base of stage 2.
Three pics. First two are the circuit. One clean one so you can read it easily, the second shows all the initial bias voltages & currents (this is when LED current = 0) to help your understanding and troubleshooting.
Third pic is plot of voltages. Note the blue trace showing LED action. It shows the LED source going from 0V->0.7V->0V, but you see it as 3.3V->2.6V->3.3V. DOn't be fooled by that. It's VCC referenced, not GND referenced, so you see the voltage as a subtraction...
Each trace is plotted at locations shown by arrows in the first pic. The yellow trace is the output of the last transistor - This is what you're interested in. You'll note my voltage rail is 3.3V, and that the output nearly goes the full requested 0-3.3V (Not quite, but hey - You're not paying me here ;) Seriously, it would be very very difficult to achieve True rail-to-rail - You're not getting that with a practical discrete transistor circuit -
This won't work exactly as is because it'll be dependent on each transistors Beta, but you can definitely tweak it. Build it a stage at a time, and reference the voltages shown. Tweak resistors to get those points to closely match.