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I am trying to further my knowledge in discrete transistor design so I am designing a CE power amplifier to drive a speaker.

What I have trouble with is determining where to bias my BJT at.

I can't really find characteristic curves for transistors on datasheets which makes it difficult. I also don't know how to determine the amount of current that will go to my load.

Say I have biased such that VC = 5 V and IC = 10 mA. If I then connect my load in parallel from the VC node how can I determine the current?

I would think my equation would look something like:

$$I_{\text{total}} = 10\text{ mA} = (5/R_{\text{load}})-I_{\text{collector}}$$

How would I determine this \$I_{\text{collector}}\$ value?

I am using a P2N2222A transistor.

Want to couple the speaker with a capacitor to the vout node.

Here is a picture (source image) for reference to the type of circuit I am talking about:

schematic

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  • \$\begingroup\$ The starting point is always a DC analysis to determine the operating point (the Q point) on the load line. Recall that capacitors block DC current. Next, perform an AC analysis by (a) setting all independent DC sources to zero, (b) replacing all capacitors with short circuits (as a first approximation analysis), and (c) redrawing the circuit in its AC configuration and performing the AC solution. Recall that Kirchoff's Current Law (KCL) states the sum of currents entering node X equals the sum of currents exiting node X. (HINT: KCL analysis might be different for DC versus AC operation.) \$\endgroup\$ Commented Mar 24, 2020 at 23:32
  • \$\begingroup\$ @G36 Thank you I will check them out! \$\endgroup\$
    – alex
    Commented Mar 25, 2020 at 20:43

1 Answer 1

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There is already a similar question, however the design procedure is still misssing there. Now to your questions:

I can't really find characteristic curves for transistors on datasheets which makes it difficult.

Although there is no output characteristic curve in the datasheet, you can find specific biasing points which might already fulfill your needs. e.g: If you want to have a 10mA collector current, your gain is more-or-less 75A/A and from there you can work out the rest.

datasheet

Another alternative, is to run a Spice simulation with models made available by the manufacturer:

enter image description here

Say I have biased such that VC = 5 V and IC = 10 mA. If I then connect my load in parallel from the VC node how can I determine the current?

The idea here (not always the case) is that you want to have the collector voltage half the supply voltage. why? If the Vc = Vcc /2 it means that the collector voltage has more headroom to swing up and down without clamping the amplified signal. The drawback of it is that this class A amplifier burns current even if it not amplifying any signal. This collector current is going to be used to determine the bias point (Vc = Vcc/2), and theoretically you should driving your load directly with this current, otherwise you would move the transistor to another operating point (meaning, collector voltage would change) e.g.

  1. Connect a load to the collector output
  2. The load will draw some current
  3. The total current is upper limited by the collector resistor
  4. Current flowing into the collector reduces
  5. Operating point changes
  6. Amplified signal might be distorted

How to solve this probblem? What you usually do, is to connect a high impedance load which can deliver the current required by the load. So, at the same time you do not disturb the operating point of your amplifier, and drive a low impedance load (loudspeaker). This is accomplished by adding a buffer to the output of your amplifier as shown in the link above and here:

buffer

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  • \$\begingroup\$ Thank you for your response and this does help alot! I printed the curves from LTSPICE and was able to get the transistor biased and simulated in LTSPICE. I think a big problem I am running into is there are so many assumptions or approximations that different things make that aren't always made clear and I sometimes find explanations to be a bit hand wavy which just confuses me. I'm now having trouble on deciding the values to choose for all these resistors. Everywhere I look I see people quoting rules of thumb but no explanation where they come from or why. \$\endgroup\$
    – alex
    Commented Mar 25, 2020 at 20:43
  • \$\begingroup\$ I am glad I could help you. Indeed there is a lot of information out there, which sometimes can be a bit confusing. Generally it comes from the designer's experience. If want to have a good walkthrough, you can check out this website allaboutcircuits.com/textbook/semiconductors/chpt-4/… \$\endgroup\$
    – vtolentino
    Commented Mar 26, 2020 at 7:30

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