CMOS gate as linear amp?
Posted: Sat Mar 30, 2024 9:33 pm
Not ideal, but if one has a left-over logic gate and a small signal linear amp is needed, the temptation is great:
Microcontroller Group, Moorabbin, Melbourne
Ref: http://melbmcu.weebly.com/cmos-gate-as- ... ifier.html
CMOS gate as a linear amplifier
By David Stonier-Gibson
There are times when you are making an essentially all-digital circuit, but need a little bit of linear amplification. Examples of this would be a microphone pre-amp for a sound triggered switch, or a photodiode pulse pre-amplifier. I have used this several times throughout my career, when a left over gate can be used instead of having to add another chip to the design. It works with any inverting gate, meaning NAND, NOR or NOT (usually called inverter). NAND and NORs have more than one input. Use one of the inputs, then tie the others to ground for NORs and + supply for NANDs.
Note: This trick will only work with unbuffered CMOS gates or inverters. That means, for 4000-series chips, the ones with a U suffix, such as 4002UB. For 74 series chips it means ones with a single 'C' in the part number, such as 74C04 (but not 74HC04. It also won't work with Schmitt trigger parts such as 40106UB or 74C14. And it certainly won't work with TTL chips like 74LS04, 7404, 74S04).
The simplest circuit
I am illustrating my circuits with an inverter, but they will also work with NAND and NOR gates (see above).
Oops, I forgot to draw the ground (power) connection!
This circuit will work well with a fairly low impedance signal source (say an electret microphone). The voltage gain will depend on the specific chip you are using, and on the supply voltage. I used a 4069UB from Texas instruments, and recorded a gain of about 40 with a 5V suppy. You can go to higher supply voltages, up to 15V with 4xxx and 74Cxx chips - more on this below
The oscillograph shows the input (bottom, red) at 50mV/div, so we have about 50mVpp. The top/green trace is the output, about 2Vpp, which is about as much as I could get out of it before getting visible distortion. I am not sure what the "fuzz" is on the input trace. I used my nScope for this, and I am pretty sure it is simply noise in the nScope. The low frequency cutoff (-3dB) point will be the capacitor versus the sum of the source impedance and amplifier input impedance. If the source impedance is very low, we can just use the amplifier input impedance of 500K/40, or about 25K, giving a cutoff of about 65Hz.
The way this works is that the 1M resistor provides DC feedback, and ensures that the circuit will stabilise itself to a quiescent voltage of about half the supply voltage. The signal to be amplified has to be couple via a capacitor, to avoid disturbing the DC operating point. It's best to not use an electrolytic capacitor, because leakage currents will disturb the operating point.
The input impedance will be roughly the feedback resistor (1M) divided by the gain of the chip (in my case, 40), so about 25K.
A little fancier... click the reference above! -This kludge can be tidied up enough to make it usefu?
Microcontroller Group, Moorabbin, Melbourne
Ref: http://melbmcu.weebly.com/cmos-gate-as- ... ifier.html
CMOS gate as a linear amplifier
By David Stonier-Gibson
There are times when you are making an essentially all-digital circuit, but need a little bit of linear amplification. Examples of this would be a microphone pre-amp for a sound triggered switch, or a photodiode pulse pre-amplifier. I have used this several times throughout my career, when a left over gate can be used instead of having to add another chip to the design. It works with any inverting gate, meaning NAND, NOR or NOT (usually called inverter). NAND and NORs have more than one input. Use one of the inputs, then tie the others to ground for NORs and + supply for NANDs.
Note: This trick will only work with unbuffered CMOS gates or inverters. That means, for 4000-series chips, the ones with a U suffix, such as 4002UB. For 74 series chips it means ones with a single 'C' in the part number, such as 74C04 (but not 74HC04. It also won't work with Schmitt trigger parts such as 40106UB or 74C14. And it certainly won't work with TTL chips like 74LS04, 7404, 74S04).
The simplest circuit
I am illustrating my circuits with an inverter, but they will also work with NAND and NOR gates (see above).
Oops, I forgot to draw the ground (power) connection!
This circuit will work well with a fairly low impedance signal source (say an electret microphone). The voltage gain will depend on the specific chip you are using, and on the supply voltage. I used a 4069UB from Texas instruments, and recorded a gain of about 40 with a 5V suppy. You can go to higher supply voltages, up to 15V with 4xxx and 74Cxx chips - more on this below
The oscillograph shows the input (bottom, red) at 50mV/div, so we have about 50mVpp. The top/green trace is the output, about 2Vpp, which is about as much as I could get out of it before getting visible distortion. I am not sure what the "fuzz" is on the input trace. I used my nScope for this, and I am pretty sure it is simply noise in the nScope. The low frequency cutoff (-3dB) point will be the capacitor versus the sum of the source impedance and amplifier input impedance. If the source impedance is very low, we can just use the amplifier input impedance of 500K/40, or about 25K, giving a cutoff of about 65Hz.
The way this works is that the 1M resistor provides DC feedback, and ensures that the circuit will stabilise itself to a quiescent voltage of about half the supply voltage. The signal to be amplified has to be couple via a capacitor, to avoid disturbing the DC operating point. It's best to not use an electrolytic capacitor, because leakage currents will disturb the operating point.
The input impedance will be roughly the feedback resistor (1M) divided by the gain of the chip (in my case, 40), so about 25K.
A little fancier... click the reference above! -This kludge can be tidied up enough to make it usefu?