amplificator

[yalect]

Thank you for your reply

I mean if you look to device that works on single cell like 9v for DC supply and contains an op-amp so it doesn’t include 0v or ground reference, is designed to work with certain voltage reference ?

[johnsold]

An op amp does not require any specific reference voltage. Many times it is easier to think about how it works if there is a reference at a convenient level.

Look closely at the data sheets for an op amp. The LM358 you are using is suitable.  I am looking at the Texas Instruments data sheet. If you have the data sheet from a different manufacturer, the details may be slightly different.

On page 4 are Electrical Characteristics for single supply (Vcc = 5 V) operation.

Common mode input voltage range: 0 to Vcc-1.5

High-level output voltage: (Vcc = 30 V) 26 V min at 2000 ohms load.

Low-level output voltage: 20 mV max.

For comparison consider the TL071 amplifier which is specified for split supply (Vcc = +/-15 V) operation.

Common mode input voltage range: +/-11 V

Output voltage range:  +/-10 V min at 2000 ohms load.

Notice that in both cases the limits approach but generally do not include the power supply voltages.  But operation near the center of the power supply ranges is always inside the acceptable ranges for both input and output.

Lynn

[yalect]

Hello, thank you for yor replies and that care

I would like to ask  you, generally What about some devices like that include op-amp like mobile phone and some receivers like radio remote control, are they designed to work with 0v and ground or they have "pseudo-ground" or reference voltage at half the power supply ?

I want go back to my circuit , if I inject signal to the base of transistor (changing it with 2n222 for high frequency) has amplitude of 1mv and frequency of 25Mhz, the signal is modulated by Data means when the Data is 1 the frequency is founded and when the Data is 0 no frequency is existed as it shown on second picture, Can that circuit work ? and how I can demodulate the signal to recover that Data after the transistor for amplifying it by op-amp ? I have to use diode or just RC circuit ? I know that some changes I have to do about the capacitors values, may you can help me about calculation and what if the coupling capacitor could be linked directly to the op-amp or I must to adapt input impedance of op-amp by resistor ?

My regards

[johnsold]

Whether to use a "pseudo-ground" reference or not is a choice that each designer needs to make. For AC coupled circuits using a midpoint bias is almost always easier. If the signals must be DC coupled and the input and output are referenced to the most negative power supply voltage (regardless of what it is called), then the situation requires careful examination of the voltages at each stage of the circuit and evaluation of errors due to limitations on signals reaching the power supply voltages.

The op amp (LM358) is not fast enough for your signal. For a signal of that type I would use AC coupling with a mid-power supply bias for the op amp.

The detector can be a diode followed by the RC low pass filter. The time constant needs to be long enough to keep the voltage close to the rf peaks while the rf is present and short enough so that the voltage decays to the point where the logic signal goes low again as soon as possible. With the frequencies shown it will be difficult to find a suitable time constant.  If the signal at the detector has very little noise, a time constant ~ one period of the rf signal is probably the best place to start. You will probably get some phase shift and duty cycle changes.

If you cannot find a suitable compromise RC time constant that gives suitable results, you will need a more sophisticated detector.

Lynn

[yalect]

Hi, I want to comment about diode detector if you observed that the signal of transistor swings at one direction means a just RC filter can’s remove the swing of the signal while the capacitor of RC filter discharge, is it ? and what about impedance of op-amp input is suitable for charge and discharge of the coupling capacitor ? and how its calculations are ?

greeting

[yalect]

Hi, might you don't reply for last post I want to talk about RC filter

I think that I have to use serial RC cause of the parallel RC resistor will force the op-amp to swing up to V+ and stay there or connecting C2 the coupling capacitor to the op-amp directly instead of the transistor collector but I don’t know if DC courant can affects on RC filter

about RC charging and discharging, don’t you think that charging capacitor passes throw the R and discharging in two way R and op-amp input impedance means the impedance of op-amp can affect on the RC constant ?

I found that it’s better to make the constant time RC equal to one period of rf frequency, less will block the signal and more rf will not allow to C enough time for charging  

what are the coupling capacitor calculations formula ? and maximum practical frequency of the op-amp according to the gain my circuit (100) attached circuit ?

My regards

[johnsold]

I apologize. I seem to have missed your previous reply.

The biasing method you use for the transistor circuit does not allow accurate prediction of the operating point because it depends strongly on the current gain and temperature of the transistor. While it can be non-linear, it does not easily go into saturation or cutoff (except for large input signals). Without knowing the operating point it is difficult to estimate how effectively and at what input levels the signal will be rectified. 

About the filter. The filter must be configured to be a low pass filter. A low pass filter allows low frequencies (including DC) to pass through while attenuating higher frequencies.  The coupling capacitor (C2) serves as a high pass filter to eliminate the DC bias point. The result is that you have two filters: The output of the transistor is high pass filtered by C2 and a combination of R3, R1 or R2 and R7. The output of the highpass filter is then low pass filtered by R1-C2 or R2-C1 and R7. Since R7 = R1 or R2, the filter is affected by both resistors.

I do not have time right now to fully analyze your filters.

Lynn

[johnsold]

Part 1. The transistor amplifier.  The operating point in that circuit can be calculated by an iterative process which is essentially what circuit simulation programs do.

Start with these assumptions:

- The currents at time t = 0 are all zero.

- The DC current gain is ~100.

- The forward drop across the base-emitter junction Vbe = 0.6 V

Iteration 1. The base current Ib = (6 V - 0.6 V)/(80.1 kohms) = 67.4 uA. The collector current Ic = 100*Ib = 6.74 mA. The collector voltage Vc = 6 V - Ic*R3 = 6 V - 6.74 mA*100 ohms = 5.33 V.

Iteration 2. Ib(2) = (Vc(1) - 0.6 V)/80000 ohms = 65.8 uA. Ic(2) = 6.58 mA. Vc(2) = 5.34 V.

Iteration 3. Ib(3) = 59.3 uA. Ic(3) = 5.93 mA. Vc(3) = 5.41 V.

Iteration 4. ib(4) = 60.1 uA. Ic(4) = 6.01 mA. Vc(4) = 5.40 V.

The changes are small so additional iterations will not get much closer.

Now, what is the ac gain of this circuit? The input current is limited primarily by the small signal input impedance, hie. For a 2N2222 hie can be from 250 to 8000 ohms for collector currents in the range of 1 to 10 mA. The small signal current gain (hfe) is 50 to 375 for the same collector current range.

If we assume hie = 1000 ohms and hfe = 100 then we can calculate the gain as follows:

Let the input voltage be vi = 1 mV. The input current ib = vi/hie = 1 uA. The collector current is ic = ib*hfe = 100 uA. The collector resistor is 100 ohms so the collector voltage vc = ic*Rc = 10 mV. Thus the small signal voltage gain is 10.

Note that the collector bias point is about 5.4 V. So a signal would need to swing about 0.5 V positive from that point before very much non-linearity will occur. The signal at the input would probably need to be > 200 mV pp before the output would look like it was being rectified.

Part 2 The filter. First consider the charging of capacitor C2 in the series case (C2 R2 R7 in series).  When the power is first applied to the circuit with no current flowing in the transistor and no current flowing through the op amp input, we have a series circuit with R3, C2, R2, R7, and R6 (ignoring C2 for the moment). We have two possibilities:

1. The inverting input of the op amp acts as a virtual short to the non-inverting input. This would hold that point to 3 V.  The required initial current is (6 V - 3 V)/(R3 + R2 + R7) = 3 V/2100 ohms = 1.43 mA. For the opamp to deliver 1.43 mA to the inverting input the output voltage needs to be 143 V more negative than the input or -140 V!

2. Since that is not possible, the voltage at the inverting input will not be at 3 V until the capacitor charges through all the resistors. The time constant is 71.3 ms. It will probably take more than 200 ms after power is applied before the capacitor charges enough for the op amp to come out of saturation.

Ignoring the coupling capacitor ( C2 = 700 nF) the circuit for the low pass filter is essentially a voltage source with a 100 ohms source impedance (the output of the transistor amplifier) driving R2, C2, and R7 and the op amp with feedback resistor R6. A straightforward network analysis of that circuit gives a transfer function:

Vo/Vi = -(R6/R7)*(1/(1 + (R3 + R2)/R7 + sC1*(R3 + R2))), where s is the complex Laplace transform variable.

The corner frequency of this network is at wC1*(R3 + R2) = 1 + (R3 + R2)/R7. For the resistor values in your circuit and w = 2*pi*f and f = 25 MHz, C1 = 12.2 pF.

No circuit simulation programs were used in this analysis of the circuit. If you want to design circuits, you should understand how to do this type of analysis.

Lynn

[yalect]

Hi, I would like to thank you for analyzing my circuit might it take a time, thank for that effort, I am thankful

I will be back for that analyze later, I want to talk about demodulation signal some use diode for detecting and other just RC filter, I think that simple transistor can detect and demodulate cause of the signal of the transistor swings at positive range (DC bias voltage is half DC voltage supply) if we consider that the input signal amplitude is 1mv and voltage gain is 10, Vcc=6, Means the output voltage will be( 3V+or-10mv )or (2.99 to 3.01v) positive so, RC circuit can’s decrease that swinging while it’s discharging from 3.01v to lower

thanks

[johnsold]

1. Note that the collector voltage in the absence of a signal is approximately 5.4 V, not 3 V.

2. For an input signal of 1 mV the output swing will be nearly symmetrical. This implies that not detection will take place.

When the input signal amplitude exceeds about 50 mV (peak) then you will begin to get some distortion at the output. As the input signal gets larger, the distortion will look more like rectification and the filter will begin to do some detection. You will not get any detection on a 1 mV signal.

3. A diode will not do much detection with that 1 mV signal either because it is too small to forward bias standard diodes.

Lynn