Hi, All
I build a simple schematic using TL431 but my transistor run hot and the output voltage isn't reach what i want. Can anyone help on this? Below is my schematic.
I had source 25v AC and i need Vout =20v Iout = 60ma
When i power up the circuit below. Vout = 16v and Transistor run hot in few sec. if i add 10uf after the transistor it come out 18v but transistor run hot as well.
Thanks if anyone can help.
I build a simple schematic using TL431 but my transistor run hot and the output voltage isn't reach what i want. Can anyone help on this? Below is my schematic.
I had source 25v AC and i need Vout =20v Iout = 60ma
When i power up the circuit below. Vout = 16v and Transistor run hot in few sec. if i add 10uf after the transistor it come out 18v but transistor run hot as well.
Thanks if anyone can help.
Attachments
The TL431 device current must be >1mA, so decrease the 2.4k resistor's value to <600R
so that it can regulate properly. A value of 200R will set the TL431 current to 3mA.
When operating as intended, the current through the 62R is about (35V-20V)/62 = 242 mA.
Power dissipated by Q ~ (20V x 0.239A) = 4.8W. That will be hot, so use a good heat sink.
Power dissipated by 62R = (15V x 15V)/62R = 3.6W. That will be hot, so use a 10W resistor.
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It really wants a cap after the rectifier; otherwise the output is pulsating NOT steady as you want.
With a "small" (100uFd) cap the output may be nearer 29V than 35V.
I get slightly lower heat numbers than rayma calculated. Depends what first cap you use, and PT parameters.
In any case, the big R is passing a LOT more than the 60mA you need. You do need "more", but 3X-4X more seems excessive. All that excess has to be absorbed by the BD140. 2 to 5 Watts is Too Much for a power transistor with no heatsink. Also the "61.9r" is around 2 Watts (while load power is only 1.2W).
Make the "61.9" much larger, over 100 Ohms. This is a balancing act; too large will not be good. 200 Ohms is too large. A second 61.9 in series may be an easy hack?
A 7815 or LM317 can be shimmed to 20V and will not have the large excess current/heat of this shunt regulator. I also suspect it has better ripple rejection here on raw DC. Shunt regulators normally go after some brute-force regulator or massive passive filter. For loads approaching a Watt, shunt regulators are always "hot".
With a "small" (100uFd) cap the output may be nearer 29V than 35V.
I get slightly lower heat numbers than rayma calculated. Depends what first cap you use, and PT parameters.
In any case, the big R is passing a LOT more than the 60mA you need. You do need "more", but 3X-4X more seems excessive. All that excess has to be absorbed by the BD140. 2 to 5 Watts is Too Much for a power transistor with no heatsink. Also the "61.9r" is around 2 Watts (while load power is only 1.2W).
Make the "61.9" much larger, over 100 Ohms. This is a balancing act; too large will not be good. 200 Ohms is too large. A second 61.9 in series may be an easy hack?
A 7815 or LM317 can be shimmed to 20V and will not have the large excess current/heat of this shunt regulator. I also suspect it has better ripple rejection here on raw DC. Shunt regulators normally go after some brute-force regulator or massive passive filter. For loads approaching a Watt, shunt regulators are always "hot".
Attachments
PRR and Rayma
Combine both of you suggestions, i had redraw the schematic please help to check whether this calculation and schematic will work in the right operating point. I plan not using the heatsink on the Q
As Q will work around 3.8-4ma if i assume the VBE 0.7v
and the Q with no load will be around 1.5w
Does this workable and did all the calculation correct?
Thanks
Attachments
Seems ok, I think the input voltage will be higher, even with the small 100uF input capacitor.
Measure it and see.
Mark, those clever net names will fly over these young whipper-snappers' heads
😀
mlloyd1
😀
mlloyd1
I agree. The 25Vac transformer is likely to give 31Vdc only when mains is at -6%.
When mains is at maximum of +6%, the average voltage on the first smoothing cap is likely to be over 35Vdc. Maybe even as high as 37Vdc.
That resistor wiil need to dissipate a lot more power. Then the Shunting Q will have to dissipate all the extra current that does not get used.
Maybe replace the 150r resistor with a 317 configured as a Constant Current Source set to ~80mA (16r in the output lead).
0.8mA through the resistor string, 2.2mA through the 431, 16mA through the shunt Q and 61mA to the load.
The load current can vary by +-15mA and the shunt will still operate.
Use heatsinks !
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