No announcement yet.

Linear Regulator Amplifier Experiments

  • Filter
  • Time
  • Show
Clear All
new posts

  • #16
    Beta multiplier


    That is not correct again. If the Cap is kept at 30V and you want to pass let say 2 amps to the battery that R4 resistor must be really big.
    John B. commented on similar circuit I posted here :
    You can see this circuit given in the datasheets of the regulators for increasing the current of the regulator but as I have researched and read the NPN darlington is much more stable. NPN is the way.

    But let's get back to the Beta Multiplier.
    Now I suppose you have learned how zener voltage regulator works. Take a look at the following link and pay attention to figure 3, and figure 7.

    Study that circuit and read carefully about the calculations. The Resistor should be calculated to supply just enough current so that the pass-transistor opens and the Zener could work properly. Once the transistor is open it will amplify that current based on the Beta. Imagine you have a Darlington arrangement with total gain of 1000. This means that every 1 mA on the base will become 1A on the output. So, the calculations are important. And passing the predetermined amount of current and voltage the transistors in the darlington should stay in their SOA (Safe Operating Area). So you should look for a transistor which is capable of passing the desired current at 30V DC.



    • #17
      thanks a ton for the find the best info i guess I'm just not as clear on what Im looking for...anyway so figure 7 is what I'm after for sure!....

      trying to figure out how to calculate Ib. I must have to have a zener across the filter cap to keep it at 30V then from there I can use Ohms law to get my resistance value if I know the zener on the base of the beta multiplier is say 15V and I know I want 1ma for Ib......R = Vfiltercap - V(zener on base)/Ib = 15V/1ma = 15kohms...I'll try and simulate this....but that will only get me 15V-0.6V = 14.5 V on I not want about 24V on Vout since I am charging a battery and need a higher voltage at that point??? but then I will need a 30V zener on the base of the darlington pair but then where does that leave the zener diode that is in paralell across the filter cap....head spinning


      • #18

        I suggest to start with the calculations from the maximum load current.
        You do not need a Zener across the filter cap. You need only the zener on the base of the beta multiplier. Actually there are two filter caps so far. One would collect the spikes and the other is indeed across the zener but on the base of the multiplier.
        Watch again the series-pass video, read through the John's post. I admit I was doing it again and again the moment I would grasp something
        My head was spinning for quite a while since I was going trough all the links and trying to filter the information to what fits the guidance of JB.
        Now I have it a more clear but still have my doubts about some things.
        I just share what I have learned so far to save the time of others who would be interested in that. After all that is why we are all here, right ? Learning, sharing, helping !

        Last edited by Lman; 09-15-2014, 03:10 AM.


        • #19
          i should be shooting for 30V-0.6V from emitter to ground correct?? Which means I need a 30V zener on the base....if i used a 15v the highest I could ever charge the battery would be 14.4V which isnt bad but I need 15V


          • #20
            i wonder how the spike accumulator stays at 30V without a zener? i cant calculate the resistor value if I don't know what the voltage will be on the collector??


            • #21

              Yes, 30V zener minus the 0.6 Base-Emitter drop would give you 29.4V out but watch the video where John shows the Linear Amplifier Regulator for SSG on Energenx channel. He has another diode going from the anode of the zener to ground (running in generator mode) which compensates also for the 0.6 drop on the transistors base emitter. So the output of the Beta Multiplier will be exactly 30V. And than you supply the amplifier with that voltage.

              About the biasing resistor there are some formulas on wiki here :



              • #22
                Lman, ok here is a zener controlled beta multiplier delivering about 5.5 Amps to a load. Seems to be working pretty well and I am alot more clear on the importance of Ib, Iz, and Ir1. I think its the simulation that goes wonky when I reduce the load impedance down to 23mOhms. When I do that, I have to increase R1 so much that its not enough to put the zener into reverse breakdown.
                Anyway...I think I'm getting the hang of just need to get the amplifier part of it in needs to adjust to that crazy low impedance some be continued....

                Circuit and test values attachedClick image for larger version

Name:	123zener-controlled-linear-power-supply.jpg
Views:	1
Size:	17.4 KB
ID:	46416


                • #23
                  R1 is the 1W 25kOhm resister


                  • #24

                    Yes, it starts to get slowly clearer. Now get the datasheet of the darlington you are simulating with and check the SOA. What happens when at 30 volts you try to pass 5 amps DC ??
                    It will be safe at 3 Amps.
                    I think this is what John means by saying that it should be chosen for the correct voltage and current.

                    Originally posted by John_Bedini View Post
                    It is simple, however in the case of the series pass device the correct one must be chosen for the voltage and current.
                    Something else he points out in the video is that the beta multiplier and the linear amplifier work together. I think here ... it is not a coincidence that he put the linear amplifier and the beta multiplier on the same heatsink. Temperature compensating probably ???
                    And also he is not using single darlington but two TO3 devices. And what we know about darlington arrangement ? The lower transistor is high gain but low power, and the upper transistor is low gain high power. Since the higher transistor will pass the most of the current it will heat more together with the Linear Amplifier. That lead me to think that there could be some temperature compensation happening while arranged like that. I may be right or wrong on that.
                    We can always learn by some mistake .

                    Let us pretend that this darlington would work at 3A. The next step is to feed the Linear amplifier with the power coming from the beta multiplier.
                    Here is the same link about the audio amp I posted over there :

                    It represents a Class A and it is configured as an Emitter follower.



                    • #25
                      I forgot to mention about the impedance of the battery. It is not 23 mOhms. John has mentioned it twice in the other thread. Once as being 0.00023 which is 230 micro Ohms and second time as being 0.0023 which is more like 2 mOhms. I suppose one was a typo.



                      • #26
                        thought i'd share this reply from some poking around i did on an electronics forum...pretty interesting...they don't know anything about the context of the radiant energy and what not i just ask electronics based questions..this reponsClick image for larger version

Name:	zener-controlled-linear-power-supply-emitter-follow-stage-added.jpg
Views:	1
Size:	18.3 KB
ID:	46428e is intense and wanted to share it...i was asking about the emitter follower driving a very low impedance load like a battery bank

                        ************************************************** **
                        No, a 3A supply cannot possibly "handle" driving a 0.0023Ω load.

                        To drive 0.0023Ω to 24V, the supply would have to be capable of delivering I=E/R = 24/0.0023 = 10435A!!!.

                        Conversely, if you can only get 3A out of a current-limited supply, the voltage across a 0.0023Ω resistor would be only E=IR = 3*0.0023 = 0.0069V = 6.9mV.

                        Where did you get the 0.0023Ω from?

                        The correct way of describing what happens when you charge a mostly-discharged 24V lead-acid battery bank is the following:

                        Initially, the charger needs to be "current-limited", mostly to protect the charger itself (from excessive heat produced in the regulator, blowing up the rectifiers, overloading the transformer, etc).

                        The initial magnitude of the charging current is chosen primarily by the components in the charger, and to a lesser extent, by the maximum charging current the battery maker allows, and finally by how long you are willing to wait for the battery to become charged...

                        The job of the "current-limiter" part of a battery charger is to hold the current at the selected value while the battery terminal voltage slowly climbs from 1.9V/Cell (~22V for a 12 cell battery) to about 2.28V/cell (27.4V). It is the current-limiter that prevents the charger from blowing up when it is connected to a discharged battery bank...

                        As the battery accumulates charge, the battery terminal voltage slowly climbs. After reaching 2.28V/cell, the battery is about 85% charged. To get it the rest of the way to full charge, at this point in the charge cycle, the charger should automatically become "voltage-limited" instead of being current-limited...

                        If the battery voltage is now held constant at 2.28V/cell (27.4V), the charging current will naturally gradually taper to a much lower value which in the steady state is determined only by the losses and leakage in the battery. The time it takes for the current to taper to the final value could be quite long.

                        There is a way to shorten the time it takes for the battery to charge the last 15% of the charge cycle, but that is beyond the level of this discussion...

                        A 3A charger is not up to the task of recharging a mostly discharged 240AH battery bank. The best you can ask of a 3A charger is to maintain (float charge) the bank near full charge while the battery bank is idle for long periods...

                        Any way my new simulated circuit with emitter follower added is attached


                        • #27

                          The formulas and what they told you about a current limiting is not wrong. A discharged battery is really hungry for current and it indeed can blow a charger. This is why most of the chargers have fuses.
                          A 3A charger can charge a 240AH battery if it is current limited but sooo slow that you will grow beard several times. That is why John has build chargers in different current and voltage ranges. You would not put a S3A12 on a 240AH battery, would you?
                          You can see thousands of battery charging circuits on the net mostly copies of one another. And not even one is close to what John is doing.

                          Looking at the circuit you posted you did not get it about the diodes. The diode to ground must be connected in series with the zener but opposite to each other. The anode of the silicone diode connected to the anode of the zener.

                          I give you here two links about the Emitter Follower :
                          On the first link start reading from where figure 14 is. Look at the phase of the wave as well.


                          Hope that you will get it now how emitter follows and what follows. And how exactly you supply a 0.0023 load with the correct current.

                          Do not forget to read again Johns posts again after that. !!!

                          Last edited by Lman; 09-19-2014, 02:48 AM.


                          • #28
                            Emitter Follower Summary

                            The links I posted in my previous post explain how Emitter Follower amplifier works.

                            1. It has voltage gain of nearly 1 but it gives us current gain.
                            2. It has high input impedance and low output impedance.
                            3. The current gain will be the current on the base multiplied by the gain of the transistor.
                            4. What about the voltage ? The emitter voltage will be always that of the base minus the base-emitter junction drop.

                            In our case to have 15 volts on the output we supply 15.7 to the base. And for the current we should again consider the SOA of the transistor at DC.

                            Last edited by Lman; 09-30-2014, 08:19 AM. Reason: removing erroneous information


                            • #29
                              have a look at this thread

                              I have done some experimenting, on this, any feed back would be great, and please correct me if I have something wrong. But I think I am very close. I hope others may find the info useful, think of it as a story, and if it make sense to you, it may help you with your experiments.

                              Everyone is thinking voltage as being a measurement to check the battery charge, charging graphs graph the voltage over time. Has anyone ever monitored the battery impedance as it was charging. If you were to measure the voltage current ratio (impedance), and graph it with a computerized battery monitor. The reason I say this because I think that The linear regulator amp regulates battery impedance.

                              What about impedance charging graphs.
                              Last edited by Nityesh Schnaderbeck; 09-19-2014, 03:50 PM.