为什么我的NPN / PNP稳压器实验爆炸？

我一直在研究从较高的电压和电流源获得低压干线的方法，实际上从线性电源（环形，桥式整流器和电解电容）中获得的电压约为53-0-53V。

我天真地以为下面的电路应该在测试负载R3上产生30V的电压，相反，我的齐纳二极管坏了，晶体管Q2产生了很好的爆炸，这有些出乎意料，令人失望。它实际上吹走了中腿，这可怜的东西。

这个想法是让+ 15V和-15V电压轨为一个或两个运算放大器供电。我期望R1，D1和R2分别下降38V，30V和38V，因此，像一对标准串联稳压器一样，Q1的发射极将稳定在15V（相对于假设的0V电压，该电压并不存在），同样Q2的集电极将处于-15V。

我做错了什么？我想知道我是否误解了通过PNP的电流，由于它们的性质相反，它们总是使我烦躁。反正我怎么了

^{模拟此电路 –使用CircuitLab创建的原理图}

更新：

齐纳二极管现在是1N4751A，30 V，8.5 mA，请参见这些规格。现在，齐纳电阻为4K7，齐纳电流约为8.5 mA。

添加电压源后，模拟运行并在齐纳二极管上产生大约+/- 2.54 V，在输出电阻上产生大约+/- 2.1V。

奇怪！要么模拟器不知道齐纳稳压器处于30 V，要么晶体管汲取了大量基极电流，但是使用如此大的负载电阻是不可能的。

您已经有未稳压的直流电源。如您所说，它是由一个桥和一些电容器构成的。显然，您在变压器次级绕组上也有一个中心抽头。所以你也有基础$\pm53V$用您的电表测量另外两个导轨。我假设它可能已卸载，因此加载时您的负载可能会少于该数量。任何人的猜测要少多少，因为这很大程度上取决于负载，您的环形设计，电容器以及其他因素。但肯定会更少。

我搜集到您正在尝试学习如何设计自己的设计 $\pm 15V$供运放使用。因此，您不一定只想购买大量的电源（这些天来它们很便宜。）由于这是关于学习的，因此它将是线性设计，而不是切换器。因此，从电源角度来说，您的电源通常效率低下。但是你很好。

也许我正在计划，但是我认为这是一个好主意。它足够谦虚，您有充分的理由获得成功。但是，有足够的知识来了解它也值得努力。我认为我的第一次学习经验是，我确实学到了很多东西，是在尝试设计自己的电源。那时，我几乎没有选择。一个年轻的少年无法获得现有的实验室用品。而且，也没有便宜的ebay供应商为基于IC的花哨开关提供服务。所以我必须自己做或不做。面对这一点，一个人学习或没有。

您的方法可能有点像从运算放大器到音频放大器的所有设备中使用的信宿/源输出驱动器。您可以采用自己所采用的方法，但是您必须将其中两种方法-一种用于$+15V$ 还有一个 $-15V$。而且它们的效率甚至更低，因为它们每个都可以从您的（+）轨道获取并沉到（-）轨道，因此您需要在AB类中运行它们。您实际上只需要从（+）来源即可$+15V$ 导轨并沉入（-）以使 $-15V$ 轨。

顺便提一句，在桥的输出端将一对泄放电阻包括在现有电容器组中可能是一个好主意。如果您关闭电源，可以摆脱储存的费用。一些$\tfrac{1}{2}W$， $10k\Omega$电阻器？那只会呈现一个$5mA$ 运行时加载。

在考虑该想法时，还应考虑尝试卸载现有的不受管制的电源，以评估其在负载下的性能。我会尝试像$\ge 5W$， $1k\Omega$ 电阻以了解有关 $50mA$负载，在存在该负载的情况下测量电压。然后我会尝试像$\ge 10W$， $270\Omega$ 电阻看看我走近会发生什么 $200mA$加载。这将测试您整个不受管制的系统，并让您了解其局限性。这些值是随机选择的。如果您已经知道了环形线圈的局限性，请尝试使用两个不同的电阻值来达到您希望支持的最大负载，而另一个电阻值可能会达到最大负载的30％。并仅注意测量的电压值。稍微放下时，了解一下不受管制的导轨会很有帮助。

我建议您从专注于一侧开始，例如创建 $+15V$不受约束的（+）导轨上的受约束的供电导轨。您还需要考虑是否也需要任何电流限制。我认为将它们包括在内会更安全。但这是您的决定。不过，为此添加一些内容并不难。而且，就我个人而言，我可能希望能够去$+12V$也一样 那么，可变输出电源可以在一定范围的输出电压范围内工作吗？

您有很大的净空！这意味着您可以使用NPN发射器跟随器，达林顿跟随器或几乎任何您想要的配置。事情并不紧张，因此您有控制结构的空间。很多房间。缺点当然是必须消散，并且电压轨足以使您必须检查数据手册，以确保器件保持在安全工作参数之内。

最后，您可能会接受必须分别设置两个电压轨值。某些电源旨在提供跟踪功能，因此，如果您设置了$+V$ 供应给 $+15V$ 然后你的管教 $-V$ 供应将跟踪并提供 $-15V$。但是我怀疑你现在可以没有那个。

如果您提出一个单独的问题，或者更好地澄清这个问题，我可能会让您开始使用三种或四种不同的离散（非IC）拓扑，以考虑自己进行分析并构建。但是，例如，我不知道您要拥有哪种当前的合规性。而且，这将有助于您知道在将未稳压电源加载到想要支持的最大电流顺应性时要测量的电压（使用高功率电阻，然后花点时间在电压表变得过热之前用电压表测量电压。 ），这将有助于更多地了解您是否确实希望在某个范围（确切地说是在什么范围内）上使用可变电压，并且，如果您只想要固定电压，则您需要多少初始精度？和我' d想知道这是否严格用于运算放大器电源（建议较低的电流兼容性），或者对于某些项目，您是否想使用它在更低的电压下实际提供更高的电流。最后，很高兴知道您拥有或愿意获得的BJT。

编辑：所以。简单的东西，不是只有当前的合规性$5mA$。让我们首先关注（+）导轨端...对于传输晶体管，可以采用NPN或PNP。完全取决于您要如何控制它。您是要从电源吸取电流，还是根据需要抽出电流？嗯 让我们尝试一下-强调简单。

^{模拟此电路 –使用CircuitLab创建的原理图}

我在原理图上写下了一些设计说明。电阻值是标准值，因此实际输出电压会略有下降。但是应该很接近。这是逻辑。

我开始使用 $Q_1$作为发射极跟随器拓扑。它是发射器的目标$15V$。所以我在那写下了“ 15V @ 5mA”。我最初估计是有用的$\beta_{Q1}=50$ 并计算 $I_{B_{Q1}}=100\mu A$ 和估计（仅从内存中） $V_{BE_{Q1}}=750mV$。由此，我决定我想要$5\times$ 来自不受管制的供应，所以我设定 $R_1=\frac{53V-15V-750mV}{500\mu A}=74.5k\Omega \approx 75k\Omega$。这意味着我需要在$400-500\mu A$ 从 $R_1$ 控制 $Q_1$在输出中的行为。这个范围很小$450\mu A\pm 50\mu A$，简单电路中的变化不会太敏感。哦，我选择了BC546$V_{CEO}=65V$。（可以将2N5551用于$V_{CEO}=150V$）

我决定在下方使用另一个NPN，其基极固定在电阻分压器上，以拉出该电流。 $Q_2$的集电极钉有电压，因此没有早期效果。精细。耗散在$Q_2$ 在下面 $10mW$，所以没有问题。（您已经知道可能存在问题$Q_1$。）二极管和电容器可提供半稳定的电压基准，因为它被馈入了相对稳定的电压 $450\mu A\pm50\mu A$当前。我估计$\beta_{Q2}=50$ （再次）并计算 $I_{B_{Q2}}=10\mu A$ 和估计（仅从内存中） $V_{BE_{Q1}}=650mV$。我也知道1N4148可以$550mV$ 在 $500\mu A$当前。所以这告诉我，除法器节点应该在$1.2V$。我也写下来了

我选择至少使分压器电流 $10\times$ 所需的最大基本电流 $Q_2$。该电路的问题之一将是环境温度，因为它们会影响晶体管的基极-发射极结。$Q_2$ （和 $D_1$），这也会影响分频器点以及几乎所有其他内容。但是添加$D_2$ 和 $D_3$在分隔线中的帮助。它提供了两个以上的温度相关结。剩下的问题是$R_3$ 以及不同的电流密度。

$D_2$ 和 $D_3$ 与约 $\tfrac{1}{5}$ 的电流密度 $D_1$ 和 $Q_2$。我碰巧记得1N4148提出了$\Delta V \approx 100mV$ 电流密度每十年变化一次，所以我想 $\Delta V = log10\left(\tfrac{100\mu A}{500\mu A}\right) \approx -70mV$每两个二极管。所以这意味着要达到$1.2V$ 在分隔线处 $R_3=\frac{1.2V - 2\cdot\left(550mV-70mV\right)}{87\mu A}\approx 2.7k\Omega$ （我用了 $87\mu A$ 作为中点当前值。） $R_3$，一个猜测。

我在分压电阻上增加了一个加速帽 $R_2$ 因此短期负载变化可能会更直接地驱动 $Q_2$。（如果$15V$ 受管制的铁路突然跳了起来，然后 $C_3$ 将立即在 $Q_2$ 使它拉走更多的驱动电流 $Q_1$，应对上升趋势。同样，在另一个方向上也是如此。）

我认为，您应该能够调拨（-）调节轨。而且要记住，你不希望加载这个东西下来太多了！您肯定会造成那可怜的TO-92严重问题。消散了$5mA\cdot\left(53V-15V\right)\approx 200mW$ 包裹里有 $\tfrac{200 ^{\circ}K}{W}$，因此可以解决 $+40^{\circ}C$已经超过环境温度 您可以看到如果有更多的电流通过，它会很快变热。您也许可以摆脱$10mA$，但不多。

概述注意：现在您可以看到一个人的过程（其他经验丰富的设计师将比我应用更多的知识），让我们花点时间从一个遥远的角度来看一下。

电路归结为：

1. 传输晶体管（$Q_1$）应该会陷入僵局 $40V$ 在不受限制的（+）轨道和所需的轨道之间 $15V$轨。该传输晶体管将需要基极电流源，以便可以将其保持在其有源区域中。它还布置成发射极跟随器配置，以便左右移动其基极电压使其发射极大约1：1移动（从基极到发射极的电压增益为$\approx 1$）
2. 我们可以使用一个简单的电阻器来解决上述（1）中的所有需求（$R_1$）到不受管制的（+）轨道。这不仅可以提供所需的基极电流，而且还非常容易控制$Q_1$，只需通过它拉或多或少的电流即可。出于设计目的，我们不希望$Q_1$的基极电流会严重影响电流，我们也正在使用它来控制基极的电压 $Q_1$. So we need to make this stream of current large, by comparison. Larger is better, and perhaps by default we might choose a factor of $10\times$. But we are also constrained by the fact that this is a $5mA$ power supply. So we might want to use something that is about $\tfrac{1}{10}$th of $5mA$ to keep it modest. This means something from $10\cdot 100\mu A=1mA$ on the one side to about $\tfrac{5mA}{10}=500\mu A$ on the other side. I decided to use the smaller value, since this is just a simple regulator and I can accept a slightly less stiff base source.
3. Something to control the current being pulled through $R_1$, based upon a voltage comparison of some kind. It turns out that a BJT is okay for something like this. (More BJTs would be better, as in an opamp, but one is sufficient here.) It has a collector current that depends upon the voltage difference between its base and emitter. So it compares its' base and emitter and adjusts a current on that basis! Practically made in heaven for this, yes? So we now stick a new BJT ($Q_2$) with its collector tied up to $R_1$ and the base of $Q_1$.
4. We need a reference voltage. Could use a real reference, like a zener or a more sophisticated IC device, but this is a simple design. Well, a diode with a fixed current density is a voltage reference. (Excepting temperature.) And guess what? We just happen to have a current we can use that is relatively stable! The very current we are using to adjust $Q_1$'s base voltage through $R_1$. So now, $R_1$ provides three services for us -- it provides base current to $Q_1$, allows us to control $Q_1$'s base by adjusting the current through it, and now that very same current can be used to stabilize the voltage of a voltage reference diode. All we do is stick that diode into the emitter of $Q_2$. And add a small capacitor across it o kill high frequency noise there. It's nice when things do multiple duties for you.
5. We have our current control collector, a voltage reference at the emitter, and now all we need to provide is a comparison voltage, derived from the output voltage, at the base of $Q_2$. It's important that if this comparison increases (the output voltage appears to increase for some unknown reason), that we will pull more current through $R_1$ to force the base voltage of $Q_1$ to decline to oppose this change. Turns out that a simple voltage divider does this job well. All we need to do is to make sure that the current through the voltage divider is a lot more than the required base current of $Q_2$, so that when $Q_2$ adjusts its collector current and needs more (or less) base current, that this doesn't affect the divider voltage (much.)

That's really the essence of it. I added those two diodes to help stabilize things vs ambient temps. But they aren't strictly necessary if you don't mind your voltage rails shifting around a little more with temperature. As it is, they may still drift around by maybe $\tfrac{25mV}{^{\circ}C}$, just doing a short loop-around bit of guess-work. But if you don't mind it being twice as bad then you can replace the resistor and two diodes with a simple resistor, instead:

^{simulate this circuit}

The actual value of $R_3$ may need to be adjusted a bit here, as we don't actually know just how much base current is needed (probably less than I guessed -- a lot less.) So perhaps closer to the $12k\Omega$ value? But you can use a potentiometer here, I suppose, to make this adjustable, too.

For one thing, a 2N2222 is only rated for 40 V. The 2907 is good for 60, but that still doesn't leave much margin for things to go wrong, particularly at startup.

I suspect the real problem is that the transistors were wired incorrectly. That could leave a direct path thru Q1, D1, and Q2. Poof!

Added about voltages on the transistors

Even when everything is working perfectly, each half of the circuit sees 53 V. The 1N4730 is a 3.9 V zener diode. That means, when everything is working perfectly, the transistor bases will be held at ±2 V. Even saying the B-E drop of each transistor is only 600 mV, the emitters will be at ±1.4 V. That means each transistor will see 52 V across it when everything is perfect.

Everything is never perfect. How accurate are the ±53 V supplies? What about startup transients? What are the real zener voltages with only half a milliamp thru them? What happens when the load draws some real current, even if only on startup to charge up a capacitor or something?

Did you look up the voltage spec for the transistors you are actually using, not just any datasheet you could find for the generic part number? There are minimum voltage specs somewhere for a 2N2222 and 2N2907, but specific manufacturers sometimes make their parts more capable. You can't use one of those datasheets to tell you the maximum a generic part is good for. To get the numbers I quoted above, I grabbed random datasheets. That means the real specs could be lower than what I quoted.

One transistor is already well out of spec, and the other is close to it. This is not good engineering.

First, Google is your friend. A 1N4730 is a 3.9 volt zener.

That said, I'm inclined to believe that you either miswired your circuit or you used the wrong values of resistors. I'm especially inclined to think that R1 or R2 might have been 100 ohms, rather than 100k. At any rate, your nominal resistor values are large enough to prevent Magic Smoke Emission, so your circuit in some way was different from your schematic.

• IF Vcemax for Q2 is 40V and beyond in secondary breakdown then Ve max is -12V

• Vb for Q2 is 1/2 of Vz (D1=3.9) or -2V approx. this Vbe = -10V while spec is -5V ABSOLUTE MAX.

• due to the catastrophic mode of failure for Vbe reverse ,

• and your careless design,
• only you are responsible for it's middle leg getting blown off, perhaps by construction errors.

This is an easier way of getting +/-15V from your rails:

^{simulate this circuit – Schematic created using CircuitLab}

R1 and R2 allow about 2.5mA to flow to the transistor bases and to the 16V zeners. The voltage at the emitters of the transistors will be about 0.7V less than the zener voltage or about +/-15.3V.

While this is a very simple and reliable circuit, note that it is not short-circuit or overload proof as a 3-terminal regulator would be.

There are a few linear regulators which can operate from your relatively high supply rails but they will not be all that cheap. Do a parametric search on a distributor or supplier web sites to find them. The negative regulator may be more of a problem, especially as your (presumably unregulated) rails might go considerably higher than 53V peak. While you can use the above circuit to drop down the voltage for a 3-terminal regulator you have to consider the worst-case conditions and how much dissipation the transistors will experience.

Reviewers rejected my latest edits to the question, and suggested to create a new answer, so:

Here is the schematic from the OP, completed with voltage sources and more appropriate zener resistors, for the recommended zener current of about 8.5 mA:

^{simulate this circuit – Schematic created using CircuitLab}

And here is the result of the simulation using the Simulate This button:

The zener is now a 1N4751A, 30 V at 8.5 mA, see these specs. Setting the correct part nr does NOT set the related zener voltage, I did that manualy in the circuit diagram editor. The zener resistors are now 4K7 for a zener current of about 8.5 mA.

After adding voltage sources the simulation runs and results in about +/- 15.0 V over the zener and +/- 14.5 V over the output resistor.

Perfect! This circuit seems to do what is expected from it.

As for the blown parts: that must be something like a wrong connection, as suggested by one of the commenters.