Search Results

What you have is a multiple output current mirror with the outputs connected in parallel. If you want a short name, you could call it a current doubler or a current multiplier.

The caption on the Wikipedia schematic says that it "[neglects] biasing details". This page has an example of a fully-biased common emitter amplifier, including a schematic: This schematic features …

I'm not a BJT master, so perhaps someone else can clarify this. 3. How can \$I_C\$ be greater than what's shown in the datasheet's Figure 1? …

It's a subscripted letter O. \$r_o\$ is the output resistance of the transistor. It represents the fact that the transistor is not an ideal voltage-controlled current source. The collector/drain volta …

There would be at least three loops since Vcc is really an ideal voltage source connected to ground. But mesh analysis is not a good way to solve this problem. You need to understand the relationship …

It's most likely a pull-down resistor to guarantee that the transistor remains off during the MCU's power-up. The value was chosen so as not to steal too much of the base current. I don't know that it …

Negative voltages are nothing to be afraid of. They work exactly like positive voltages, just with a minus sign. I'll walk you through the first part, which is to figure out \$R_1\$. You know the vol …

I'm seeing a gain of about 330 when I simulate that circuit with the default 2N3904 model at 1 kHz. Are you sure you're not dividing by two somewhere? This is a bad circuit design. Here are some probl …

The differential amplifier is what gives you the common-mode rejection. Differential gain implies common-mode rejection. The high input impedance also comes from the input stage. If FETs are used, th …

It looks like the problem is the values of R6 and R7. The voltage divider ratio is \$\frac{R7}{R6 + R7}\$, which works out to \$\frac{1.8k}{18k + 1.8k} = 0.091.\$ That gives you about 45.5mA, which is …

It's easier if you redraw the circuit to make the three-terminal BJT a pair of two-terminal components. You can do a large-signal analysis; it doesn't have to be small-signal. …

Your electronics class has probably taught you the hybrid-pi model and given you some complex (yet accurate) formulas for gain, input resistance, and output resistance of the various amplifier topolog …

If you draw the AC equivalents right and leave off most of the biasing, CC and CE amplifiers look almost identical. (Remember, ideally no current flows through the signal source!) simulate this …

I found a thread on dyiAudio that discusses this. User ubergeeknz asks: In most older (pre 1980) designs that I've seen, mostly British or British based designs, the output stage is either a totem po …