I want to have a square wave that goes from 0 VDC to 5VDC go through some sort of amplifier and that amplification takes that same wave ( at a frequency of 1200Hz but could also be adjusted later ) but now makes it 0VDC to 13VDC (at least 13VDC, it could be 13.4 VDC, but not lower than 13VDC). The original wave would be generated from a function generator, but I do not want to exceed 5VDC. How would I got about doing this?
Electronic – Square Wave Amplifier – 5VDC -> 13VDC
amplifierdesigndigital-logicsignalwave
Related Solutions
If the frequency for both waves is going to be 100 kHz with the same amplitude, you could construct a narrow bandpass filter at 200 kHz to put the signal through. In theory a pure squarewave should only have odd harmonics, so there should not be much output at the second harmonic frequency. On the other hand, a sawtooth wave has booth even and odd harmonics, so you will get a greater output. The peak amplitude for the second harmonic of a sawtooth wave will simply be \$ \frac{2A}{\pi} \$, where A is the peak amplitude of the input sawtooth. If wish you can then follow up the output of the bandpass filter with a peak detector and some kind of comparator.
A diode is series is one way, but your diagram shows two problems with that setup. First, you have the diode backwards. Second, you have provided no load for the diode. The scope is has high impedance, so the negative part is either being capacitively coupled thru the diode or via its leakage.
Here is how to use a diode correctly:
All negative input voltages will result in 0 output voltage with a impedance of 1 kΩ. Note that due to real diodes not being perfect, during the positive part of the input, the output will be about 700 mV less. Put in mathematical terms:
Out = max(0, In - 700mV)
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Best Answer
There are many methods to approach this. You can adjust your power supply to 13V and use an opamp as a comparator (Out1) or use a two stage npn amplifier (Out2) to amplify your signal. If you want to adjust gain but not your power supply, then use a non-inverting opamp gain (Out3).
How each works:
When your signal is above VDD*R2/(R2+R1) = 13*1k/(1k+10k) = 1.18V, the opamp with its open loop gain is saturated. You're essentially making a switch, either on or off.
Your signal comes in. The cascaded stages of npn yield high gain. You can get away with one as the beta, gain factor, of the transistor is around 100.
Vo = Vin*(1 + R7/R10). Just turn the pot to adjust.
If you need high current output, then use a power opamp like a LT1210 from linear technology.
simulate this circuit – Schematic created using CircuitLab