Output problem of triode common emitter amplifier circuit

S3S4S5S6

Output problem of triode common emitter amplifier circuit [Copy link]

 

VCC is equal to 5V, and the input Vi is a small sinusoidal signal (for example, the signal is 0~1V). How to adjust this circuit (if one transistor is not enough in the circuit, you can add another one) so that the amplitude of the amplified signal Vo can reach (0~5V) similar to the rail output,

Because of the voltage drop problem of the transistor, the maximum output of the circuit in the figure is only about 4 volts.

chunyang

It's simply impossible, don't even think about it.

S3S4S5S6

chunyang posted on 2021-10-11 16:02 It is impossible to do it, don't even think about it.

Suddenly I thought of how others made rail-to-rail op amps

maychang

S3S4S5S6 posted on 2021-10-11 16:24 Suddenly I thought of how others made rail-to-rail op amps

The rail-to-rail op amp output stage is a push-pull output, but your circuit is a single-ended output resistor load, so it cannot achieve "rail-to-rail".

maychang

S3S4S5S6 posted on 2021-10-11 16:24 Suddenly I thought of how others made rail-to-rail op amps

The power supply voltage is 5V, and the output peak-to-peak value reaches about 4.5V, which is pretty good.

If this circuit wants to output a peak-to-peak value greater than 5V, the easiest way is to increase the power supply voltage.

S3S4S5S6

maychang published on 2021-10-11 18:53 The rail-to-rail op amp output stage is a push-pull output, but your circuit is a single-ended output resistor load, so it cannot achieve "rail-to-rail".

maychang Is the basic circuit of push-pull output composed of two transistors?

S3S4S5S6

maychang published on 2021-10-11 18:53 The rail-to-rail op amp output stage is a push-pull output, but your circuit is a single-ended output resistor load, so it cannot achieve "rail-to-rail".

Can you give me a practical circuit?

maychang

S3S4S5S6 Published on 2021-10-12 08:04 maychang Is the basic circuit of push-pull output composed of two triodes?

"Is the basic circuit of push-pull output composed of two transistors?"

yes.

If it is rail-to-rail output, it is mostly composed of two MOS tubes.

maychang

S3S4S5S6 posted on 2021-10-12 08:10 Can you give me a practical circuit?

Can you give me a practical circuit?

This can often be found in the rail-to-rail op amp manual.

But if it is constructed using separate parts, it would be quite troublesome.

S3S4S5S6

maychang posted on 2021-10-12 08:29 "Can you give me a practical circuit?" This can often be found in the manual of the rail-to-rail op amp. But if you want to use discrete parts to construct it, then...

Well, I thought I could still rescue it, but this road is blocked again. Ah, it’s harder

chunyang

S3S4S5S6 posted on 2021-10-11 16:24 Suddenly I thought of how others made rail-to-rail op amps

You can look at the internal circuit diagram of a rail-to-rail op amp, but it is of no use and cannot be used as a reference. If you need rail-to-rail output, just use the relevant op amp directly. Don't even think about the Class A common emitter circuit.

PowerAnts

Capacitor coupling, the average output voltage is 0V. If there is a positive, there must be a negative. So where does the non-negative output come from?

埋土书生

Maybe you can consider a differential output structure . The differential voltage of a 5V power supply should be able to reach more than 7V.

S3S4S5S6

chunyang posted on 2021-10-12 15:58 You can find the internal circuit diagram of the rail-to-rail op amp, but it is useless and cannot be used as a reference. If you need rail-to-rail output, just use the relevant op amp directly, A...

For this type of follower, with a power supply voltage of 5V, can the output be close to 5V? For example, the output is 4.9V

@chunyang

@maychang

maychang

S3S4S5S6 Published on 2021-10-21 08:11 For this type of follower, with a power supply voltage of 5V, can the output be close to 5V? For example, the output is 4.9V @chunyang @maychang

For this type of follower, if the input signal amplitude is large enough (larger than the power supply voltage amplitude), the output amplitude can reach about 4.6V~4.7V. It may be quite difficult to reach 4.9V.

maychang

S3S4S5S6 Published on 2021-10-21 08:11 For this type of follower, with a power supply voltage of 5V, can the output be close to 5V? For example, the output is 4.9V @chunyang @maychang

Bipolar transistors are different from field effect transistors. Bipolar transistors have a saturation voltage drop, while field effect transistors do not. When fully turned on, the field effect transistor is approximately a resistor. As long as the current is small enough, the voltage drop is also small enough.

The saturation voltage drop of a bipolar transistor is about 0.1~0.15V for a common silicon low-power transistor. Therefore, the amplitude of the push-pull output is 0.2~0.3V smaller than the amplitude of the power supply voltage.

maychang

S3S4S5S6 posted on 2021-10-21 08:11 For this type of follower, with a power supply voltage of 5V, can the output be close to 5V? For example, the output is 4.9V @chunyang @maychang

The circuit on the 14th floor, because there are two diodes at the input end, the base current of the upper transistor is supplied by the 10 kilo-ohm resistor during the positive half cycle of the input signal. When the base potential of the upper tube is equal to the positive power supply, the base current is zero, so the output end can only reach a maximum of 0.7V lower than the power supply voltage (actually it can only reach about 1V lower than the power supply voltage).

To make the output voltage 0.1-0.2V lower than the power supply voltage, the base resistor must be connected to a power supply with a higher voltage than the power supply. Since there is a higher voltage power supply, the two output bipolar transistors also use this higher voltage power supply. This is a bit wasteful for the power supply, but there is no need to use two power supplies, one high and one low.

Using only a 5V supply, the output amplitude can be slightly increased if a bootstrap circuit is used.

chunyang

S3S4S5S6 Published on 2021-10-21 08:11 For this type of follower, with a power supply voltage of 5V, can the output be close to 5V? For example, the output is 4.9V @chunyang @maychang

The emitter follower cannot do this. Your diagram is called a Class AB push-pull amplifier, which is used to eliminate crossover distortion and is mainly used for current amplification in the output stage of a power amplifier. It is not suitable for your application.

S3S4S5S6

maychang posted on 2021-10-21 10:37 S3S4S5S6 posted on 2021-10-21 08:11 For this type of follower, the power supply voltage is 5V, can the output be close to 5V? For example, the output is 4.9V @chu ...

In the figure in the first post, Rb and Rc share a common power supply Vcc. Is it possible to use two power supplies V1 and V2=Vcc to supply Rb and Rc respectively?

Because I hope V1 is a constant value and is not affected by power supply fluctuations. As mentioned in the previous post, the most direct and effective way is to increase the power supply voltage. The actual situation is that the power supply voltage of 12V fluctuates greatly and can reach 3V. If the base is powered by another separate stable power supply, is it feasible?

maychang

S3S4S5S6 Published on 2021-10-22 14:36 In the first post, Rb and Rc share the same power supply Vcc. Can two power supplies V1 and V2 = Vcc be used to supply Rb and Rc respectively? Because I...

You can use two DC power supplies to supply power separately, but as long as the power supply for the collector in the first post is 5V, the output amplitude can only reach 4.5V. Supplying power separately cannot increase the output amplitude.