amplificator

[yalect]

Hi,

Please, can you tell me what the better way I can link between the transistor and op amp as shown on attached picture and if I inject a signal modulated by digital data means 1or (5v) there is frequency and 0 or (0 v) no frequency founded, and how I can demodulate the frequency for getting  the data after the transistor so, from collector of the transistor ?

My thanks  

[Tien_P]

Have a look at this link to learn more about modulation and demodulation.

https://sites.google.com/a/asu.edu/signals-and-systems/home/am-modulation-and-demodulation-example

[yalect]

Hi,thank you for your reply

what about better way for  linking between op amp and transistor ?

thank you

[Tien_P]

Please elaborate, I am not sure what you want to do.

[yalect]

Hi, thnk you for your reply

my goal is to Link between transistor and op-amp as it’s shown on attached picture but I think I have to match the  impedance of transistor output with op-amp input, because impedance of transistor output low and impedance of op-amp input high

Therefore I make resistor between the negative DC supply and inverting input

what do you say ? and that circuits can works ?

[johnsold]

Typically you only need to match impedances when you are concerned about power transfer between stages. In this circuit you are only concerned about the voltage so impedance matching is not necessary.

You do need a resistor at R5 to provide a path for current through C2. 

The circuit as shown is not an amplifier. Or, more correctly, it is an amplifier with very high voltage gain and no feedback on the output stage. The result is likely constant at ~5.5 V for small input signal amplitudes and a square wave at higher input amplitudes, assuming a sinusoidal input signal.

Lynn

[yalect]

Hi, thank you for your reply, you mean that circuit is suitable for amplifying digital signal which has two state only (0 or 1) not for variant amplitude

greeting 

[johnsold]

That circuit (without feedback around the op amp) operates at the open loop gain of the amplifier. For the LM 358 the typical open loop gain is 100000. Considering that the output swing is from ~0 to Vcc -1.5 V, or ~4.5 V for Vcc = 6 V, the voltage range at the inverting input of the op amp for which the output is within the linear range is approximately 2.999970 to 3.000015 V. The signal at the inverting input is ground referenced due to R5. So, any time the signal from Q1 exceeds 6 Vpp, you will get a pulse out of the op amp. For smaller signals the op amp output will stay saturated at the maximum output voltage of about 4.5 V.

Lynn

[yalect]

Hi,

In  previous circuit I forgot the resistors of feedback but even with resistors of feedback I marked that voltage of biasing v +input is fixed means every time the input signal is small the difference between inverting and non-inverting input will be large than will be amplified by ratio of feedback resistors, may that causes saturating output voltage ?

for example if v – input = 100mv, v +input = 3v

with Gain = 100, the result is : (3 - 0.1) * 100 = 290v !

 you said, I do need resistor at R5 to make current through C2, means cause of the high impedance of op-amp input It’s butter to make resistor at R5 to allow for current passing at low resistance and to create a voltage can be produced to op-amp input ?  

greeting

[johnsold]

Because you are using a single ended power supply, you need to bias the inverting input to approximately half the power supply. Replace R5 with a 1 megohm resistor. Add a 1 megohm resistor from the R5-C2 junction to the+ power supply. That will set the DC level at the inverting input to half the power supply and keep the time constant with C2 the same.

Lynn