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I am trying to make some simulations of the BC547, but I have some problems.

I adjusted the hfe (beta) to 320 and measured the base current and the collector current as below:

enter image description here

The RMS value of the base current is 33.552uA. RMS value of the the emitter current is 1.17mA.

If we multiply base current by hfe, the result is 10.73mA but Multisim shows it is 1.17mA.

Why are calculation and simulation so different? Are my calculations wrong ?

If I wait some time, the base current becomes 68uA and the emitter current 1.4mA.

The values of the components are as below:

enter image description here

JRE
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Piko
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  • β only applies when the transistor is in forward-active mode. Your transistor is in saturation. – Hearth Dec 26 '20 at 17:47
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    Try probing the collector and emitter voltages of Q1 (doing transient analysis). – user57037 Dec 26 '20 at 17:56
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    You should verify the dc bias conditions of your circuit. Make sure that you changed beta correctly. – Elliot Alderson Dec 26 '20 at 17:56
  • @Hearth is not saturation current 12/3.9k = 3.076mA ? – Piko Dec 26 '20 at 17:57
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    @Piko in order to get 10mA rms through a 3.9k resistor, you would have a voltage swing of 39V rms. But your supply is only 12V. You are not in saturation at DC, but during the input voltage swing you are hitting saturation. Please look at the collector voltage waveform. It will be illuminating. – user57037 Dec 26 '20 at 18:01
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    As @mkeith says, look at the collector waveform. Then knock the input level down to 10% of its current value and try again. –  Dec 26 '20 at 18:04
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    The time constant of R4 and C1 is 1000 * 0.1 = 100 seconds. So the DC operating point may take a very long time to arrive at its final DC steady state (several minutes). This is kind of inconvenient for transient analysis simulation. You may want to consider reducing C1. Maybe 10uF would be OK in this case. – user57037 Dec 26 '20 at 18:17
  • @mkeith Yes, absolutely, I had checked the waveform and it was not like an amplifier. I increased the Vcc, 12V to 39V even 60V but result is same. I reduced the amplitude of the signal but result is same again. – Piko Dec 26 '20 at 18:20
  • @mkeith After I replaced C1 with 10uF, the circuit was worked like an amplifier. However, the base and collector current is not right. Base current is 855nA, collector current is 1.341mA so, hfe is 1567. – Piko Dec 26 '20 at 18:28
  • Maybe change V1 into a current source or add a series resistance so you can get the AC input current down to a low level. Also, even though I think the DC bias point is OK, like Elliot said, you should double check that. Once you confirm DC bias is OK, start with 1uA AC current input and go up from there. – user57037 Dec 26 '20 at 18:29
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    Make sure you are looking at peak to peak current. RMS may possibly include the DC component of current (bias current). – user57037 Dec 26 '20 at 18:40
  • @mkeith I have to look at RMS value because my laboratory document says it. I don't know why, my laboratory document says C1 should be 100mF. – Piko Dec 26 '20 at 18:44
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    OK, well, do AC analysis then (not transient analysis). Hopefully your simulator will do the DC analysis first, and then do AC analysis at the DC operating point. I am not that knowledgeable about simulation tools and how they work. I usually only do transient analysis on simple circuits. Maybe try to talk to the professor or an assistant if possible. – user57037 Dec 26 '20 at 18:49
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    "I don't know why, my laboratory document says C1 should be 100mF" - this is probably a typo. In the old days 'microfarads' was often abbreviated as 'mfd', so 100mF would be written as '100,000mfd'. – Bruce Abbott Dec 26 '20 at 18:51
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    I agree with @BruceAbbott. Also, the point of that capacitor is to make the emitter resistor disappear at high frequencies. Since your signal is 400 Hz, we really only need the cutoff frequency of RC to be way, way below 400 Hz. 100uF is more than enough for that purpose. – user57037 Dec 26 '20 at 22:27

1 Answers1

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If I wait some time, the base current becomes 68uA and the emitter current 1.4mA.

This suggests a capacitor in the circuit is taking some time to charge up to a stable DC voltage. In your schematic C1 = 100 mF. This is a very large value (1/10th of a Farad!). 100 μF would be more reasonable.

I simulated your circuit in LTspice using a BC547B (which has HFE of ~300). Here is a plot of Emitter and Collector voltage during the first 200 seconds after switch on.

enter image description here

It takes over 20 seconds for the Emitter voltage (green line) to rise enough to get the transistor out of saturation, and over 100 seconds to reach a stable voltage. The Collector voltage (which looks like a solid area due to the high signal frequency) shows that the input level is too high, causing the output to bang against the supply rails as the transistor turns fully on and off.

The plot below shows Base current (green), voltage across C2 (red) and Collector voltage (blue) over a few cycles of the signal:-

enter image description here

The input signal is so high that the Base-Emitter junction is acting as a rectifier, charging C2 with negative current which reduces the Base bias voltage from ~2 V to ~1.5 V. The output waveform is strongly clipped on both positive and negative peaks, causing massive distortion.

After reducing the input level from 0.125 V to 0.0125 V, it looks much better:-

enter image description here

Now the Base current is continuous (with negligible rectifier action), Base bias voltage has returned to normal, and the output waveform only shows a small amount of distortion.

Why are calculation and simulation so different? Are my calculations wrong ?

Your calculations are based on a small signal level which doesn't cause significant distortion or change the DC bias level. It also assumes that all capacitors are charged to their final DC values.

When simulating a circuit with potentially large AC voltages it always pays to inspect the waveforms. Average, rms, peak and DC values do not tell the full story and can be hard to interpret.

If some elements may cause significant time delays then simulate for long enough to ensure that the circuit has stabilized. If this takes too long then either change component values to reduce the delay, or set the required initial conditions to skip it. be aware that simulating a small time period may give misleading results.

Bruce Abbott
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    Thanks for taking the time to write this all up. Very well done answer IMO. Hopefully it is helpful for the OP also. – user57037 Dec 26 '20 at 22:29
  • Your answer so celar and helpful, I got a lot of point from your answer. Thank you so much. I have also one question more. When I measure the output and input voltage in proteus, I see up point is 1.7V and down point is -2V. Mustn't they be same ? İn multisim, they are same. Do you think it is about proteus ? I adjuster all values like yo did. Here the picture https://hizliresim.com/cfdg5W – Piko Dec 27 '20 at 11:54
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    Different simulators have different models, and some are more accurate than others. The output has some distortion due to the Base having an exponential I/V curve, so the voltage gain increases as output voltage decreases. This causes the tops of the sine wave to be slightly flattened while the bottoms are elongated. The result is an asymmetrical AC output that goes down more than up. Seems that Proteus models this more accurately than Multisim. – Bruce Abbott Dec 27 '20 at 12:20