Load regulation help.

[jep_pogi]

mga masters,

tinignan ko po pala datasheet ng TIP31, and yung gagmitin kong pamalit sa kanya na MJE15032.
okey naman yung capacity ng Vceo at Vcbo nya.
yung Vebo yung mejo problematic kasi 5V lang yung maximum rating nya dun.

sabihin natin na 20 x 4 = 80Volts yung nasa base ng TIP31 kunware, e may time na umabot yung output ko sa 85V +.. so ang magiging Veb nya is 85V(or more) - 80 Volts. which is pwedeng maging > 5V. pwede kayang yan yung dahilan ng pagkasira ng TIP31 ko?

publema ko kasi e 5V lang din ung rating ng MJE15032
reply po sana kayo masters

[labgruppen]

mga masters,

tinignan ko po pala datasheet ng TIP31, and yung gagmitin kong pamalit sa kanya na MJE15032.
okey naman yung capacity ng Vceo at Vcbo nya.
yung Vebo yung mejo problematic kasi 5V lang yung maximum rating nya dun.

sabihin natin na 20 x 4 = 80Volts yung nasa base ng TIP31 kunware, e may time na umabot yung output ko sa 85V +.. so ang magiging Veb nya is 85V(or more) - 80 Volts. which is pwedeng maging > 5V. pwede kayang yan yung dahilan ng pagkasira ng TIP31 ko?

publema ko kasi e 5V lang din ung rating ng MJE15032
reply po sana kayo masters

not a master here, but i would like to ask you some question. do you really think that the emitter voltage of an emitter follower circuit will exceed the voltage at its base at normal operating conditions? sorry to be blunt, but do you understand how an emitter follower circuit works? baka masyadong matalas yung tanong ko, pero since this is your thesis, you must at least know this basic info for your defense.. time to review your electronics101.
unless may ibang voltage source na magdrive sa output ng psu mo, the output should never exceed the voltage of your zener string under normal operation. simply analyze this, if the emitter voltage of q1 is higher than its base, will there be forward base current? will q1 conduct? if q1 is at cut-off, will q2 conduct?  if you have doubts, simply put protection diodes on emitter-base and base-collector terminals of q1. the polarity of the protection diodes should be pretty evident.
i hope the masters would pitch-in and explain it in a simpler form.  i've always been a poor explainer.

bale yung sa collector part nya diba po ay ung casing nya, tapos pag hineatsink, yung screws na maghohold ng MJ15025 sa heatsink ay shorted na sa collector, e di po xa nasosolderan. so parang pinulupot ko lang yung wire from the PCB papunta dun sa screws which is ung collector ng MJ15025.

use the proper to3 mounting hardware. may nabibiling set sa mga electronics stores complete with mica insulator, nylon washers, bolts and nuts and eyelet washers.  solder your collector wire to the eyelet washer to ensure good electrical contact. wag basta-basta ipulupot lang yung wire sa screw. the way you connected your collector increases the probability that my hunch is true.

[jep_pogi]

sir sorry po.
alam ko po yung operations ng emitter follower na ung diode drop lang na Vbe yung difference ng input sa output voltage. nagwonder lang po ako.. pasensya na po.

ung sa pag mount ng TO3, okey po gets ko na po. mali talaga yung paraan ng pag connect ko sa kanya. kaya malaki po tlaga yung chance na tama yung hunch nyo.
salamat po.

[jep_pogi]

bale construct ko na muna po ng maayos yung circuit this week. natapos ko na rin yung layout nya ifafab ko nlng this week and solder.

tetest ko kung anu yung output nya at different load currents tas magdecide nlng po ako kng baguhin ko ba yung zener string kung masyado mababa yung output voltage ko.

[jep_pogi]

palitan ko rin po pala yung TIP32C ng MJE15033. kasi mataas din po Vce ng sa constant current source, may time na dadaan yung galing sa output ng bridge.
nakita ko rin po pala na kailangan na nasa FORWARD ACTIVE region lang dapat yung dalawang transistors sa output.
e kapag full load, mga 27mA and 13mA yung kailangang base drive ng Q2 at Q1 respectively. e kung ganyan po nasa saturation region na xa e, which is hindi maganda. pero naisip ko naman po na yung Vce naman ay parang pinipilit na greater than Vce@saturation.
Enough na po ba yan para masabi na FA yung transistors?
wala po kasi Ic vs Vce curve at Ic vs Beta curve na kailangan ko e para maverify talaga yung beta at that region.. kasi assume lng ako ng mga worst case beta para jan.

[TinTopHack]

jep-pogi - I am not a master either but let me try if I can help here.

Below I have redrawn your schematic into a simplified one. I removed the filter caps and their corresponding balancing resistors and replaced your current circuit with a current source symbol.
I replaced Q1 and Q2 with Qs to help in the explanation that follows.


As mentioned by labgruppen, the output circuit is just a voltage follower. So with Qs shown, it is clear that we have a voltage follower. As a follower the output at the emitter is almost equal to the input at the base. Vout = 71.4V is practically equal to Vin-base = 72V. The base of Qs is pegged (naka-pako) at 72V by the zener chain no matter how the supply voltage input varies (85 ~ 120VDC). The Ve of Qs will approx. be 0.6V less than its Vb so it will be at 71.4V. The Vce of Qs will be 13.6V at a Vsupply in of 85VDC. Under this condition, Qs is operating in linear mode (you call it forward active region) because it is not fully off and not fully on (not in saturation). What forces this condition - operation in the linear region - to occur? Negative feedback.

How does negative feedback happen here?
Imagine that the output voltage at the emitter of Qs starts to rise for whatever reason. Since the base of Qs is pegged at 72V, its Vbe has no choice but to decrease. As Vbe starts to decrease (drop below 0.6V) its base current decreases too causing the conduction of Qs to decrease (collector current decrease) also. This decrease in conduction of Qs is like Qs increasing its resistance between collector and emitter. If the resistance of Qs increases, the voltage drop across it will increase. This voltage drop across Qs will cause the voltage across the load (same as the Vout voltage at the emitter of Qs) to drop.Thus the rise in output voltage is opposed. This feedback mechanism will insure that the emitter voltage of Qs will never be higher than its base voltage> This means the condition that your output voltage will be higher than your zener voltage can not happen (unless there is something wrong in the circuit).

As you can see on the right side of the drawing, Qs is represented by a variable resistor in series with the load. That is exactly what Qs (series pass transistor) does - to act as a variable resistor in series with the load. Another way of looking at it is > Qs and the load forms a voltage divider. In this voltage divider, the resistance of Qs is controlled automatically so that the voltage drop across the load is kept constant. What produces this automatic control? The zener chain that keeps the base voltage of Qs constant and the negative feedback just described.

Why not just use one transistor Qs instead of having Q1 and Q2? Because the base current needed for Qs will be quite large. If your output current is 2.25A max and the hfe of Qs is 15 only for example, the base current needed will be 150mA. You will need a larger current source and you will need high wattage zeners to absorb the surplus current when Qs needs less base current at no load conditions.

Below is your circuit with Q1 (NPN) and Q2 (PNP) shown (but still simplified to make it easier to understand).

So in this case the large base current of Q2 (PNP transistor) is supplied by Q1. The smaller base current of Q1 is supplied by the current source. As as result the surplus current at no load is small enough to be absorbed by the zeners. Both Q1 and Q2 also operates in the linear region unless your Vsupply_in drops too low - approaching 72V - under this condition they could saturate. Just remember Q1 and Q2 acts like one big transistor - like Qs. At this point I leave the rest of the details for you to figure out.

[jep_pogi]

sir, thanks po ng marami thanks po nag effort pa kayo magdrawing para sa akin.
naiintindihan ko na po salamat sa inyo ni sir labgruppen pati sa iba pang naka2long skn.
nafab ko na yung bagong ckt ko nilakihan ko yung board pati inayos ko ung pagconnect ng MJ15025.
bukas mag solder na ako tapos test na rin, may E-load na sa lab ko na pwede ko magamit anytime. inaaral ko na po ngayon panu gamitin ung e-load hehe.

pati binabalak ko po palitan yung zener string kung mejo mababa ung output at gusto ko taasan ng konti. bale 84V max lang naman ang limit ko po.

thank you sirs!

[jep_pogi]

tinest ko na po sa eload yung circuit ko mga sirs.
okey po siya, nasa 76 to 78 volts lang output nya from 0Amps to 2Amps.

umiinit nga lang po yung power transistor at yung transformer mismo kapag mejo matagal sa 2Amps.
di ko pa tinatry itest xa ng matagal tlga (1hour).
nasa 20minutes to 30 minutes plng yung pinakamatagal ko na test.
 tapos matagal bago lumamig ulit ung transformer.
maglagay daw po ako ng fan.
balak ko dalawang fan sana.

[labgruppen]

normal lang uminit yung transformer lalo na yung transistor. the transistor has to dissipate the power due to the voltage drop across it(Vce) and the load current. so if the voltage at the smoothing capacitor is at 110V and the output is at 78V 1.2amperes, the dissipation of the series pass transistor is approximately 38.4Watts.
at 30mins with constant load, the power supply should have reached already a thermal equilibrium imho. so it should not get significantly hotter than what you have observed in your test. anyway, placing a fan is a good idea.
you can buy some "surplus" 220VAC box fans in ROKS (surplus pero maganda quality, equipped with sealed bearing pa) complete with guard grilles.
+1, kudos on your successful test.
BTW, how's the load regulation?

[jep_pogi]

ay sir eto po yung outputs:

@0A: Vout = 80V
@2A: Vout = 79V

and chineck ko kung Forward Active sila, FA naman po. bale fan nlng ung idadagdag ko muna. then i thnk okey na po.

[studyante]

uhm mga master patanong lang. san sa circuit schematic mo pwedeng i-"tap" yung fan para mapower-up mo siya?

kung 220VAC fan ang bibilhin nya, bale ipaparallel nya yun sa primary side ng transformer tama ho ba ako? o mali?