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Wire the circuit shown:



  • R1 = 68KΩ
  • R2 = 10KΩ
  • R3 = 10KΩ
  • R4 = 200Ω ( two 100Ω in series )
  • R5 = 2.2kΩ
  • C1 = 0.1 µF
  • C2 = 0.1 µF
  • Q1 = 2N3904

enter image description here

  • Set the sig. Gen to 50mv RMS.  With frequency equal to 10KHz.
  • Record vo for the frequencies listed and calculate Av for each frequency:

f: 10kHz, 25kHz, 50kHz, 100kHz, 500kHz, 1MHz, 2MHz, 3MHz

This is what I got when I attempted to do the problem. However, I don't think it is correct because frequency does not affect my answers. Can someone tell me if my answers should be different and if frequency affects voltage gain?

vo 3.34m 3.35m 3.37m 3.45m 5.39m 9.02m 16.52m 23.42m

f 10kHz 25kHz 50kHz 100kHz 500kHz 1MHz 2MHz 3MHz

Av 0.066 0.067 0.067 0.069 0.107 0.180 0.330 0.468


Okay, so I did my circuit all over again, but I am even more confused now because I feel like my wiring is okay. However, this time, the output voltage is not changing at all when I increase the frequency. It stays at a constant 6.72V. I don't know what I'm doing wrong :(

enter image description here

enter image description here

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    \$\begingroup\$ Maybe your lab TA? We don’t do homework problems without any attempt at a solution. \$\endgroup\$ – John D May 10 '18 at 3:04
  • \$\begingroup\$ I did it already, but I don't think is correct because there is no effect from frequency in my answers. So I guess my real question would be is there any effect on voltage gain from frequency? \$\endgroup\$ – Carlos May 10 '18 at 3:06
  • \$\begingroup\$ @Carlos Thanks for putting in values, schematic, and what you measured as results. Not a lot of folks do anywhere near that much. +1 \$\endgroup\$ – jonk May 10 '18 at 3:12
  • \$\begingroup\$ @Carlos What do you already know about this particular design? Can you do any calculations on paper to make estimates of the quiescent state of the amplifier? Or are you just following instructions right now? \$\endgroup\$ – jonk May 10 '18 at 3:14
  • \$\begingroup\$ To answer this, it depends on the capacitor choice. Usually they will be sized in such a way that they won't damp the desired frequency range (they all act as high pass filter in that circuit). From your answer, it seems that the gain Av increases when the input frequency increase. However, keep in mind that the capacitor C2 will short R5 and removing it from your gain equation. In that circuit, the theoretical Av gain is approximately -R3/R4 if the frequency causes no attenuation of the input signal. \$\endgroup\$ – Simon Marcoux May 10 '18 at 3:15
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I'm going to assume you were just handed this schematic and asked to perform some tasks. The results you got do not sound correct for the circuit. (It's not a great design. But it's enough to get by.)

The first thing to check is the voltage of the power supply. If this is a \$9\:\text{V}\$ battery that was sitting on a lab shelf, definitely check it out with a meter -- using a \$1\:\text{k}\Omega\$ resistor across its terminals while you use the meter to measure the voltage. If this is a commercial power supply, check it anyway. It doesn't hurt to verify that your voltage is what it is supposed to be.

The next thing to check is the resistor values. Make sure they are near the values you are told they should be. Again, a few moments spent here can save you lots of time, later. It's only five resistors, so it shouldn't take a lot of time.

It's a good idea to have checked the datasheet for the specific BJT you are using. The 2N3904 comes in a variety of packages and I wouldn't be surprised to find different pinouts even for the same TO-92 package. It's worth a moment to make sure you know where your pins are located. I won't suggest that you spend time checking out the BJT, since you will be doing that soon enough when you plug it into the circuit. But if you do have a transistor checker available, please do use it.

Finally, check your wiring two or three times over. I make mistakes. So you can make them, too. Check things over several times. Might save more time, later.


I won't explain, but I'll do a few quick calculations for you. These will be added here so that you can check this out with a voltmeter before running your tests.

Assuming that the BJT isn't saturated, the base current should be:

$$I_\text{B}=\frac{V_\text{TH}-V_\text{BE}}{R_\text{TH}+\left(\beta+1\right)\:R_\text{E}}$$

Here, \$R_\text{E}=200\:\Omega+2.2\:\text{k}\Omega=2.4\:\text{k}\Omega\$, \$R_\text{TH}=\frac{R_1\:R_2}{R_1+R_2}\$ and \$V_\text{TH}=9\:\text{V}\frac{R_2}{R_1+R_2}\$. The value for \$V_\text{BE}\$ will vary a little, but I can already tell that it's closer to \$600\:\text{mV}\$ than to \$700\:\text{mV}\$. So I'll just pick something just under the middle of that. This gives me about \$800\:\text{nA}\$ for the base current (using a random guess for \$\beta\approx 260\$.) And therefore a collector current of only about \$210\:\mu\text{A}\$. About a \$2.1\:\text{V}\$ drop across the collector resistor.

So the next thing you do is do NOT put a signal up to your circuit. Just leave it open for now. Apply power and measure the output voltage. It should be in the area of about \$6.5\:\text{V}\$ to \$7.3\:\text{V}\$. Hopefully nearer the middle of that area. If the voltage is outside that range, stop everything and go measure the base voltage. This should be:

$$V_\text{B}=V_\text{TH} - I_\text{B}\cdot R_\text{TH}$$

In this case, that works out to about \$1.1\:\text{V}\$ to \$1.2\:\text{V}\$. If the base voltage also is too far outside that range (a few tenths of a volt above or below it), then just stop. Something is wired wrong. Get help and/or check things out.

These are basic checks.

If these things seem okay, then there is a good chance you have it wired correctly. Go back and verify the voltage source, just in case. Make sure it's still \$9\:\text{V}\$.

If all that is in line, then go ahead and add the capacitors and then apply your starting signal. You definitely should be getting more than a few millivolts (at the collector, relative to the [-] terminal of the power supply.)

Which reminds me -- polarity is important. Make sure that the emitter of the BJT is towards the [-] side of the power supply and that the collector of the BJT is towards the [+] side.


My guess is that you didn't hook up the power supply correctly, got a PNP instead of an NPN, or really flubbed up the wiring somehow. Check things. Then check them again.

Also, this particular circuit will have varying gain -- and some distortion as a result. (If you actually get it built up correctly.) But for these purposes, that should be fine enough.

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  • \$\begingroup\$ Thank you for your awesome post @jonk ! You make really good points and I will check my circuit once I get back home from work. I will let you guys know what I come across. \$\endgroup\$ – Carlos May 10 '18 at 14:07
  • \$\begingroup\$ @jonk quick question for you. I usually don't bother using the Rth/Vth model to calculate this specific transistor circuit. I usually simply go the much simpler Vb, Ve, Ie, Ic, Vc route (assuming Ve is greater than 1 and that Ib is much smaller than the Ir1/IR2). In what conditions should I take the Rth/Vth approach? I also know how to derive the constant beta exact solution, but with two pages worth of equations, it is almost never useful. \$\endgroup\$ – Simon Marcoux May 11 '18 at 0:35
  • \$\begingroup\$ @SimonMarcoux I'd never use \$R_\text{TH}\$ and \$V_\text{TH}\$ to design this kind of CE circuit (which I don't ever use, anyway -- at a minimum I'd add bootstrapping or I wouldn't even bother with it.) I use that to analyze the circuit later once parts are selected to figure out what to actually expect and measure, or when looking at someone else's circuit where they don't disclose their design process, or when wanting to know sensitivity to some variable of interest: $$\frac{\frac{\text{d}y}{y}}{\frac{\text{d}x}{x}}$$. \$\endgroup\$ – jonk May 11 '18 at 0:46
  • \$\begingroup\$ @Carlos if his answer was helpful, then upvote it! \$\endgroup\$ – Flávio Alegretti May 11 '18 at 0:47
  • \$\begingroup\$ @FlávioAlegretti I did, but it does not show because my reputation is less than 15 \$\endgroup\$ – Carlos May 11 '18 at 0:49
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The circuit looks OK theoretically. At least all component nominals provide reasonable working point of the amplifier.

enter image description here

The gain at 10 kHz should be about 56. If the circuit was theoretically calculated, then there must be some mistake in formulas/calculations. If it was really wired and measured with a 50 mV signal generator, it is likely that either the transistor pinout was wrongly determined, or the particular transistor sample is dead.

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