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Home hobbyist here, just trying to learn. I'm trying to calculate voltage drops across the various components, and I'm stuck on R4. I calculated R1 and R2 no problem (I think), using:

R1 Voltage drop = 10 * (R1/(R1+R2)) = 7.85
R2 Voltage drop = 10-7.85 = 2.15

Please correct me if I'm wrong up to this point. I don't know enough as to whether R3 and R3 would affect the voltage divider here.

For R3, I figured there was ~0.6 V drop through the BE junction, so I have 1.55 V across R3.

For R4, however. I'm completely stuck. My measured voltage was 0.86 V, but I have no idea how or why that is the case. The voltage drop across CE was measured at ~8.46 V.

First, do I have the R1, R2, R3 calculation correct, or am I forgetting something, and second, how is R4 calculated?

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  • \$\begingroup\$ In this circuit, the drop across the collector resistor will be (12.1/22) of what you computed for the emitter resistor. Since you figured 1.55 V for the emitter resistor, then there will be 12.1/22*1.55 or approximately 0.85 V across the collector resistor. (This assumes the BJT is in active mode so don't apply this idea in every case you see.) \$\endgroup\$ Feb 20, 2023 at 11:33

3 Answers 3

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You're good up to the calculation of the voltage across R3, which you correctly found to be 1.55V. Later, I'll explain the justification for ignoring the transistor, R3 and R4, when calculating the potential at the junction of R1 and R2.

From there you can calculate the current \$I_E\$ through R3, which is emerging from the transistor's emitter, with Ohm's law:

$$ I_E = \frac{V_{R3}}{R_3} = \frac{1.55\text{ V}}{22\text{ kΩ}} = 70\text{ μA}$$

To understand the next step, first you need to understand the relationship between \$I_E\$ and the other two currents in the transistor's terminals, \$I_B\$ and \$I_C\$. Here's the circuit with those currents annotated:

schematic

simulate this circuit – Schematic created using CircuitLab

This relationship is embodied by Kirchhoff's Current Law (KCL), which essentially says that total current entering something must equal the total current leaving it. For this transistor, currents entering it are base current \$I_B\$, and collector current \$I_C\$. KCL tells us that their sum must equal the only current leaving, emitter current \$I_E\$:

$$ I_B + I_C = I_E = 70\text{ μA} $$

The transistor's role here is amplify the base current \$I_B\$, by multiplying by it some factor β, which will be the resulting collector current \$I_C = \beta \times I_B\$. If β is large, we can make the approximation that \$I_B\$ is very small compared to \$I_C\$, which simplifies the calculations somewhat.

For example, current gain β for a typical signal transistor like the 2N3904 is well over 100, meaning that base current will be less than 1% of collector current. By making the approximation \$I_B \approx 0\$, we can rewrite the above KCL equation:

$$ I_C \approx I_E \approx 70\text{ μA} $$

Now you have a value for current \$I_C\$ through R4, and you can use Ohm's law and Kirchhoff's Voltage Law (KVL) to find the remaining voltages.

You could consider the transistor's base as a path via which current is "leeched" away from the potential divider formed by R1 and R2. If that current \$I_B\$ were significant, compared to current in R1 and R2, then you may not ignore \$I_B\$ in your calculations for base and emitter potentials. However, as long as \$I_B\$ is small compared to currents in R1 and R2, we can again use the assumption \$I_B \approx 0\$, and disregard it here too.

As we found above, base current will be less than 1% of collector current \$ 70\text{ μA}\$; that's well under \$ 1\text{ μA}\$. Current in R1 + R2 across 10V potential difference will be:

$$ I_{R1} = \frac{10}{12k + 3k} = 670\text{ μA} $$

Current drawn away by T1's base is very small compared to that, which is why in this case it's safe to ignore T1 (and everything else) in your calculation of base potential.

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  • \$\begingroup\$ Very nice. Looking at the datasheet for the gain, am I to be looking at hfe, which shows "small signal current gain" between 100-400? \$\endgroup\$
    – LarryBud
    Feb 20, 2023 at 12:56
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    \$\begingroup\$ Yes, that's the parameter. \$\endgroup\$ Feb 20, 2023 at 17:54
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You are most of the way there. You know the voltage across R3, so Ohm's Law gives you its current. That current can come from only 2 places, the base and the collector.

If you assume the transistor has infinite gain, then all of the R3 current comes from the collector through R4. You know the R4 value (ohms) and current, so ...

Don't worry about that infinite thing. Once you get the basic answers, you can use the transistor's gain (from the datasheet) to determine how much of the R3 current is coming from the collector, and how much from the base.

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  • \$\begingroup\$ Walk me through this: Calculated current at R3 would be 1.54 / 22000 = 0.07 ma. Do I have step 1 right? \$\endgroup\$
    – LarryBud
    Feb 20, 2023 at 0:34
  • \$\begingroup\$ And then 0.07 * 12100 = 847 mv, or 0.847 volts across R4, would that be right? \$\endgroup\$
    – LarryBud
    Feb 20, 2023 at 0:36
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    \$\begingroup\$ Your measured R4 voltage differs from calculated by less than 2%. Smells good to me. \$\endgroup\$
    – AnalogKid
    Feb 20, 2023 at 5:03
  • \$\begingroup\$ Just wanted to make sure I was doing the correct logic. Thanks! \$\endgroup\$
    – LarryBud
    Feb 20, 2023 at 12:11
  • \$\begingroup\$ With such a low voltage across R4 then the transistor is almost cutoff and will clip positive output peaks. The rough voltage gain of output/input is 12.1k/22k= 0.55 times so your transistor attenuates instead of amplifying. \$\endgroup\$
    – Audioguru
    Feb 22, 2023 at 1:13
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Home hobbyist here, just trying to learn.

If so, you need a lot more than a formal calculation. You also need to understand what these 4 resistors and a transistor mean, what they do and why they do it. Understanding means seeing old familiar things (ideas, functional blocks, sub-circuits) in the new unfamiliar circuit. Understanding is generalization - seeing forest behind the trees, cities behind houses, circuits behind devices.
I will use this specific circuit solution to show how we can make unfamiliar circuits familiar in three possible ways - by understanding, building and inventing them.

Understanding the circuit

Preparing the schematic. The first thing we should do is to redraw the circuit in CircuitLab (if it has not already been done). This will help us check our assumptions by changing the parameters and observing the voltages and currents. Then we start looking for familiar things by examining the schematic in the usual direction in which signals flow - from left to right.

STEP 1: Voltage divider seen. First we see that resistors R1 and R2 form a voltage divider; let's circle it with a dashed line and label it VD. This device reduces the supply voltage Vcc; thus another "voltage source" is made. Its voltage is lower than the supply voltage because the voltages inside a circuit with active elements are within the limits of the supply voltage. It is imperfect because, as they say, it has a significant output (Thevenin) resistance and when measuring its output voltage, we see that it does not exactly correspond to the calculated... Aha... the base current of the transistor is the reason...

schematic

simulate this circuit – Schematic created using CircuitLab

STEP 2: Emitter follower seen. Next we see that the transistor Q is connected as an emitter follower circuit; so its emitter voltage should be equal to the base voltage. Yes, but not quite... The transistor input is not "ideal" like an op-amp input (0 V) but has a forward voltage of several hundred millivolts. However, this buffered voltage divider is a much better constant voltage source than the humble voltage divider.

schematic

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STEP 3: Negative feedback principle seen. The transistor creates the emitter voltage copy by the help of the ubiquitous negative feedback principle: making Y = X by changing Y and keeping X - Y -> 0.

schematic

simulate this circuit

By passing its collector current through R3, the transistor adjusts its emitter voltage (Y) so that to make it (almost) equal to the input voltage (X). For this purpose, the transistor changes its collector-emitter "resistance" until (almost) zero its base-emitter voltage (the voltage difference).

STEP 4: Voltage-to-current converter (current source) seen. Since a constant voltage is applied across the constant resistor R3, a constant current I = V/R (Ohm's law) flows through it. As though, the constant input voltage is converted to a constant output current. So the combination of the power supply Vcc, voltage divider VD, emitter follower EF and emitter resistor R3 acts as a constant current source (V-I converter).

schematic

simulate this circuit

STEP 5: Current-to-voltage converter seen. Next we see that the (collector) current flows through the (collector) resistor R4. Aha...Ohm's law again but written in its other form V = I.R. This was the role of the collector resistor - to convert the current into voltage.
So, in this arrangement, R4 is supplied by a current source and the voltage across R4 can be used as an output voltage. The problem, however, is that it is not referenced to ground...

schematic

simulate this circuit

Building the circuit

Step-by-step exposition. Once we have understood the circuit we feel a desire to explain it to others; by explaining it, we understand it even more. The best way to explain the circuit is by building it step by step. We do not have to do it in the same sequence as when understanding but it usually happens like this.

Preparing the schematic. It is convenient to draw the individual schematic stages (steps) from back to front - first the last stage is drawn and then, by gradually erasing parts of the schematic, the intermediate stages are obtained.

STEP 1: Assemble the voltage divider by connecting the two resistors R1 and R2 in series. Then we supply it through the Vcc power source and measure its output voltage by an "ideal" voltmeter. We see that it matches the calculated voltage.

schematic

simulate this circuit

STEP 2: Buffer the voltage divider by an emitter follower EF to create a copy of the input voltage. We initially connect the collector to Vcc and some load (e.g., a real voltmeter) in the emitter. Interesting... Why did the voltage decrease?

schematic

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STEP 3: Apply the copy voltage across a constant resistor R3 to obtain a proportional current I = V/R (Ohm's law). Thus a voltage-to-current converter V-I is obtained.

schematic

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STEP 4: Insert another constant resistor R4 to obtain a proportional voltage V = I.R (Ohm's law in its other form). Thus a current-to-voltage converter I-V is obtained.

schematic

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STEP 5: Use the voltage drop across the resistor R4 as an output voltage. But a problem appears - the load is ungrounded. If it is a "floating" voltmeter (VR4), no problem; but if it is a next transistor stage?

schematic

simulate this circuit

STEP 6: Use the complimentary (to Vcc) voltage drop Vout = Vcc - VR4 as a grounded output. An additional property of this configuration (aka "common-emitter amplifier stage") is that it is inverting.

schematic

simulate this circuit

"Inventing" the circuit

Of course, it is most interesting and exciting... and extremely useful for understanding to reveal how the circuit was invented. Let's do it then!

STEP 1: The problem. The idea is neither op-amp nor transistor nor tube. It is simply electric and could have been implemented as early as the 19th century. We can formulate it like this: Make a voltage divider with transfer ratio Vout/Vin = R2/R1. In comparison, the transfer ratio of the ordinary voltage divider is Vout/Vin = R2/(R1 + R2). If it does, it would be a "true voltage divider" that can work as an attenuator (R2/R1 < 1), follower (R2/R1 = 1) or amplifier (R2/R1 > 1). In comparison, the ordinary voltage divider can only work as an attenuator since R2/(R1 + R2) < 1.

STEP 2: Resistors in series. We know from electrical circuits that if we connect two resistors (R3 and R4) in series, the same current flows through them and the voltage drops across them (VR3 and VR4) are in the same ratio as the resistances (VR4/VR3 = R4/R3). We can change the supply voltage (Vcc) or the middle resistor R as an input quantity and take the two voltages as output quantities. The problem is that we want something different - one voltage (VR3) to be input and the other (VR4) to be output. But how do we do it? We cannot connect the input voltage source directly to R3 because the current will only flow through it but not through R4.

schematic

simulate this circuit

STEP 3: Applying a negative feedback. Here the ubiquitous principle comes to our aid. Using the power source, we make a copy of the input voltage Vin across the resistor R3 as follows: we compare the two voltages (Vin and VR3) by connecting them contrary in series and observe the difference using a zero indicator ZI (a sensitive voltmeter).

schematic

simulate this circuit

As a result, the output voltage VR4 is proportional to the input voltage Vin (VR4 = VR3.R4/R3 = Vin.R4/R3).

schematic

simulate this circuit

To simulate this conceptual experiment, double click on Rvar to open the attributes and begin changing its resistance observing at the same time the zero idicator. Stop when it shows zero difference voltage.

STEP 4: Transistor implementation. Now it is only necessary to replace Rvar with a transistor (Q) and the circuit is "invented". Now the transistor will change its collector-emitter "resistance" until it (almost) zero its base-emitter voltage (the voltage difference). As a result, the transistor "resistance" will be 68.57 k as in the conceptual electric circuit above (i.e., it will cause the same voltage drop across itself).

schematic

simulate this circuit

STEP 5: Op-amp implementations. I added this step to show the power of basic ideas. An op-amp inverting amplifier is also such a true voltage divider with a transfer ratio equal to -R2/R1. This is because it uses a variation of the same idea. We can see it even in the op-amp non-inverting amplifier.

Applications

If we add a few capacitors and an AC input voltage source, we will get the classic circuit of an AC transistor amplifier.

schematic

simulate this circuit

AC amplifier

It would be interesting to connect another capacitor in parallel with R3 and see how the signals change dramatically... but that is another story...

Real experiments

Of course, simulations cannot replace real experiments. I have developed a set of such experiments about this circuit for "home hobbyists" and posted a story on my blog. I have used potentiometers to set the input voltages, an LED to visualize the collector current and multimeters to measure the voltages.

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    \$\begingroup\$ Thank you. I'm far from going through a formal education. Software dev by trade, have built a bunch of electronics with microprocessors, assembly, etc, just trying to go "back to the basics" and learn the analog stuff! \$\endgroup\$
    – LarryBud
    Feb 23, 2023 at 2:25

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