Linear regulator question

[budm]

Re: Linear regulator question

"Now, question: you say a transistor should read around 0.7V between the emitter and base ? Why ? Does the datasheet tell you something I'm not seeing ?"
This where you need to know what the device is on how it basic function is.
https://cdn.badcaps-static.com/pdfs/...102e6e50a7.pdf
http://www.learningaboutelectronics....f-a-transistor
Use you DMM in Diode mode and see what the reading on the B-E (forward bias) of the Transistor shows (you are reading the junction Voltage).

[Dannyx]

Re: Linear regulator question

have a play with this ..
http://www.falstad.com/circuit/e-npn.html

Damn...I need to go back to school again Don't have the patience to read anything right now...will check it out someday maybe - my head kinda spins

[budm]

Re: Linear regulator question

""Now, question: you say a transistor should read around 0.7V between the emitter and base ? Why ? Does the datasheet tell you something I'm not seeing ?"

See page 3 for example. 2n3906

See page 2 for MJE13007 which has very high Vbe.

[Dannyx]

Re: Linear regulator question

""Now, question: you say a transistor should read around 0.7V between the emitter and base ? Why ? Does the datasheet tell you something I'm not seeing ?"

See page 3 for example. 2n3906

See page 2 for MJE13007 which has very high Vbe.

I knew about the VBEsat parameter, but my MJE13007 doesn't have a minimum value - just says 1.2VdcMAX, so naturally I got confused as to where the hell those 0.7 came from. Ok, so now we know VBE, so perhaps my figures were right after all, as 9.4-1.4 = 8 which is exactly what I was reading...slowly starting to get the hang of things. hFe always confused me as it has no value (I HATE that !) so it messes up my calculations big time when I forget where to place the "dot" (as in mA or A ? W or mW ?)....whew :|

[budm]

Re: Linear regulator question

So when you indicate that you only get 3V at the output with heavy load, that means the Vdrops of the Zener is not at 9V but a lot less.
Your Transistor is not bias on hard enough to supply the required load current you want.
Since Beta of that Transistor can be as low as 8 then it will need much lower Base resistor value which in turn will cause more dissipation on the Zener diode, so you need to get better pass Transistor for this circuit.

[Dannyx]

Re: Linear regulator question

So when you indicate that you only get 3V at the output with heavy load, that means the Vdrops of the Zener is not at 9V but a lot less.
Your Transistor is not bias on hard enough to supply the required load current you want.
Since Beta of that Transistor can be as low as 8 then it will need much lower Base resistor value which in turn will cause more dissipation on the Zener diode, so you need to get better pass Transistor for this circuit.

You hit the nail on the head with that one, as that's exactly what I was thinking: too low of a gain requires more base current, which requires a lower value resistor, hence the diode will burn more power also. Ok, got it all figured out now A darlington transistor has better gain from my knowledge, but will also require rethinking the values again....hope I'll still know my formulas by then

[PeteS in CA]

Re: Linear regulator question

In basic electronics classes, a diode or B-E junction are said to drop .6 or .7 volts. And beta is treated as constant. It's done to simplify the teaching and calculations, but classes really should say more about datasheet curves and various technologies (e.g. Schottky and ultrafast recovery). I think my wake-up moment was seeing that the 2N6547, while rated for 15A, was really only useful for 10A or 12A.

In a darlington, the betas of the two internal transistors are multiplied, resulting in a really high "beta". The price is that the "B-E" drop is double, and the saturation voltage will also be higher.

[Dannyx]

Re: Linear regulator question

Another formula I found VERY useful for a math bum like me is the one for calculating the Ib of a transistor: hFe * Ib = Ic. Since we know hFe (gain) and Ic (required output current) I rewrote it as Ic / hFe = Ib which makes it easier for me. Again, it's the math that's killing me and I have to write stuff like I was taught in school which was decades ago In my case, 0.5A/8beta=0.0625 A, so a current of 0.6mA would be required to "open" the transistor enough for those 0.5 amps to flow through....correct me if I'm wrong.

[budm]

Re: Linear regulator question

Another formula I found VERY useful for a math bum like me is the one for calculating the Ib of a transistor: hFe * Ib = Ic. Since we know hFe (gain) and Ic (required output current) I rewrote it as Ic / hFe = Ib which makes it easier for me. Again, it's the math that's killing me and I have to write stuff like I was taught in school which was decades ago In my case, 0.5A/8beta=0.0625 A, so a current of 0.6mA would be required to "open" the transistor enough for those 0.5 amps to flow through....correct me if I'm wrong.

Your figure is off again: 0.0625A = 62.5 mA

[Dannyx]

Re: Linear regulator question

This is what I was saying earlier about messing up the values...I lose track of the multiplier. So 0.6A would = 600mA and 0.06A would = 60mA (used round numbers again, but I get the gist).

[momaka]

Re: Linear regulator question

Another important point (and this was briefly mentioned by redwire and myself): watch those power ratings! If you have 12V going into the circuit and 9V coming out, and the running current is 1A, then your linear regulator circuit will be dropping 12 - 9 = 3V. At 1A, the power the circuit will have to dissipate is
V * I = P .... or 3V * 1A = 3 Watts.

Now 3 Watts may not seem like a lot of power, but on a fairly small surface area, such as the tab on a TO-220 transistor (like your MJE13007), even 3 Watts will be enough to burn it out if it doesn't have a heatsink.

This is where you have to read in the datasheets the stated Total Power Dissipation, Pd. Moreover, you need to look whether Pd is stated for case temperature (Tc) or ambient temperature (Ta), and what that temperature is. Typically, most datasheets use 25 degrees Centigrade, which is more or less room temperature. Ta is used to show how much power the device can dissipate *without* a heatsink, whereas Tc states the maximum power the device can dissipate even if it is connected to an "arbitrarily large" heatsink.

Thus, if you don't want to fry/burn out your components, always make sure that you are staying well within their power-handling limits.

For a TO-220 transistor, I personally don't recommend going over 1W of power dissipation if you don't have a heatsink.

Another formula I found VERY useful for a math bum like me is the one for calculating the Ib of a transistor: hFe * Ib = Ic. Since we know hFe (gain) and Ic (required output current) I rewrote it as Ic / hFe = Ib which makes it easier for me.

Actually, you need not worry about that formula at all if your transistor has high enough gain.

For making a simple linear regulator circuit like this, all you need to know is that the voltage at the transistor's Emitter will always be the voltage at the Base minus a diode drop (due to Vbe). In other words, Vc = Vb - Vbe (with all voltages measured with respect to ground).

Thus, if you set Vb to be 9V with a 9V Zener diode, then the Emiter voltage should be approximately 9 - 0.65 = 8.35 Volts.

But again, all of that above is provided that you are supplying enough current to the Base of the transistor. And like you found out with your MJE13007, when the gain is too low and/or you are not proving enough current to the Base, the output voltage at the Emitter will drop.

So that's where a Darlington transistor will indeed come in handy. Pete pretty much nailed it what you need to take into consideration when doing the calculations (that is, Vbe will be equal to approximately two voltage drops, or 1.2 to 1.4 Volts).

So with a Darlington, if you build that same circuit, the output will be 9 - 1.4 = 7.4 Volts. To raise that output voltage, you can add two regular diodes in series with the Zener diode to make the total voltage at base 9V (the Zener's voltage drop) plus 1.4V (two diodes with 0.7V drop in series) equal to about 10.4V.

Finally, if you don't want to buy a Darlington transistor, make your own!
That might sound awfully complicated, but it actually isn't. All you will need, besides your MJE13007, is another smaller transistor (a NPN TO-92 should work) and two resistors. Just look up the datasheet of the TIP120/121/122 or TIP125/126/127 Darlington transistor, and copy the internal schematic of the transistor.

And again, don't forget to heatsink the MJE13007 if you do use it, because it WILL need it.

Again, it's the math that's killing me and I have to write stuff like I was taught in school which was decades ago

Well, there shouldn't be any shame in that. I write stuff down myself all the time.

Just keep going at it, and it will slowly start to make more sense. Really, the best way to learn something is to spend time reading *and* experimenting.

[Dannyx]

Re: Linear regulator question

Another important point (and this was briefly mentioned by redwire and myself): watch those power ratings! If you have 12V going into the circuit and 9V coming out, and the running current is 1A, then your linear regulator circuit will be dropping 12 - 9 = 3V. At 1A, the power the circuit will have to dissipate is
V * I = P .... or 3V * 1A = 3 Watts.

Now 3 Watts may not seem like a lot of power, but on a fairly small surface area, such as the tab on a TO-220 transistor (like your MJE13007), even 3 Watts will be enough to burn it out if it doesn't have a heatsink.

This is where you have to read in the datasheets the stated Total Power Dissipation, Pd. Moreover, you need to look whether Pd is stated for case temperature (Tc) or ambient temperature (Ta), and what that temperature is. Typically, most datasheets use 25 degrees Centigrade, which is more or less room temperature. Ta is used to show how much power the device can dissipate *without* a heatsink, whereas Tc states the maximum power the device can dissipate even if it is connected to an "arbitrarily large" heatsink.

Thus, if you don't want to fry/burn out your components, always make sure that you are staying well within their power-handling limits.

For a TO-220 transistor, I personally don't recommend going over 1W of power dissipation if you don't have a heatsink.


Actually, you need not worry about that formula at all if your transistor has high enough gain.

For making a simple linear regulator circuit like this, all you need to know is that the voltage at the transistor's Emitter will always be the voltage at the Base minus a diode drop (due to Vbe). In other words, Vc = Vb - Vbe (with all voltages measured with respect to ground).

Thus, if you set Vb to be 9V with a 9V Zener diode, then the Emiter voltage should be approximately 9 - 0.65 = 8.35 Volts.

But again, all of that above is provided that you are supplying enough current to the Base of the transistor. And like you found out with your MJE13007, when the gain is too low and/or you are not proving enough current to the Base, the output voltage at the Emitter will drop.

So that's where a Darlington transistor will indeed come in handy. Pete pretty much nailed it what you need to take into consideration when doing the calculations (that is, Vbe will be equal to approximately two voltage drops, or 1.2 to 1.4 Volts).

So with a Darlington, if you build that same circuit, the output will be 9 - 1.4 = 7.4 Volts. To raise that output voltage, you can add two regular diodes in series with the Zener diode to make the total voltage at base 9V (the Zener's voltage drop) plus 1.4V (two diodes with 0.7V drop in series) equal to about 10.4V.

Finally, if you don't want to buy a Darlington transistor, make your own!
That might sound awfully complicated, but it actually isn't. All you will need, besides your MJE13007, is another smaller transistor (a NPN TO-92 should work) and two resistors. Just look up the datasheet of the TIP120/121/122 or TIP125/126/127 Darlington transistor, and copy the internal schematic of the transistor.

And again, don't forget to heatsink the MJE13007 if you do use it, because it WILL need it.


Well, there shouldn't be any shame in that. I write stuff down myself all the time.

Just keep going at it, and it will slowly start to make more sense. Really, the best way to learn something is to spend time reading *and* experimenting.

I scrounged up a 317 from an UPS board I found around the shop and used that instead - not because I gave up on this circuit (I actually got it going as well) but because I needed to save some space inside the router's case. Part of the reason why my initial circuit wasn't working is because even after getting my calculations right, the initial power adapter I chose could only put out 500mA and that meant that even when it was turned all the way up to 12v (it's one of those adjustable chinese things which does 3/4.5/7/9/12v), once hooked up to the circuit, it dropped to 4-5v which of course messed things up big time....the same thing happened with the 317: got it to 9v open circuit, but attach the router to it and it dropped to 4v again. I now switched over to a 12v 1A SMPS and it works with both circuits. The things I've learned here will hopefully come in handy for building that adjustable supply some of you may know I wrote about in another post of mine....the one with the UA741 op amp...some things make a lot more sense now.

[PeteS in CA]

Re: Linear regulator question

Well, now. If you want an adjustable linear P/S, use a TL431 in the pass transistor base circuit. A resistor and a pot give you an adjustable zener with good temperature stability. Just make sure you provide enough current to the TL431 or it will oscillate (BTDTGTTS).

The TL431 is one of those ancient parts that just is too good to go obsolete.

[Dannyx]

Re: Linear regulator question

Well, now. If you want an adjustable linear P/S, use a TL431 in the pass transistor base circuit. A resistor and a pot give you an adjustable zener with good temperature stability. Just make sure you provide enough current to the TL431 or it will oscillate (BTDTGTTS).

The TL431 is one of those ancient parts that just is too good to go obsolete.

So you're saying I should replace the Op-amp idea for the tl431 ? I actually have some of those in my junk drawer Actually no, I think you're saying I should use it for the reference voltage on the non-inverting pin of the op amp.