Hi I'm building the power supply in the attachment however i want to change kt818 to a pair of parallel п210б (p210b) transistors and change the bd139 to something bigger my question is how to calculate the change to R20
the schematic came from hobbyhour although by the style and components it looks like a scan from an old soviet era magazine
Why do you think you need to change R20 ?
on the site the comments say to upgrade volatage and current change the bd139 the kt818 and recalculate r20
So you have to tell us what you are trying to upgrade the current too ?
It will likely involve emitter resistors to ensure the pass transistors share properly, more drive current and bigger heatsinks as well
well the transformer i got is capable of 20A although il never need that amount the transistors can handle 20A so putting 2 in parallel should keep them cool although i intend to install a thermistor controlled fan on a separate transformer the bd 139 im looking at c5591 than can handle 20A as well the elec cap im gonna use a bank of 6 200v 820uf hooked in parallel its not in the scematic but ive got a 15a bridge diode so yep components are all being up rated
Sorry but you still did not say how many amps you are trying to uprate the output to ?
Say for example its 20A and you want that down to 5V then the power dissipation will be at least 500Watts
Are you sure you have enough heatsinking for that ?
If thats an N2106 russian transistor it's only rated at 45W
hi yep bucket full of heatsinks like i said never gona use 20A but the transformer can handle it so i wanna make it push 20 for experiments sake i was gonna try a 160A transformer i got layng here but i dont wanna kill myself its like the devil gets into me and i cant help but play lol
Like I said you will never do it with N2106 transistors there simply not man enough for the job.
Normally for something with that sort of output a switching pre-regulator such as a pre-transformer triac or post transformer controlled rectifier would be used to keep the series pass dissipation within reason. To do it in one hit as a linear series pass is gonna cost you a fortune in transistors and a seriously better circuit design in my opinion but there maybe someone with a better idea
yep quite correct 12A 45w 65v so 3 of em in parallel
yep quite correct 12A 45w 65v so 3 of em in parallel
Ever heard of transistor SOA (Safe Operating Area) and specifically for BJTs, secondary breakdown?
At high Vce, (e.g. if the output is shorted) you'll be lucky if your transistors can handle even 2.5A each even if you put them on a water-cooled heatsink held below 25 deg C.
I cant find a good datasheet for your transistors, but lets consider a
TIP3055, which is nominally 15A, 60V, 90W. At 30V Vce, its only good for 3A steady state, *IF* you can keep its mounting surface below 25 deg C. (see datasheet fig. 2). That's the highest voltage at which you can get the rated wattage out of it. Above that voltage secondary breakdown limits the permitted current more stringently so you cant get the nominal wattage.
As fourtytwo42 has pointed out, to implement a high current linear regulator capable of 30V out would require a massive array of pass transistors on an absolutely ludicrously large heatsink, probably with forced air cooling. Getting enough base drive without either slugging the bandwidth resulting in very poor transient response, or running into massive instability issues is going to be very challenging, especially in a LDO topology.
R20 is a current shunt and is used to sense the current in the circuit, not sure why you would change it since it's easier to adjust R14 for example and you would get the same result.
Also even if the transistor can supply 20A that does not mean you can exceed the 45W max power dissipation. At 20A the best your transistor could do is a 2.25V drop from your supply rail.
For example if you get 50V after the bridge rectifier then the best you can get at 20A is 47.75V on the output, lowering the voltage any lower than that will destroy the transistor.
Connecting the transistors in parallel makes them prone to instability and thermal runaway, which is very relevant in your situation if you're trying to push big amps through the transistors.
Any difference in voltage between your supply rail and the output of the power supply multiplied by the current draw will result in power dissipated by the transistors.
It is written near the output '0-30V 0-3A'. So we have 3A max with 0.1R shunt. That means we'll have 6A max with 0.05R shunt.
Thank you how did you get that calculation
Simple proportion and Ohm's Law: Halve the current sense resistor and you'll double the current required to get the same voltage across it.
after taking a reality check i have read the advice given here and have settled for a tip42c and wont go above 5A thank you for all your help