# Emitter-follower as regulator with low voltage difference

I'm trying to figure out emitter-follower voltage regulator for my solar panel. Panel outputs almost 7V on open circuit and 1.2A at short circuit. I want to use it as phone/powerbank charger -> 5V output.

I would like setup like this (but with floating input voltage): Since input voltage will vary anywhere up to 6V (under load) I think standard regulators won'tdo it for me (high voltage drop, doesn't work near 5V input).

I have some question I couldn't find answers to though.

1. In emitter-follower setup is it possible to have input voltage lower than zener diode voltage and still get some output (even if the output would be lower than 5V)? Or is the transistor closed once input reaches zener diode voltage?

2. Is there some other thing I need to consider besides zener diode voltage (which should be 5.6V for my setup) and bipolar transistor Ice (which should be at least 1.5A for my setup)?

3. Is it even feasible? If "yes", what transistor would you recommend me? What should the resistance of R1 be? Does it even matter (100 ohm, 1k or 100k)?

I think I'm a little experienced in electronics but it's still a hobby for me and I learn something new almost everyday.

• Since input voltage will vary anywhere up to 6V (under load) I think standard regulators won'tdo it for me (high voltage drop, doesn't work near 5V input). That's a confusing sentence as you might mean that the voltage can go down to 6 V and you fear that that might not work as the voltage drop to 5 V will then be too low. – Bimpelrekkie Jun 1 '18 at 8:04
• 1. yes, no. 2. dissipation. 3. yes. some cheap to-92 w/ heatsink (but really you should use a DC boost+buck). R1 needs to be enough to provide enough current to the bjt when the zener's vf is subtracted, basically r1=(1.5/hfe)*(vIn-vf), to provide 1.5a to the charger, but that won't really happen, so focus on not burning the zener by dissipating too much. – dandavis Jun 1 '18 at 8:48
• Bimpelrekkie: Actually I think that input voltage will not get above 6V. Solar under heavy load will not probably get higher. In fact I think that it will operate somewhere between 4.5V and 6V at best... What I’m afraid is how the emitter-follower will behave once the input drops to voltage of zener diode or lower - what should I expect at the output in this state. This is what dandavis actually answered in hist comment - 1. It will run at lower output (than 5V) – Pavouk106 Jun 1 '18 at 15:36

Take the OnSemi D44H11 as an example. It's datasheet shows $\beta=120$ at $1.5\:\text{A}$ and at $-40\:^\circ\text{C}$. But that's "typical" at that cold temperature (worst case) and the actual value does vary from part to part. Let's say you decide to use this part and make a reasoned decision that you can expect to see at least $\beta=100$, over part and temperature variations. (You decide this is safe for your needs.) Then this means that you need a base current of $\frac{1.5\:\text{A}}{\beta=100}=15\:\text{mA}$. Of course, if your load needs less, then this figure would be smaller. And if your load requires more, then it would be higher.

The next step is to examine the zener diodes you might use. In this case, you make a reasoned choice for a $5.6\:\text{V}$ zener. Here's the Vishay datasheet on zeners. Here, you see that the 1N4734A is the right designation. It specifies two test currents: $45\:\text{mA}$ and $1\:\text{mA}$. (In general, these ratings are specified for $\frac{1}{4}\:\text{W}$ operation. So if you multiply the larger test current times the rated zener voltage, you should find that this is about $250\:\text{mW}$.)

With $45\:\text{mA}$ through the zener itself, a variation of base current up to $15\:\text{mA}$ might not be too bad to try. You could consider supplying $55\:\text{mA}$ total; with let's say $40\:\text{mA}$ through the zener and another $15\:\text{mA}$ through the base of the D44H11 BJT. If the BJT has a very light load, then at worst you'd be sinking all $55\:\text{mA}$ through the zener. And if the BJT has a heavy load, requiring say up to $20\:\text{mA}$ base current, then the zener might be starved down to $35\:\text{mA}$. That's not too far from the specification, and considering the dynamic resistance this would only mean anything from $5.55\:\text{V}$ to $5.65\:\text{V}$ at the base of the D44H11 BJT. A variation of $100\:\text{mV}$ might be acceptable to you, over load variations.

So let's design this for the D44H11 BJT. At this point, I need to take note of the fact that you have a supply rail that may vary from $6\:\text{V}$ to about $7\:\text{V}$ (unloaded.) (My apologies for earlier not reading this detail.)

Here's the circuit: simulate this circuit – Schematic created using CircuitLab

On the output side I've added a bleeder resistor and an electrolytic capacitor. The bleeder resistor performs two functions: (1) it provides a minimum load on the design to keep it in an active mode; and, (2) it quickly bleeds charge from the output capacitor when the power is turned off. Both functions are useful.

Noting above that $R_1=6.8\:\Omega$, the worst minimum case current through $R_1$ is $\frac{6\:\text{V}-5.65\:\text{V}}{6.8\:\Omega}\approx 51.5\:\text{mA}$. The worst maximum case current through $R_1$ is $\frac{7\:\text{V}-5.55\:\text{V}}{6.8\:\Omega}\approx 210\:\text{mA}$.

This range of currents is a problem because it means the zener diode has too wide of a potential variation. Recalling from the zener datasheet, there is about $5\:\Omega$ impedance at $45\:\text{mA}$. Assuming this holds, this means the zener would range something over $800\:\text{mV}$ of variation, which I'm sure isn't acceptable.

Lowering $R_1$'s resistance would perhaps make the variation less, but at the expense of an increased output voltage and very excessive power dissipation in the zener. Raising $R_1$'s resistance would make the variation worse and might starve the BJT. Trapped between a rock and a hard place, so to speak. This is due to the fact that you do NOT have much "regulation headroom" here. You only have as little as about $400\:\text{mV}$, but this can vary to over $1.4\:\text{V}$, too. So a factor of 3.5X!!

Dealing with that kind of voltage variation with a simple resistor alone is what's causing the huge variation and making it difficult to manage the situation. You'll need to find another way.

Let's say you want to keep this a linear (not a switcher) design. The hard part here is the voltage headroom, as just mentioned. But there is a way to do this. By allowing one of the BJTs of a current mirror to operate in saturated mode. It's harder still, because I want to deal with both $V_\text{BE}$ variation (most important of the two) and $\beta$ variation.

Here's the schematic: simulate this circuit

That should work. The smaller saturated BJT used in the current mirror will not exceed about $100\:\text{mW}$. But it may vary from $10\:\text{mW}$ to $100\:\text{mW}$ over the solar powered voltage range. The other smaller active mode BJT used in the current mirror will be about a constant $50\:\text{mW}$ regardless of the solar powered voltage range. I've added some resistors to deal with the differential vagaries of the temperature variations (as much as $\pm 10\:^\circ\text{C}$ differential) between the parts and part variations of $V_\text{BE}$ and $\beta$ (equivalent to another $\pm 7\:^\circ\text{C}$ variation in temperature.) Ambient temperature variations are "common mode" so to speak, so the circuit should be fine on that point.

The details on the mirror resistor values are a little more complex and are also a matter of debatable judgment. Different designers may arrive at somewhat different values. (Not a lot different, though.) I'll avoid the long discussion and hope you can simply accept that there was some thinking behind their choice.

Note that it takes some extra work to deal with the 3.5X variation in headroom. But at least it's possible to manage the situation. The upshot is that you cannot really use your schematic (or the one I started with, at the outset.) You have to take an extra step as shown above to put the circuit into reasonable management.

• This answer doesn’t account that I will have input voltage of maximum 6V and that means I will be dropping much lower energy which should be pretty much ok even with small heatsink. BUT I have learned a lot from it and now I know how to set up the whole thing. – Pavouk106 Jun 1 '18 at 15:42
• @Pavouk106 Sorry I missed that detail. I must not have been reading well. My apologies. Your lower voltage will be a problem, though. – jonk Jun 1 '18 at 16:06
• I didn’t want to sound rude. In fact you example showed what could ve problem when dropping from higher voltage.What do you mean that my lower voltage will be a problem? – Pavouk106 Jun 1 '18 at 17:12
• @Pavouk106 Is it a regulated 6V? I get it that it comes from a solar panel. But what's the range here? Can it be as high as 8? Or 9? or... How low can it be, as well, before you'll accept a failure of operation? (It was my failure to read -- I looked only at the diagram and failed to read your text -- so that's entirely my fault and none of yours. So you weren't being rude in any way.) – jonk Jun 1 '18 at 17:46
• @G36 I bet you would. I didn't want to write it up here as it is a matter of trade-offs, which means I'd have to write even more. For now, let's just say that I selected about $100\:\text{mV}$ for the voltage drop across the emitter resistors based upon my estimate of $\pm 35\:\text{mV}$ variation between the two BJTs taking into account all variational vagaries. The base resistor itself needs to be about twice that, plus about another 2 times little-re or so. Then you have to go back and rework boundaries to make sure you didn't fall outside the box. I did that. – jonk Jun 1 '18 at 21:14