Don't understand you question. What do you mean "voltage gain of collector to base bias circuit"?hi
how to calculate the voltage gain of collector to base bias circuit using a pnp transistor ? iam using BC 558, Rb= 1M, Rc= 4.7k , Vcc= 9V.
No. Maybe you did not read my last reply.1st transistor - RC coupled amplifier- iam using BC 548, hfe= 224, without byepass capacicot ,CE. therfore gain is 2.2k/680= 3.2 . is it correct?
Every tutorial about the voltage gain of a transistor has the equation.Vcc=9V, VRc= 2.55.
2nd transistor- BC 558 , mention gain will be 4.7k/26 ? how we calculate this 26? any equation?
Because it is driving the non-linear low impedance base-emitter diode of the second transistor.in the above circuit, input to RC coupled is 200mv to 60mv. after the first stage 500mv to 150mv. but the sine wave is not perfect. wide in the negative half cycle.
Of course the output is saturated. The input signal level is too high or the gain is too high. In my simulation I turned it down then measure a gain of 225 times. With the input at 200mV peak the output is trying to be (0.2V peak x 225= 45V peak but it can't since my supply is only 9V.after the pnp stage, output will be 9v to 5.8v. but sine get saturated when the input is 200mv to 160mv. i think gain is high.but i have to obtain a perfect sine wave. to reduce the gain i need an equation for collector to base bias circuit? i am trying to reduce 4.7k to 1k.. and 1M to 220k.. but still saturated
The 2N3906 has almost the same spec's as the BC558. But Fairchild says its input impedance at 1mA is only 500 ohms and my 1969 Philips databook says that a BC178B has an input impedance of 6k ohms at 1mA.in ur previous reply, mention the input impedance of a PNP 2N3906 transistor at 1mA. It is only 500 ohms. The 1M base resistor applies a little negative feedback which reduces the input impedance to about 480 ohms.
but iam using BC 558. is it same 1mA and 480 ohms for both?
My simulation has an input of 200mV peak and the output was saturated. With an input of 10mV peak the output was a good sine-wave and the gain was 225 times.i am doing obstacle detection. when obstacle comes closer, reciver setion picks 200mV peak to peak sine wave. it is fed to RC coupled and output is 500mVp-p. then it is given to pnp stage, output is 9Vp-p. but it is saturated.
input after pnp
200mV 9V saturated
180mV 8.8V sat
160mV 8.6V sat
100mV 7.6V sat
60mV 5.8V better sine
when obstacle moves far ie, 15cm distance, input is 60mV.
u mention gain about 101. therefore the above values are satisfying ?
Because it did not have enough base current for a 1k collector load rersistance.i tried to change the Rc of pnp stage from 4.7k to 1k. therefore output is saturated in negative cycle. why saturation only in negative cycle?
RE adds negative feedback. It reduces the gain because gain= Rc/Re.in ur last reply, mention about adding Re. actually what happens when Re added?
You have an opamp trying to saturate then it feeds the clock of the CD4017.this sine wave given as clock to divide by 10 counter 4017 , it divides 40kHz to 4kHz. is it required square wave to apply as clock? or any wave can be used as clock?
Because I wrongly looked at the datasheet of Fairchild's 2N3906 similar transistor that must have an error about its input impedance (it shows 500 ohms).in ur previous reply, mention about gain is 101 times by calculation. but in simulation why it is 225 times?
therefore actually gain is 225 or 101?
The transistors must be re-biased if you change their Rc. But adding an emitter resistor reduces the gain better because it adds negative feedback that reduces distortion while it reduces gain.Rc of pnp and npn actually controls gain. but if i vary both, still the signal get sturated. why? without 1.5k as Re.
It cuts high frequency gain to reduce radio interference.in pnp section, it is collector to base bias circuit, actually there is no capacitor across base and collector? why a 10pF capacitor used?
The output will have an average voltage of 0V when it has an output coupling capacitor feeding a resistor to ground. Then the signal swings positive and negative. Your opamp has a coupling capacitor so the output signal swings positive and negative.supply is 0 to 9V. therefore after pnp stage, amplified value goes to below zero? is it clipped becoz of that?
Correct.u mention about 45V, actually this wave has both negative peak and positive peak or 0 to 45v or -22.5 to 22.5.
i know supply 9V, therefore it will saturate, thats y we cant get 45v.
My simulation shows that with an input level of 10mV peak then the output of your transistor circuit has low distortion and swings from +3.0V to +7.6V.but i want to know Vp-p value.
A 741 opamp is 45 years old. It has a minimum output of +1.5V and a maximum output of +7.5V at low frequencies when it has a +9V supply. At 4kHz its low slew rate changes a square-wave into a triangle-wave.in the circuit, i am using 741 opamp as amplifier at final stage.
The 741 opamp is spec'd to drive a load that is a minimum of 2000 ohms, not 8 ohms. The inductance of an 8 ohm speaker causes its impedance to be about 40 ohms at 4kHz. The typical maximum output current of a 741 opamp is 25mA peak so the maximum output to a 40 ohm load is only 1V peak. Then the power in the speaker is only 12.5mW. Headphones are louder.opamp 741 used to drive a speaker of 8ohm. iam using inverting amplifier. therefore output voltage swing between which values?
You calculated when the transistors have "typical" current gain. When the current gain is higher or lower then the bias point and load line will be different.RC coupled - npn section.
i have to draw the dc load line. i got Ic as 1.29mA. Vce= 50% of Vcc = 4.5v , hfe= 224. is required Q point will be at 4.5v? in hardware, i got Vce= 5.24v. is this Ic and Vce correct to show the bias point?
The gain is 9.25 times in my simulation.after adding Re = 1.5k, i got gain as 10. is that correct?
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