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enter image description here I made a discrete voltage regulator circuit and constructed it. There input comes from a bridge rectifier fed by a 15V 10A transformer. The output current is limited to 3.3 Amperes. Why is that, how can it be corrected?

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  • \$\begingroup\$ Can you rotate the image and upload again? What was your output design specification? \$\endgroup\$
    – Transistor
    Commented Oct 11, 2020 at 13:59
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    \$\begingroup\$ @ Hearth. To say that a 2N3055 beats another transistor is astonishing ! The MJ13007 is a poor choice, It's designed as a SWITCHING transistor. I never fail to be amazed at the clumsy choice of components by simulators who are then puzzled why their circuit isn't 'real world' capable ! \$\endgroup\$ Commented Oct 12, 2020 at 2:04
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    \$\begingroup\$ ALL Y'ALL - assistance and experience based advice given to the OP so far is much appreciated. A bit less flagellation at this stage would be appreciated. Wisdom and suitably tempered experience based anecdotes once the problem is addressed may be apposite. \$\endgroup\$
    – Russell McMahon
    Commented Oct 12, 2020 at 2:27
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    \$\begingroup\$ Determine the MAX current Q2 is rated to give. Multiply this by the Beta of the MJE13007 in the desired current range and see if Beta xI_Q2_max is anywhere near Iout_desired.' \$\endgroup\$
    – Russell McMahon
    Commented Oct 12, 2020 at 2:29
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    \$\begingroup\$ @arjun looking at the data sheets for all devices you use is mandatory if you are going to design any circuit. Not all switching transistors gave very low beta, but you happen to have chosen an extreme example. It's forte us with very high voltage use. Low beta does not matter as base cct power is low compared to switched power. \$\endgroup\$
    – Russell McMahon
    Commented Oct 12, 2020 at 7:13

3 Answers 3

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Your output transistor is a high-voltage part and has exceptionally low hFE, as low as 5 at 5A.

If the hFE is actually, say, 10, at 5A out, the poor BC547 is being called upon to deliver 0.5A with around 13V across it. That's 7.5W in a TO-92.

Reminds me of a problem I had recently (actually it was metalworking rather than electronics, but the principle is the same) that caused a couple of my devices to stop unlocking from the fingerprint sensor for a month or so.

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  • \$\begingroup\$ May I ask, if it has a very low hFE at 5 Amps, how can it be used in half bridge SMPS? Won't the drive circuit will have to provide large current even for a very little time? Also lets assume BC547 is providing .5A for like 5 seconds before it smokes, how will it create the problem of low current output? \$\endgroup\$ Commented Oct 11, 2020 at 15:53
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    \$\begingroup\$ Yes, the drive circuit in a bipolar output SMPS has to provide large current- that's why the large drive transformer is used. The BC547 can't really provide that kind of current- the hFE of it drops precipitously above about 150mA, so the 2K series resistance will limit your output current. Please search out and link the datasheets for those two transistors (edited into the question, not in a comment). \$\endgroup\$ Commented Oct 11, 2020 at 16:12
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    \$\begingroup\$ This circuit is 'workable' with careful attention to detail as I'm sure you know. All it needs is some refinement, especially with regard to device choice, also maybe operating currents. How have we come to this ? Why are 'kids' today unaware of the issues ? Is this problem 'cultural' in some way ? Are we too afraid of being critical (in a positive way) for fear that we're 'discriminating' ? What is needed of us 'seniors' to correct this ? Over to you. [Edited: RMc] \$\endgroup\$ Commented Oct 12, 2020 at 3:27
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    \$\begingroup\$ The last paragraph sounds entirely unrelated. \$\endgroup\$ Commented Oct 12, 2020 at 14:54
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Overview

Re-drawn in more readable format:

schematic

simulate this circuit – Schematic created using CircuitLab

The whole thing now basically makes sense to me. The way you drew it is more like what I see in a magazine, somewhere.

Also, I notice that you are using an MJE13007. Did you source that power BJT from a failed compact CFL? It's the kind of BJT that may be needed in such devices (and other mains-powered switcher-derived light bulbs.)

I am preferring the D44H11, these days, in TO-220. They have relatively much higher guaranteed \$\beta\$. In your application, I don't think there's any question which of the two is better:

Comparison:

enter image description here

Note that the D44H11 doesn't handle the same high voltages that the MJE13007 can support. But your application doesn't have to handle mains voltages. So the MJE13007 HV capability is wasted on your application. Note also that the D44H11 has a minimum guarantee for \$\beta\$ that is an order of magnitude larger. (It's Early Effect is terrible, but your application doesn't care about that fact.) Also, the D44H11 is at least as capable, if not more so, in handling high collector currents. And in fact, is rated for \$10\:\text{A}\$, continuous, which the MJE13007 is not. (Not that you should be normally running either of these devices right at, or exceeding, their Maximum Ratings.)

Analysis

Before we go anywhere, let's work out a rough estimate for the output voltage.

Assuming about \$V_{D_{_1}}=700\:\text{mV}\$ and another \$V_{\text{BE}_{_3}}=700\:\text{mV}\$, we can find that the output voltage should be around \$\left(V_{D_{_1}}+V_{\text{BE}_{_3}}\right)\cdot\left(1+\frac{R_4}{R_5}\right)\approx 4.5\:\text{V}\$. This suggests about \$\frac{15\:\text{V}\cdot\sqrt{2}-4.5\:\text{V}-1.4\:\text{V}}{1\:\text{k}\Omega+1\:\text{k}\Omega}\approx 7\:\text{mA}\$ for the current in \$R_3\$.

(Also, given the worst case \$\beta=200\$ for the BC547B, this might require as much as \$8.5\:\text{mA}\$ of base current for the BC547B. And we've already calculated that we don't have that much. So yet another possible problem.)

Next, let's look at the following figure from the MJE13007 datasheet:

enter image description here

Here, we can easily see that we need about \$1.3\:\text{A}\$ of base current for a collector current of \$8\:\text{A}\$. Let's assume that works all the way up to a collector current of \$10\:\text{A}\$ (beyond the maximum specifications.) This would imply that we need support for a base current that could be as high as \$1.7\:\text{A}\$.

But the BC547B is specified as having an absolute maximum collector current of \$500\:\text{mA}\$. So we are already well beyond the BC547B maximum capability. So this may already be a problem here.

This is why I'm suggesting that you consider the D44H11, instead. In this case, the BC547B would at best only need to support about \$170\:\text{mA}\$ of base current. And that is well within the specifications for the BC547B.


I see that an answer has been selected already, so at this point I'll hold short of the analysis. It was touched on by Spehro and there's no need to flog this horse, further.

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  • \$\begingroup\$ Thank you for your input. I did not have a "normal" power transistor readily and I was reluctant to go out and buy one at this time of lockdown. So I had to use one I brought for replacing in a ATX Power Supply. I absolutely forgot that such devices won't perform well in linear mode. \$\endgroup\$ Commented Oct 12, 2020 at 5:43
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The first capacitor --- 10,000 uF --- is too small.

That only supports 1 amp drain at 1 volt ripple at 120 Hertz.

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  • \$\begingroup\$ I'm aware of that. But even at 5V output, I can't draw more than 3.3 Amperes. \$\endgroup\$ Commented Oct 11, 2020 at 15:01

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