PSU Question - (Newbie)

Thread Starter

CCinBR200

Joined Oct 4, 2020
80
That regulator is going to be dissipating nearly 7 watts and will cook itself to death in short order. Are you familiar with what happens to an IC when you let the magic smoke out?
In my short hobby career, I have made smoke, yes! This wasn't intentional. (lol) From the advice I got, I expected the voltage to be reduced to about 20V coming out of the 1N4004. Don't understand why that's not the case.
Thanks for any help!
 

Thread Starter

CCinBR200

Joined Oct 4, 2020
80
No your drawing is using the Bridge rectifier, my diagram doesn't, it uses one diode.
Ahhh. But one concern. The main power circuit is working perfectly and it comes after the bridge rectifier. So, I tried doing your circuit without the main circuit connected.

Getting closer!

at anode of 1N4004: 14V
at pin 1 of L7812: 14V
at pin 3 of L7812 (no load): 5.4V
at pin 3 of L7812 (fan connected): 2.8V
Obviously, fan didn't spin.

Tried with a different L7812CV. Same results.
 

Dodgydave

Joined Jun 22, 2012
9,936
If you have 14V on the input then you should be getting 12v on the output of the 7812, if not then you have it wired wrong OR it's blown.

Ideally post a picture of the working circuit board.
 

Thread Starter

CCinBR200

Joined Oct 4, 2020
80
If you have 14V on the input then you should be getting 12v on the output of the 7812, if not then you have it wired wrong OR it's blown.

Ideally post a picture of the working circuit board.
I don't understand what happened?? Now I'm getting:

at anode of 1N4004: 14mV (!!)
at cathode of 1N4004: 13.8V
at pin 1 of L7812: 13.8V
at pin 3 of L7812 (no load): 5.4V
at pin 3 of L7812 (fan connected): 1.8V

I'm starting to hope I'm not the worst hobbyist in this board's history.IMG_20201120_143446616.jpg
 

Thread Starter

CCinBR200

Joined Oct 4, 2020
80
It's the base-emitter junction that needs to be turned on, so the calculation is
\(R=\frac{V_{BE}}{I_{D1}}=\frac{0.7V}{9mA}=77.7\Omega\)
The closest standard 5% value would be 75 ohms.
Thank you.

You're missing a connection dot on the cathode of C2 (critical) and the anode of C3 (cosmetic). The secondary voltage should be on the secondary, not by the switch. We don't wrap circuits like that. The preferred direction is left to right and top to bottom. There are exceptions to that rule, but this isn't one of them.
Thank you. I knew about one of those and really struggled with how to do it on Scheme-It. It's part of the learning curve for me.

EDIT: 1N4745 is a 16V zener, so you're not understanding the circuit.
Man, I have no argument with that statement! I'm trying to learn as fast as I can. I don't want to ask you and the others to teach me everything so I try to take every suggestion and learn as much as I can about it on my own and try to come up with a solution.

As you may have seen, I am working with some ideas suggested using a halfwave diode circuit. I do have the parts on order to follow your suggestion for using a Zener diode if I can't make that idea work. Thank you for your generosity. I will be a better hobbyist for it.
 

Thread Starter

CCinBR200

Joined Oct 4, 2020
80
I only have Fridays for my hobby. So my workshop is closed. Next Friday is a holiday so I'll be back in 2 weeks. Thanks to all the generous folks who are trying to help me despite my limitations.
 

dl324

Joined Mar 30, 2015
12,786
I'm trying to learn as fast as I can. I don't want to ask you and the others to teach me everything so I try to take every suggestion and learn as much as I can about it on my own and try to come up with a solution.
clipimage.jpg
I sometimes forget what it's like to be at the bottom of the pile of things you need to know to be able to design things.

This circuit drops the input voltage by Vz+1V. I'd normally use 0.7V for the base-emitter voltage drop, but NatSemi was taking in to consideration that this would be a power transistor passing significant current, so they used 1V.

When the transistor is fully on, 1V will be across R. If you want 9mA in the zener,
\(R = \frac{V_R}{I_R}=\frac{1V}{9mA} = 111\Omega\)
The nearest standard 5% value would be 110 ohms.

But, there's a complication. The beta for power transistors is low, so we'd use more like 15 for a current of 1A (TIP29). That would require a zener current of more like 75mA. Depending on the zener voltage, that might require too much current. In that case, you'd use a darlington configuration to get more current gain and not have to use a power zener diode anyway.
clipimage.jpg
Now the beta is going to be more like 250+, so a 1A load current would only require 4mA and the zener current could be 9mA and still have a reasonably stable zener voltage.

EDIT: corrected diode orientation.
EDIT 2: Just noticed that I used the symbol for TIP31C (closest Eagle had). That transistor has a slightly higher minimum beta (at 1A) and is rated for 3A.
 
Last edited:

Thread Starter

CCinBR200

Joined Oct 4, 2020
80
View attachment 222924
I sometimes forget what it's like to be at the bottom of the pile of things you need to know to be able to design things.

This circuit drops the input voltage by Vz+1V. I'd normally use 0.7V for the base-emitter voltage drop, but NatSemi was taking in to consideration that this would be a power transistor passing significant current, so they used 1V.

When the transistor is fully on, 1V will be across R. If you want 9mA in the zener,
\(R = \frac{V_R}{I_R}=\frac{1V}{9mA} = 111\Omega\)
The nearest standard 5% value would be 110 ohms.

But, there's a complication. The beta for power transistors is low, so we'd use more like 15 for a current of 1A (TIP29). That would require a zener current of more like 75mA. Depending on the zener voltage, that might require too much current. In that case, you'd use a darlington configuration to get more current gain and not have to use a power zener diode anyway.
View attachment 222926
Now the beta is going to be more like 250+, so a 1A load current would only require 4mA and the zener current could be 9mA and still have a reasonably stable zener voltage.

EDIT: corrected diode orientation.
EDIT 2: Just noticed that I used the symbol for TIP31C (closest Eagle had). That transistor has a slightly higher minimum beta (at 1A) and is rated for 3A.
View attachment 222924
I sometimes forget what it's like to be at the bottom of the pile of things you need to know to be able to design things.

This circuit drops the input voltage by Vz+1V. I'd normally use 0.7V for the base-emitter voltage drop, but NatSemi was taking in to consideration that this would be a power transistor passing significant current, so they used 1V.

When the transistor is fully on, 1V will be across R. If you want 9mA in the zener,
\(R = \frac{V_R}{I_R}=\frac{1V}{9mA} = 111\Omega\)
The nearest standard 5% value would be 110 ohms.

But, there's a complication. The beta for power transistors is low, so we'd use more like 15 for a current of 1A (TIP29). That would require a zener current of more like 75mA. Depending on the zener voltage, that might require too much current. In that case, you'd use a darlington configuration to get more current gain and not have to use a power zener diode anyway.
View attachment 222926
Now the beta is going to be more like 250+, so a 1A load current would only require 4mA and the zener current could be 9mA and still have a reasonably stable zener voltage.

EDIT: corrected diode orientation.
EDIT 2: Just noticed that I used the symbol for TIP31C (closest Eagle had). That transistor has a slightly higher minimum beta (at 1A) and is rated for 3A.
Hello again! I got all the parts and wired this up with a 1W 150ohm resistor. When I plugged it in, my bridge rectifier burned up. What do you think I'm doing wrong?

Thanks!
 

Thread Starter

CCinBR200

Joined Oct 4, 2020
80
Hold onto your answers. I'm pretty sure I caused a short.
View attachment 222924
I sometimes forget what it's like to be at the bottom of the pile of things you need to know to be able to design things.

This circuit drops the input voltage by Vz+1V. I'd normally use 0.7V for the base-emitter voltage drop, but NatSemi was taking in to consideration that this would be a power transistor passing significant current, so they used 1V.

When the transistor is fully on, 1V will be across R. If you want 9mA in the zener,
\(R = \frac{V_R}{I_R}=\frac{1V}{9mA} = 111\Omega\)
The nearest standard 5% value would be 110 ohms.

But, there's a complication. The beta for power transistors is low, so we'd use more like 15 for a current of 1A (TIP29). That would require a zener current of more like 75mA. Depending on the zener voltage, that might require too much current. In that case, you'd use a darlington configuration to get more current gain and not have to use a power zener diode anyway.
View attachment 222926
Now the beta is going to be more like 250+, so a 1A load current would only require 4mA and the zener current could be 9mA and still have a reasonably stable zener voltage.

EDIT: corrected diode orientation.
EDIT 2: Just noticed that I used the symbol for TIP31C (closest Eagle had). That transistor has a slightly higher minimum beta (at 1A) and is rated for 3A.
Ok. Redid it without the short :) and no more smoke. But... I'm getting 40V after the TIP31. What do you think I'm doing wrong?
 

Thread Starter

CCinBR200

Joined Oct 4, 2020
80
Which circuit are you using? Are you certain you have things wired correctly? What are the voltages on the TIP31 terminals?
I'm using the TIP31 circuit you suggested. I'm not sure of anything :) but I've checked it a few times and it seems right. The voltages at the TIP31 are:
1) 41.1
2) 40.7
3) 41.1

Thanks!
 

dl324

Joined Mar 30, 2015
12,786
The voltages at the TIP31 are:
1) 41.1
2) 40.7
3) 41.1
If the terminal numbers correspond to the datasheet, 1=base, 2=collector, 3=emitter; the base emitter junction looks like it's shorted. Don't see how the collector voltage could be lower than the emitter.

Are you certain of the terminals and connections?
clipimage.jpg
 
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