Rail to ground "digital" output from common emitter amplifier? - Page 1
Forum section: electronics / beginners

author: Infraviolet
author_rank: Super Contributor
timestamp: October 18, 2023, 08:22:09 pm
content: Is there any way this can be done?I was trying to use an NPN common emitter amplifier to amplify an analogue signal centred on a voltage in the 1V to 3.5V range, using 5V and Gnd as the power rails. But while the output signal, when the resistance in series with the emitter to ground capacitor is small enough, can amplify a small signal so that the positive parts go high enough to clip close to the 5V rail, the closest to outputting ground it can get is about 1.3V. When the ratio between the resistance in series with the emitter-to-gnd cap and the collector-to-5V resistance is large it also ends up drawing quite a lot of current from the signal source, so often needs a voltage follower stage placed before it to give a lower impedance source to feed in to this amplifier's base.Is there a way to get the output voltage a lot lower? to 0.7V or even to somewhere <0.5V?I'm trying to amplify the incoming AC signal (the exact voltage it will vary about is constant, but could be at different centre levels depending on what exact circuitry I've got before this amplification stage) up to a clipped square wave large enough to drive 74HC series logic with 5V power. 1.3V is unpleasantly close to the highest value which HC logic counts as LOW, so I want to get down to somewhere comfortably below 1V for amplifying the lowest troughs of the waveform.Thanks

author: Ian.M
author_rank: Super Contributor
timestamp: October 18, 2023, 09:14:41 pm
content: When I had a similar problem many many moons ago, I had to regenerate a 0.6V pk-pk interrupted clock signal at several MHz to normal TTL levels without excessive propagation delay.A 74HCU04unbufferedhex inverter can be used as a linear amplifier by AC coupling the input and applying negative feedback via a high value resistor to stabilise its input DC bias. Depending on your input signal level you may need two AC coupled stages to get enough gain. Finish squaring it up by putting it through more inverters, with DC coupling.See:https://www.eevblog.com/forum/beginners/how-to-design-with-cmos-inverter-amplifiers/Of course you can get similar results with enough discrete BJTs but its far more of a PITA for high speed signals . . .

author: Infraviolet
author_rank: Super Contributor
timestamp: October 18, 2023, 09:24:02 pm
content: Yes, I've used the HCU trick, it is a nice one to know, that was my thread about them infact. But it turns out to be quite current hungry, particularly when the input signal is small. Hungry enough you only dare use about 2 channels of a hex chip (grounding the other 4, or using them in strictly digital-input applications) lest you go beyond the 74HCU04's max total current rating. That's why I was hoping there might be a transistor way.

author: magic
author_rank: Super Contributor
timestamp: October 18, 2023, 09:58:12 pm
content: The only concrete information you gave is the average DC level, which is likely irrelevant if AC coupling is acceptable to shift it to whatever the amplifier needs.And something about peak output voltages observed from an unspecified circuit.

author: fourfathom
author_rank: Super Contributor
timestamp: October 18, 2023, 10:00:49 pm
content: Why do you have an emitter resistor? In this type of application typically the emitter is connected directly to ground, and the input signal connects to the base through a series resistor. The collector resistor is fairly large, just small enough to compensate for load capacitance and impedance (at the frequency of interest). Since your analog input range is 1V to 3.5V you will either need to AC-couple this to a base-bias network (at about 0.25V) or otherwise shift the DC level down to the transistor will shut off at the low-voltage excursion range.What frequency range is your input? Is this a stable 1V - 3.5V signal, or can the Voltage-swing be smaller than this?As for your question regarding the emitter and collector resistors, the best you can do with that is to consider the transistor as a short-circuit (or perhaps a 0.1V diode). What then is the voltage at the collector-tap of the resulting divider? That will be the lowest DC level you can achieve.A CMOS schmitt-trigger may be a better "slicer", but your 2.5V input swing may be too small for reliable operation. A comparator is the most flexible and reliable solution. Ian.M's suggestion is also a good one.