Voltage Regulator Verification

[ACharnley]

Hi,

Can you verify this is the best design and advise whether a proper voltage reference (LM4040 etc) is the way to go.

Criteria -

1. Input: 4V - 100V
2. Current: 20mA
3. R31 must be small. The design shown R31/47Kis a 1206 resistor, I'd prefer a 0603 but that requires going up to 100K, fine for the darlington configuration but I'm already pushing the zener way below the minimum current threshold.

The circuit works fine at present. The voltage varies a lot but doesn't go beyond ~5v. I'm concerned however that it's relying hugely on zener tolerance.

Cheers,

A.

 

[ronsimpson]

1. Input: 4V - 100V

Really? Input voltage anywhere from 4V to 100V. Output is 5V?

 

[AnalogKid]

1. With a linear regulator, you cannot get a 5 V output with a 4 V input.

2. Over a 4-100 V range, the current through the zener diode will vary by over 20:1. That's a lot if you want any kind of precision output.

3. Why would going to a 0603 size mandate changing to a higher resistance value? 47K is available in all resistor sizes.

4. For a 5.6 V zener, 20 mA means a power dissipation of over 0.1 W. Not dangerous, but I wouldn't run it any higher. Also, with a darlington output buffer there is no reason for such a high static current. I think the LM4040 is good down to 100 uA or something like that. --- Unless you mean 20 mA of output current - ?

With a darlington current amplifier, the output voltage will be approx. 1.2 V below the zener voltage, not 0.6 V. You will need a minimum input of 6.2 V for a 5.0 V output.

ak

 

[Pommie]

Is the input AC?

Mike.

 

[ACharnley]

1/2. Input is DC via the diode. Circuit happy from 3V, but 5V is maximum, thus precision isn't a concern.
3. --> 95^2/47000 = 1206 to handle current dissipation
4. 20mA at the darlington output.

5.6v works for about 4.7v output, likely because the minimum current through the zener is not being met.

 

[Pommie]

I'll ask again, is the input ac 100v or dc between 4 and 100v?

Mike.

 

[ronsimpson]

Input is DC via the diode.

So it is AC above the diode. Half wave rectified. The input voltage is only there part of the time. With out capacitors the output will only be there part of the time.

 

[Nigel Goodwin]

The entire scheme looks dubious and dangerous, not to mention a lack of information.

It's not going to be nice when the transistors go S/C.

 

[ACharnley]

Dubious and dangerous? How did you come to that conclusion, or should I say, assumption?

If you'd asked, I'd have told you the transistors are 160v rated.

 

[alec_t]

Is 100V the RMS AC voltage or the peak voltage?

 

[ACharnley]

Peak, clipped by TVS. It's actually more like 95v, a little less than the diode rating in the schematic.

My focus point is specifically on the zener and whether I'm relying too much on zener manufacturing tolerance or should I play it safe with a LM4040 or similar. There is a 5V one which is a little low but would suffice.

 

[AnalogKid]

National (now TI) has several adjustable voltage references. LM3xxxxxx.

If the load has a known minimum current, you could use a dropping resistor and a 3-terminal regulator like an LM317.

ak

 

[Nigel Goodwin]

Dubious and dangerous? How did you come to that conclusion, or should I say, assumption?

If you'd asked, I'd have told you the transistors are 160v rated.

I'd presumed they were high voltage, but it's a REALLY poor circuit, and pretty well any failure will be catastrophic. And as you're massively over heating the transistors they WILL go S/C in a very short time.

You're really asking a lot wanting an input range of 4-100V, and a crude analogue design like this isn't really the way to go.

 

[ronsimpson]

It looks to me like the Zener current goes from 0 to 2mA. Here is a graph of low voltage Zeners. Note the voltage for current from 1mA to 1uA. Then our output voltage will be two diode drops below that. (2D=1.3V) It is a lot to ask a Zener to do 10:1 in current and hold the voltage. Beyoun 10:1 the Zener voltage is more unpredictable.

I also worry that at 5V_in you don't have enough base current to keep the transistors turned on.

 

[ACharnley]

OK a bit more info which may help Nigel. First, the transistors are rated at 0.2W, well below 2W which would be passed at full voltage. To keep it simple I cut off the rest of this part of the circuit but here it is to explain.

The circuit is a bootstrap circuit, the AC ramps up slowly and may eventually reach a theoretical 95V (say 15 seconds). First it starts an MCU (at ~4v) and the MCU quickly turns on a charging circuit which drives the AC down somewhere below 10V. The stored charge is buck/boosted to 5V and feeds back into the bootstrap output. The voltage is higher than the bootstrap (4.7v with a lot of variance as discussed earlier) so it shouldn't pull on the bootstrap, however there is a fail-safe whereby the MCU shuts the bootstrap off. This all occurs in milliseconds.

The bootstrap is required because initially the main rectification/buck circuitry is disconnected from the AC. This is because once the storage is charged the load on AC would be minimal and would rise beyond 20V, which tends to be the point at which several other components in the system will smoke.

 

[ACharnley]

That's what I needed Ron. So at minimal input voltage I can expect 4 - 1.4 = 2.6v which the LDO is happy with, and the MCU will fire up. The only potential issue then is as you say the zener leakage current at the low voltage point. It's looking like I may need a LM4040 after all?..

 

[ACharnley]

100k is ideal, a smaller resistor and zener (space critical design). LTSpice says it'll work but I don't know how far it models zener characteristics.

 

[ronsimpson]

Part of what bothers me: The 4 to 100V is only there half time. Half wave rectified. With a 0.1uF cap the voltage may collapse for half of the cycle.
I do not see any 100uF cap.

 

[ACharnley]

Sorry that's in the aforementioned buck/boost circuit which feeds/shares into this bootstrap. Currently 1000uF.

 

[Pommie]

At peak voltage your darlington is dissipating close to 2W.

Mike.