hey guys, in this video, we will look at a very important aspect of bandicam design, which is startup. In the last video, we discussed this circuit. We said that we want to make voltages at V1 and V3 and current in these two branches equal, and for that, we put a feedback around this circuit. Now, they have seen that there will be a point where these two conditions will be met, and that will give us our desired treated current into these branches. But there exists another point where voltages V1 and V3, as well as current I1 and I2, will be equal, and at that point, all these four values will be 0. Think about it. If there is no current in these two branches, that means currents are 0. These two currents are equal, which is a perfectly valid solution for this circuit. Now, if currents are 0, voltages at these nodes will also be 0, and since voltages at both nodes are 0 and hence equal, the amplifier will be perfectly happy and

will be sitting there doing the thing, meaning to say it is a stable operating point. And this is precisely the existence of this second stable operating point that we need to think about startup. Now let's look at these two operating points in some more detail and see what lies in between. We know that because of the presence of the diode, the IV relation in these two branches is nonlinear. One simple way to solve these kinds of nonlinear relationships is to do it graphically. For that, we plot the IV curves of these points on the same graph, and wherever those curves meet is the solution. So let's start with the diode d1. So this plot is a typical diode IV plot, except in the standard form. V is on the x-axis, and I is on the y-axis. Now let's plot the same curve for diode d2. This v2 is the same type of curve as v1, except the two is always somewhat less than v1. And this is because d2 is n times bigger than d1. So if you just look at these two curves

they meet only at one point, that is at the origin. In other words, the only way where voltage across and current in 2D two diodes are equal is to have no current at all. Now let's plot the IV curve of R1, which we already know will be simply a straight line passing through the origin. To get the IV curve at V3, we simply add the IV curves of D2 and R1. So V3 also begins at the origin and initially remains close to V2 but diverges more and more as current increases. So we see that V3 meets or even at two points, one at the origin, which is the undesired solution, and again at some higher value of current, which is our desired solution. Any other operating condition except these two is not a stable circuit. If the circuit finds itself in one of those unstable operating points, then the loop will move the circuit in one of the stable points, but which of the two stable points now? The fact is if the circuit is anywhere in this region, it

will always move to the desired operating point. Now, to understand how this happens, we need to look into the loop dynamics in some more detail. Now we know that this is a feedback loop, but which kind is it? A negative feedback or a positive? So, if we trace the loop starting from v3, the amplifier acts as a non-inverting part because v3 goes to the positive terminal of this amplifier. But there is an inverting stage in the form of p2, which is a common source amplifier. So, in all, this looks like a negative feedback loop. But if we start with v1, then there are two inversions in the loop. One is caused by the amplifier because now v1 goes to the negative terminal of the amplifier, and the other is a common source p1. So, this in fact makes it a positive feedback. As a matter of fact, there are two loops in this circuit: one is negative and the other is positive feedback. But overall, this feedback loop can only be either positive or negative feedback; it can't be both. And it is the gain of

these two paths, which will ultimately decide which one it will be. So, if the gain of the positive feedback path is higher than the negative feedback path, then it will be a positive feedback system. Otherwise, it will be a negative feedback system. Now, in each of these two feedback paths, there are two stages. One stage is this amplifier, which is common to both the paths. The other gain stage is found by these common source prima stages and these diode and diode resistor elements. Now, again, this prima stage is the same for both the paths, so the difference in the gain eventually boils down to the difference in impedance of this diode and this diode resistor combination. The impedance of the asari is simply its value r1. In order to find the small signal resistance of these diodes, let's write their VI equations again. Now, to find a small signal impedance, we simply differentiate the voltages with their respective currents. So, we find that the impedance of these diodes is simply VT over current into it. But since currents are equal because of the presence of these current mirrors.

that means these impedances are also equal, and from this argument, it looks like that the impedance of the negative path will always dominate because there is an additional r1 in this path. Now, these impedances may be equal at the higher value of current, but this cannot be true for the whole range, and we can appreciate this fact by just looking at these two plots of the diode voltages. If the impedance of these two paths is exactly equal and they start with the origin, these two plots cannot be different. As a matter of fact, this VI relation is only an approximate equation; a more precise equation includes a minus 1 factor in it. So, let's bring that in. So, after using a more accurate equation, we see that the impedance equation of two diodes is indeed different, and the reason for this difference is that d2 is and tan bigger. Now, since I1 and I2 are equal currents, the impedance of d1 is always somewhat bigger at a higher value of current. We can practically ignore the contribution of

is and is term and hence the 2 impedance appear to be equal at extremely low value of currents. These impedances would be very high in mega ohms, already in Giga ohms. So, as a result, at such very low value of Carine, even after the presence of arwen in the negative feedback path, it is the gain of positive feedback which dominates. To get a better field, let's assign some numerical values to these components. So, here we have assumed I used to be one picoampere and to be eight and temperature to be 25 degrees Celsius. Clearly, Rd one is much bigger than Rd 2. So, even after the presence of r1, which is of the order of few tens of kilo ohm, it is clearly the positive feedback gain which is higher. Now, this is an extreme example of the current one Pico ampere. As we increase the current, this difference in impedance reduces drastically. For example, at 1 micro ampere current, both diode impedances are around 25 kilo ohm. Now, this Ireland

clearly makes a difference, and at this value of current, it is, in fact, negative feedback gain which dominates the loop. So, we see that at an extremely low value of current, the loop has positive feedback, and for the higher value of current, it has negative feedback. Now, positive feedback is a regenerative feedback. This kind of feedback is often used in digital latches or the comparators. The effect of regenerative feedback is to cause a runaway action, so any difference in the input will be continuously amplified until the circuit saturates. In relation to this architecture, what it means is that any tiny difference in the current at the origin will be amplified. So, we see that both the current started at zero, but then they start to diverge, that means become different. On the other hand, negative feedback is known as degenerative feedback. Its action is to attenuate any disturbances around the desired operating point. So, for example, if any disturbance causes the desired operating point to move right, the negative feedback will bring it back to the original point. So, we see that in that sense, these two solutions to this

circuit are not symmetrical. The solution at the origin has positive feedback in its vicinity, while the other solution has negative feedback. So, in that sense, the other solution is more stable. To get a better intuitive feel, we can compare this situation to the ball in the hill and the valley. Imagine an uneven surface containing one hill and one valley, similar to a bandgap circuit. There are two stable points in this surface. A stable point in the surface means points where the ball can stay still. The first stable point is at the top of the hill, and the other stable point is at the bottom of the valley. Now, here also, these two operating points are not symmetrical. If we disturb the ball at the bottom of the valley, it will come back to its stable point. In that sense, it is a degenerative stable point. On the other hand, if we give a slight push to the ball at the top of the hill, it cannot come back. So, the top of the hill is akin to the stable point at the origin, and the bottom of the valley is equivalent to the desired operating

point now. If you give a little thought, you may say that in that case, we don't need the setup circuit at all because there will always be some tiny difference in these two currents because of noise or because of mismatch, and that would cause the degenerative action at the undesired operating point and move it to a desired operating point. Now, this is true in some sense, but there are two issues with this approach. The first problem is, although the regenerative feedback will move the operating point from an undesired location to a desired location, the time taken in this process is not well defined. So, left on its own, it may come up very quickly, or it may take a very long time to start. In practical circuit designs, you would like to have a better handle on this startup time. The second problem is a more serious problem, and it is caused by the offset of this amplifier. To understand that, let's consider again how this startup process works. In this curve, there will be a suitable point where positive feedback turns into negative feedback. So, let's assume that that

switchover occurs at this line, so loop feedback is positive on the left and negative on the right. Now, job a positive feedback is to move the operating point from a point where the difference between v1 and v3 is zero to a point where there is a finite difference between v1 and v3. And this difference must have a particular polarity that is v1 must be higher than v3. And after that, negative feedback takes over and moves the point to the desired operating point. Now, any amplifier has a finite offset, and that offset being random can have any polarity. Now, if the amplifier happened to have a large offset in such a way that v3 is higher than leaver near the origin, then this loop will tend to sustain this operating point. Since the amplifier has high gain, a high enough value of v3 will make the output of the amplifier so high that p1 and p2 will be effectively off. In that sense, it is similar to the situation where there is a small local value on the top itself, and it will take

big enough push that is big enough startup current to get things started in real circuit is a third reason which makes startup circuit a necessity in many chips. Then gap is the first circuit to start in the chip, so this whole circuit needs to be self-bias. That means amplifier bias comes from band gap circuit itself. In absence of any current, this circuit is likely to stay at the origin, that is, zero current and zero voltages. So now that we have made a good case for the sort of circuit, let's see some of the ways a startup can be implemented in the band gap circuits. The sole purpose of startup circuit is to inject high enough current into these two branches. We are particularly interested in injecting current into this branch. Now this injection can be done either in one-shot way or in a continuous manner. In one-shot sort of V, inject the current only once when V is done on the circuit. For example, in a very crude one-shot startup circuit, we can have capacitive coupling between enable signal and v1 node. Although one-shot startup circuits are

simple, and they may not require a static current consumption. In my view, they are not very reliable. This is because this hyper startup doesn't continuously monitor the bandgap. So, if Bank A fails to start, or worse, if it comes back to the undesired point, we are in trouble. And for that reason, always-on startup circuits are preferred schemes for most of the bandgap circuits. Always-on startup schemes also inject current at appropriate points in the bandgap circuit. Let's briefly consider two such points. One obvious point is to inject current into V1. Now, this current itself can be generated by a very simple diode-resistor circuit, which is guaranteed to start. A second point can be the output of the amplifier. We want to pull the VG down so that P1 and P2 turn on and inject current in these two branches. Now, these startup currents need not be very precise, but there are some important design considerations. If startup current is too low, then the amplifier may not start at all. On the other hand, a very high startup current

causes an overshoot and ringing in the output. So, in this video, we have seen the necessity and mechanism of startup in bandgap circuits. In the next video, we will expand this amplifier into the real circuit, and we'll consider the implementation of the startup circuit in some more details. I hope you found this discussion useful, so post your comment below, and thanks for watching.

#25 Band-gap Reference Circuits - Start up

#25 Band-gap Reference Circuits - Start up