# current mirrior circuit

Discussion in 'Homework Help' started by anhnha, Aug 14, 2012.

Apr 19, 2012
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2. ### bretm Member

Feb 6, 2012
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It's just ohms law. If you reduce the resistance of Rbias it will increase the current, and that will increase the current through the diode since they're in series.

3. ### anhnha Thread Starter Well-Known Member

Apr 19, 2012
866
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Thank for help.
But can you analyse the circuit under diode equation, it mean that use the Id = Is * [e^ (qVd/NkT) -1].
"The voltage dropped across the diode probably won't be 0.7 volts exactly. The exact amount of forward voltage dropped across it depends on the current through the diode, and the diode's temperature, all in accordance with the diode equation. If diode current is increased (say, by reducing the resistance of Rbias), its voltage drop will increase slightly, increasing the voltage drop across the transistor's base-emitter junction, which will increase the emitter current by the same proportion, assuming the diode's PN junction and the transistor's base-emitter junction are well-matched to each other. In other words, transistor emitter current will closely equal diode current at any given time. If you change the diode current by changing the resistance value of Rbias, then the transistor's emitter current will follow suit, because the emitter current is described by the same equation as the diode's, and both PN junctions experience the same voltage drop."
In this the author don't consider the voltage drop in diode is constant. According to this, when reduce resistance of Rbias then the voltage drop in diode will increase slightly. I get stuck to explain it when use diode equation.

4. ### WBahn Moderator

Mar 31, 2012
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You have two components in series, a resistor and a diode that have a fixed voltage across the combination. Because of this, you know that whatever current flows in the resistor must be the same current that flows in the diode. You also know that the sum of the voltages across the two components must equal the total voltage applied to the overall pair.

Thus, you have two different ways to compute the relationship between the voltage at the junction of the two components and the current flowing in them. Let's pick some example values so that we can work with actual numbers. I've attached a circuit diagram with the specifics. I've also attached a chart with the load lines for three different resistor values.

The curved (blue) line is the relationship that has to exist between the voltage at Node A and the resistor/diode current in order for the characteristic equation for the diode to be satisfied. The three straight lines are, likewise, the relationships that would have to exist between that same voltage and current (the voltage at Node A and the resistor/diode current) in order to satisfy the characteristic equations for three different resistors {510Ω, 1kΩ, 5.1kΩ}.

If the bias resistor is 1kΩ, then the green curve is the one that matters. Since we can only have a single voltage at Node A and a single current flowing, we know that the operating point of the circuit must satisfy both equations simultaneously and, hence, is where the two equations intersect.

Unfortunately, you can't just solve for the voltage and/or current directly because you have a nonlinear equation and we aren't lucky enough for it to yield a nice, simple closed-form solution. So you have to solve it either graphically (which is what the load-line curves do) or iteratively. But without doing that, you can see that the voltage across the diode doesn't change very much and, hence, assuming that the voltage is about 0.7V will get you in the ballpark (and would serve as a good starting point for an iterative approach). But you can also see that, while the voltage doesn't change by much, it DOES change. As you would expect, both intuitively and from the diode equation, if the current goes down then the voltage goes down and, similarly, if the current goes up then the voltage goes up.

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5. ### anhnha Thread Starter Well-Known Member

Apr 19, 2012
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Thanks,
It solved my problem.In the picture if the resistor R decrease from R2 to R1 then the current through diode increase from I2 to I1.
But there is still a problem I feel confused.In the current mirror circuit (attached below) it also have involved of Ibase. If I ignore base current Ibase I am sure that if Rbias decrease then the current through diode Id will increase.But If we have to take Ibase into account, how can I get the conclude above that if Rbias decrease, the current through diode Id will increase.

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6. ### WBahn Moderator

Mar 31, 2012
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There are a number of ways to come at it, but the most direct and conclusive (IMNSHO) is the following:

If we reduce Rbias, the voltage across the diode will increase. Period.
If we increase the voltage across the diode, the current will increase. Period.

Now, if there is another device (namely the base-emitter diode junction of transistor), then it's current will increase to, by exactly the same reasoning.

Another way of looking at it is like this:

Imagine we have the resistor in the circuit with the diode and transistor and it is sitting at a particular base voltage and resistor current (some of which is going into the diode and the rest into the transistor base). No image turning down the resistance a little bit. If the same current were to continue to flow, then the base voltage has to rise and, if the base voltage rises then both the diode current and the base current must increase. But this means that we have to have more current through the bias resistor, and so the base voltage will drop back down (but not all the way to where it started). A new, higher equilibrium point will be achieved.

Another way to look at it. Imagine you applied a voltage source directly to the base (removed the bias resistor entirely and connected the voltage source to the base and the diode anode. As you raise the voltage, more current will flow in both the diode and the transistor base. At any give time, you can replace the supply with a bias resistor whose value is determined by the voltage difference between Vcc and the base voltage and dividing that difference by the current that is flowing. As you turn up the voltage supply, the voltage difference gets smaller and the current gets bigger, then the resistor value that you calculate has to get smaller.