0
\$\begingroup\$

I want to design amplifier in 20kHz - 200 kHz range. I have designed following circuit using BC547 transistor.my circuit

I have set VCE voltage to 6v and set Q point to the middle of the load line. But when I supply 20 kHz 200mv peak to peak sinusoidal signal, the output signal shape in two half cycles are not identical. oscilloscope output

How can I solve this problem?

\$\endgroup\$
5
  • 2
    \$\begingroup\$ Put a scope trace on the collector of the BJT. Now do you see what the problem is? \$\endgroup\$
    – The Photon
    Commented May 31, 2020 at 5:41
  • 1
    \$\begingroup\$ What is the voltage gain of your amplifier? \$\endgroup\$
    – G36
    Commented May 31, 2020 at 6:13
  • \$\begingroup\$ voltage gain is 57 \$\endgroup\$
    – jan411
    Commented May 31, 2020 at 14:37
  • \$\begingroup\$ @Jan You want to achieve a voltage gain of 57? Why that number, in particular? Are you willing to accept other values near that number? (It's hard to make it an exact value in reality.) Would other values much less than that be okay? Say, 15 or 20? \$\endgroup\$
    – jonk
    Commented May 31, 2020 at 18:12
  • \$\begingroup\$ My 5 minute hand calc of your circuit resulted in a gain of 153. In my simulation i saw a gain of 114. How did you arrive at a value of 57? Even your simulation suggests that when unclipped, the output should be roughly 11Vpp, a gain of 110. \$\endgroup\$
    – Yet Another Michael
    Commented Jun 1, 2020 at 7:56

1 Answer 1

Reset to default
2
\$\begingroup\$

Let's look at your design step by step.

First, the current through R1 and R2. Because those resistor values are fairly small, I will assume that the BJT base current is a tiny percentage of the R1 current, and will ignore it: that is I_R1 = I_R2. From that, we can find the DC voltage at the base of the transistor using a simple voltage divider. $$V_B = 12 * \frac{8k}{8k + 22k}$$ $$V_B = 3.2V$$

Nice, so we have the BJT base voltage, and we know that there is a base to emitter diode, so we can expect the emitter to be roughly 0.7V lower than the base voltage. My assumption here is V_BE = 0.7V, but with a quick simulation you should see similar values.

So now we have the DC voltage at the collector terminal. The resistor R4 is the DC conduction path to ground, so we can calculate the current using ohm's law. Solving 2.5V / 270Ω gives a whopping 9.26mA! This is a huge current btw, in later revisions I advise reducing this current by a factor of at least ~10.

Finally, we will assume that I_R3 = I_CE = I_R4. I called the base current negligible again. Using ohms one last time:

$$V_C = 12V - (430Ω * 9.26mA)$$ $$V_C = 8.018V$$

This voltage only gives you 4 volts of swing toward the positive rail. That's why you are clipping.

However, your problems go beyond that. Your input voltage is way too large! You have emitter resistor bypassed with (huge) cap, so:

$$gm ≈ I_C / V_t$$ $$I_C = 9.26mA,_. V_t ≈ 26mV $$ $$gm = 0.356$$

which when multiplied by the output resistance (≈R3) gives a voltage gain of 153. When you multiply your 100mVp input by this, you get over 15Vp! Your rail will never support this even if you did fix the collector voltage.

Try your simulation again with a input voltage of 20mVp. You will see some distortion due to linearity violations, but that should get you a roughly 3Vp output (20mVp * 150V/V = 3Vp).

Also, your 1000uF cap can be swapped out with a value of ~10uF. Your wallet and footprint will thank me later.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.