How does the LT1055-based 125V output circuit work? - Page 1
Forum section: electronics / projects

author: tree
author_rank: Regular Contributor
timestamp: September 13, 2013, 02:52:42 pm
content: I've been trying to figure out how this circuit works but have trouble understanding what a few of the components do.Here is as far as I've gotten with the help of others:The 1N965 Zener diodes provide a +/- 15V reference to the op amp.The op amp is configured for a gain of 10 (inverting).I believe the 2N2222 and the 2N2907 are there for current limiting.I'm not quite sure what the 1N4148 diodes are doing (maybe compensating for the Vbe drop of the output transistors, or also used as a type of current limit for the output since the Vbe drop + the voltage drop across the 27ohm resistor cannot exceed the voltage dropped across the diode).The 510ohm resistor provides biasing to the first stage 2N5415 and 2N3440 such that the voltage drop across it must be equal to the voltage drop across the 330ohm + Veb of the 2N5415. As a result the current available to drive the base of the output transistor is chosen via the 330ohm resistor in this case approximately 2 to 3 mA (depending on what you use for Veb of the 2N5415).I have no idea what the true purpose of the 1N4148 diodes, and the 2N2222 and 2N2907 transistors is.Furthermore, why is the base of the 2N2222 2N2907 driven with a 1kOhm resistor?What do the 1MegOhm resistors do?Why is the output tied to all those points? (possibly for symmetry)

author: Bertho
author_rank: Regular Contributor
timestamp: September 13, 2013, 04:15:36 pm
content: Quote from: tree on September 13, 2013, 02:52:42 pmThe 2N2222/2N2907 are current limiters to the output. The max. output current is set at 25mA (as indicated on the image). When the current reaches 25mA, then the 27Ohm resistor drops ~0.7V and the transistor starts to conduct, effectively quenching the base-current of the output transistor(s). The 1k series resistor is just a limiter to assure that the transistors will not be killed by output transients.The 1N4148 diodes are to compensate for the Vbe drop to linearize the output.The 1M resistor bias the second stage setting up ~125uA base-current. The input ladder combined with the 1M resistors set the bias current and amplification of that stage. The points are connected because the output is symmetric.

author: Marco
author_rank: Super Contributor
timestamp: September 13, 2013, 04:22:28 pm
content: I knew I saw that before.http://cds.linear.com/docs/en/application-note/an18f.pdfThe diodes are simply there to bias the output transistors base emitter voltages to reduce crossover distortion and Q5/Q6 are indeed for current limiting.PS. there isanother wayto create a composite high voltage opamp.

author: tree
author_rank: Regular Contributor
timestamp: September 13, 2013, 04:27:09 pm
content: Quote from: Bertho on September 13, 2013, 04:15:36 pmThat makes sense.Quote from: Bertho on September 13, 2013, 04:15:36 pmJust as I suspected.Quote from: Bertho on September 13, 2013, 04:15:36 pmWhat exactly is the second stage? 125uA base current for which transistor? Amplification of what stage?Quote from: Bertho on September 13, 2013, 04:15:36 pmI tried disconnecting them in a simulation and the circuit still worked so I'm not exactly sure why they have to be connected.

author: dannyf
author_rank: Super Contributor
timestamp: September 13, 2013, 04:30:33 pm
content: QuoteThe discrete output stage has voltage gain.

author: tree
author_rank: Regular Contributor
timestamp: September 13, 2013, 04:34:13 pm
content: Quote from: dannyf on September 13, 2013, 04:30:33 pmI should have been a bit more clear in my original post. Sorry.I not only meant what they do, but how.

author: dannyf
author_rank: Super Contributor
timestamp: September 13, 2013, 04:45:46 pm
content: Think of it as a V-I converter: the voltage output from the opamp causes the voltage over the 510ohm resistor to change, which in tern will cause the current going through the 330ohm Re resistor to change. That changes Ic of the transistor, which changes the Ib of the output follower -> voltage output on the load changes.

author: dannyf
author_rank: Super Contributor
timestamp: September 13, 2013, 04:48:01 pm
content: This approach differs from the traditional bootstrapping: the bootstrapping is essentially a variable power supply whose output floats with the input signal.Here, the output swing of the opamp is amplified by the OPS.

author: Bertho
author_rank: Regular Contributor
timestamp: September 13, 2013, 04:48:56 pm
content: Quote from: tree on September 13, 2013, 04:27:09 pmThe first stage is the opamp, the second stage is Q1/Q2 and the third (output) stage is Q3/Q4.The 1M ladder together with the 510+50k+50k+510 ladder set the gain at the second stage. According to the linked pdf above: "The local 1M-50k feedback pairs set stage gain at 20,...".Disconnecting the 1M resistors will surely work, but the gain will be different. Probably also has an influence on linearity and response-timing too.

author: tree
author_rank: Regular Contributor
timestamp: September 13, 2013, 04:59:26 pm
content: Quote from: Bertho on September 13, 2013, 04:48:56 pmI am assuming Q1/Q2 are the transistors whose bases are tied to the 1M resistors and the output transistors are Q3/Q4. I can do the math 1M/50k = 20. The question is why do you need that gain there, in what way is that gain and where do you see that gain? I thought the gain is set by virtue of feedback (100k and 10k resistors).

author: tree
author_rank: Regular Contributor
timestamp: September 13, 2013, 05:01:50 pm
content: Quote from: dannyf on September 13, 2013, 04:45:46 pmI understand that. But I don't understand the reason for other components like the 1M resistors and I still have no idea why the output has to be connected to them.

author: tree
author_rank: Regular Contributor
timestamp: September 13, 2013, 05:13:20 pm
content: In that PDF the output is not connected to the diodes whereas in the picture they are.

author: Bertho
author_rank: Regular Contributor
timestamp: September 13, 2013, 05:15:51 pm
content: Quote from: tree on September 13, 2013, 04:59:26 pmYou have an outer-loop-gain, which is set with the 100k/10k. Then you have an inner-loop-gain set by 1M/50k.The inner-loop-gain is there to catapult the +/-15V first stage to allow the output to drive +/-120V; that needs amplification (it is a voltage amplifier). The outer-loop-gain sets the relation between ultimate input and output, but it does influence the inner loop as such.

author: tree
author_rank: Regular Contributor
timestamp: September 13, 2013, 05:19:07 pm
content: Quote from: Bertho on September 13, 2013, 05:15:51 pmWhere is that inner-loop? What nodes in the schematic should I be looking at?

author: Bertho
author_rank: Regular Contributor
timestamp: September 13, 2013, 05:49:05 pm
content: Quote from: tree on September 13, 2013, 05:19:07 pmThe inner-loop is stage 2 and 3 feeding back on itself through the 1M resistors.--> opamp --> Q1/Q2 --> Q3/Q4 -+-->| | inner || +-------------------+| outer |+-----------------------------+

author: tree
author_rank: Regular Contributor
timestamp: September 13, 2013, 05:52:31 pm
content: Quote from: Bertho on September 13, 2013, 05:49:05 pmOk. So how does that exactly work?

author: Bertho
author_rank: Regular Contributor
timestamp: September 13, 2013, 06:35:04 pm
content: Quote from: tree on September 13, 2013, 05:52:31 pmThe sum of the currents at the intersection of base of Q1/Q2, the 50k and 1M resistors (and also the 510Ohm resistor) should be zero. That is the loop control. For example, when the output voltage is higher than the setpoint (as controlled at the opamp output), then the current through the Q1/Q2 base will change (lowered) through the current flowing through the 1M resistor and correct the output. If the output is too low, then the 1M resistor does not divert enough current and that will flow through the base of Q1/Q2, in effect boosting the output to higher level.

author: tree
author_rank: Regular Contributor
timestamp: September 13, 2013, 06:58:53 pm
content: Quote from: Bertho on September 13, 2013, 06:35:04 pmSorry. Can't see it. I have been trying to work an example but it's failing me. If I assume that I want my output voltage to be 110V and the current output is 109V, what will happen?I guess I understand how the negative feedback works. Let's work an example with a positive output voltage. If the output voltage decreases, then more current is drawn from the base of Q1 through the 1M resistor, thus increasing base drive to the output transistor, which increases the drop across the load (the voltage at the output rises). As a result the voltage drop falls across the 1M resistor, which decreases base current through Q1, which decreases the base drive to the output transistor, which then decreases the current through the load at which point the voltage drop across the load decreases.I just don't understand how the value of the 1M resistor matters.

author: Bertho
author_rank: Regular Contributor
timestamp: September 13, 2013, 07:26:07 pm
content: Quote from: tree on September 13, 2013, 06:58:53 pmIf the output is too low, then the current through the 1M resistor will increase (the current through the 1M resistor is, for all practical intents, proportional to the positive supply minus output voltage). More current through the 1M resistor will draw more current through the base of Q1, which therefore sends more current through the collector of Q1, directly into the base of Q3 and opening Q3 further, increasing the output voltage; loop complete.

author: tree
author_rank: Regular Contributor
timestamp: September 13, 2013, 07:31:51 pm
content: Quote from: Bertho on September 13, 2013, 07:26:07 pmI guess you wrote a response just as soon as I figured out how the negative feedback works. But like I stated in my previous post, why does the resistance (the 1M) matter?

author: Bertho
author_rank: Regular Contributor
timestamp: September 13, 2013, 07:35:03 pm
content: Quote from: tree on September 13, 2013, 06:58:53 pmThat is the proportion of the input amplifier vs. the output amplifier. The 50k is the input and the 1M at the output. The relation of the two make a 1:20 voltage gain amplifier. The LT1055 has an output-swing which is much lower than the output swing of the output stage. All these amplifiers work on the simple principle that the sum of all currents at the control-node (where 50k and 1M come together) is zero. If the output is not at the right position, then an error-current will flow to correct for that imbalance. The error-current, in this case, is a modulation of the Q1 base-current. For the currents to be equal with such different voltages you need different resistors because I = U/R will hold at any time.

author: tree
author_rank: Regular Contributor
timestamp: September 13, 2013, 07:46:09 pm
content: Quote from: Bertho on September 13, 2013, 07:35:03 pmCan you give me an example again, similar in nature to the one you gave earlier regarding the negative feedback mechanism?

author: Bertho
author_rank: Regular Contributor
timestamp: September 13, 2013, 07:54:34 pm
content: Quote from: tree on September 13, 2013, 07:46:09 pmThat would not give you any additional information because it would be the same again.You have the circuit in simulation, so I suggest that you measure the currents through all components that sum into the control-node (base of Q1/Q2). Graph the currents and see how they change with varying input and output/load. You can put a pulse on the input to see the step-response behaviour (including possible over/under-shoot). That should give you a feel for how the currents behave at various conditions.

author: tree
author_rank: Regular Contributor
timestamp: September 13, 2013, 08:04:28 pm
content: Quote from: Bertho on September 13, 2013, 07:54:34 pmI did that, more so in a manual way. I changed input voltages and the value of the resistor (1M) to various other values. All the numbers show is how they change and that KCL works. I still have no intuitive understanding of what's happening.

author: Bertho
author_rank: Regular Contributor
timestamp: September 13, 2013, 08:33:02 pm
content: Quote from: tree on September 13, 2013, 08:04:28 pmWell, the setup is, seen at a high level, an example of a closed loop control system (https://en.wikipedia.org/wiki/Control_theory) where the circuit's inner loop circuit is a simple P-type controller where the control signals are currents). The fact that it is not intuitive may be because you have not worked with these systems that much. Analog circuits can be a bit daunting.My best advice is to follow the currents and apply Kirchhoff's law (https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws). It is not "easy", but it does tell you what happens. It took me many years to get an intuitive feel for analog electronics...I know that it does not help you as such, but I have no cluewhereyour line-of-thought fails to come to the right understanding of the circuit. Without me understanding your problem, I cannot give you the definite pointer.

author: tree
author_rank: Regular Contributor
timestamp: September 13, 2013, 08:37:07 pm
content: Quote from: Bertho on September 13, 2013, 08:33:02 pmI understand everything up until how the ratio of the two resistors changes the output voltage swing. It's actually counterintuitive for me. I would have though that raising the resistor value (1M) would have lowered the gain. But it seems like raising it actually increases the gain.I'm trying to draw an analogy.If we consider an inverting op amp configuration with resistors Rf and Rg and the ratio, Rf/Rg, setting the gain, does the same principle apply in this case? In other words, Rf is the 1M resistor and the 50k and the 510 resistor represent Rg? Considering the node that connects all three of those to the base of Q1 has roughly the same voltage at all times and the base current not change by much can we apply the inverting op amp analogy to this?

author: Bertho
author_rank: Regular Contributor
timestamp: September 13, 2013, 08:45:38 pm
content: Quote from: tree on September 13, 2013, 08:37:07 pmBut it worksexactlylike an opamp. The 1M is the feedback resistor (Rf) and the 50k the input resistor (Rg). The gain is 1M/50k. Increasing the 1M resistor increases the gain. The control-node is the virtual zero-point and the voltages over Rf and Rg must be proportional.

author: tree
author_rank: Regular Contributor
timestamp: September 13, 2013, 09:21:29 pm
content: Hmm. Still doesn't make any sense.I simulated the circuit (although I changed some of the transistors) with a 1M and a 100k resistor and got the results you see attached. The input is set to -4V and as can be seen the base currents differ by 1uA and the current through the 50k change by 10uA and the largest current change occurs in the 510 and 1M resistor. I lowered the 1M by one order of magnitude while the current increased by two orders of magnitude and the voltage decreased by an order of magnitude. Mathematically that makes sense. Still no intuition.

author: Bertho
author_rank: Regular Contributor
timestamp: September 13, 2013, 11:13:48 pm
content: Quote from: tree on September 13, 2013, 09:21:29 pmYou did a DC operating point analysis. However, you didnotsimulate the entire circuit. You took the opamp out of the equation and that does not work as you expect. The output voltage is controlled by the outer loop and not by the inner loop. The inner loop is there as a voltage trampoline. The actual control loop for the output is the outer loop.When you put DC -4V on the opamp, then you will see that the system correctly adjusts the output. The opamp's output will regulate its output to the point where the second-stage input (at the 50k) is at the optimal point. The inner loop is by no means as linear as you would hope. When you do change the -4V DC to a sine and simulate AC, then you'll see that the top and bottom values are quite a bit different. The transistors are not a complementary pair I guess.

author: tree
author_rank: Regular Contributor
timestamp: September 13, 2013, 11:27:06 pm
content: Quote from: Bertho on September 13, 2013, 11:13:48 pmI know that they are not linear when I do that. I just wanted to examine how the gain changes if the input remains the same, while the resistance values change. I understand that the op amp regulates its output in order to satisfy the outer loop. My problem is with the inner loop gain. Why does lowering the 1M resistance result in a smaller output swing.Quote from: Bertho on September 13, 2013, 11:13:48 pmHow does the inner loop behave like a voltage trampoline?

author: Bertho
author_rank: Regular Contributor
timestamp: September 13, 2013, 11:44:54 pm
content: Quote from: tree on September 13, 2013, 11:27:06 pmFor a balance in the system, the current you pull out through 1M will be exactly what is needed so that Ib retains a current for a stable point. That current is nearly constant because the voltage over the 510 Ohm resistor dictates the major part of the potential at the base and therefore the balance of currents at the control-node..Changing the feedback to 100k will keep the base voltage at about the same point (it is forced there by that 510 Ohm). Therefore, the current through the 100k resistor will beabout the sameas would be with the 1M. The voltage drop across 100k is 10 times lower and therefore the output will be about 10 times lower (give or take some linearity problems).

author: Bertho
author_rank: Regular Contributor
timestamp: September 13, 2013, 11:49:54 pm
content: Quote from: tree on September 13, 2013, 11:27:06 pmYou are controlling the 0..120V output voltage with a 0..12V input.

author: tree
author_rank: Regular Contributor
timestamp: September 13, 2013, 11:52:47 pm
content: Quote from: Bertho on September 13, 2013, 11:49:54 pmI know. I understand that the output voltage swing is dictated by the ratio of 1M to 50k. But how does that ratio dictate the output voltage swing?

author: tree
author_rank: Regular Contributor
timestamp: September 13, 2013, 11:55:11 pm
content: Quote from: Bertho on September 13, 2013, 11:44:54 pmThe current through the 100k resistor is two orders of magnitude higher than through the 1M. The simulation proved that.

author: tree
author_rank: Regular Contributor
timestamp: September 14, 2013, 01:17:39 am
content: So if we assume that the input can swing between -10V and +10V, I have simulated the output swing using 1M and 100k resistors. I only changed the resistor values and I want to know why/how the output swing is larger using the 1M resistors.

author: tree
author_rank: Regular Contributor
timestamp: September 14, 2013, 01:56:38 am
content: I finally figured it out. I forgot to take into account the quiescent current and the associated change in said current when the input drops. Because 510 is 1% of 50k, the increase in current through the 510 is negligible, so when the input drops, the current through the 1M has to decrease by roughly the same amount. When you use a lower value resistor the voltage change is not as significant because more quiescent current flows and a small change in current doesn't produce a large change in voltage. However when the resistor is large, the quiescent current is small and a small change in current changes the output voltage by a larger value.

author: Bertho
author_rank: Regular Contributor
timestamp: September 14, 2013, 11:33:24 am
content: Quote from: tree on September 14, 2013, 01:56:38 amThe input voltage is setting up a current through the 50k resistor to pull current from Q1 (amplifying the output). The feedback resistor, 1M, "steals" current from Q1 so it opens less. Going from 1M to 100k steals proportional more current from Q1 and thereby reducing the output.