Strange transistor driver circuit

"Philip Pemberton" wrote in message news: snipped-for-privacy@dsl.pipex.com...

Hi.

Those transistors are not Darlington connected. (But I have never heard of a "complementary Darlington", so maybe you're right.)

I suspect a current setting or limiting resistor is missing somewhere. Either the emitter or collector of the PNP must have a resistor in series, otherwise the NPN base current is poorly determined and likely too high.

If the circuit was really as you've drawn it, then there would be a contest between the zener being over-driven and the NPN base being over-driven. There must be a another resistor.

I think the zener orientation you've drawn is correct. I'm not sure why the designer chooses to dissipate the stored energy from the solenoid in the transistor rather than simply grounding the zener and dissipating it there. Is it a small zener and a larger NPN? Maybe the NPN had to be large anyway to carry the necessary current, and that connection allowed a smaller zener to be used.

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Larry Brasfield
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"Philip Pemberton" wrote in message news: snipped-for-privacy@dsl.pipex.com...

Thanks for that bit of education. I've used that configuration without knowing it had a name. One nit is left to pick here, however. The pair would have to form a three terminal cutset to be properly named a "complementary Darlington" or "Sziklai pair". Your circuit's pair still has four distinct terminals.

Strange. I don't see how the current sharing is effected, unless the NPNs are expected to split the current equally enough by virtue of limited mismatch or their own series resistance. And if they share the resistor, they must be connected together.

It looks like a smaller zener was the motivation. I wonder how the NPN bases that share the 150 Ohm limiting resistor manage to split the feedback thru the zeners during solenoid shutoff. That alone makes me suspect the sharing is not quite as claimed.

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Larry Brasfield

"Philip Pemberton" wrote in message news: snipped-for-privacy@dsl.pipex.com...

No, the NPN collector is regulated, via feedback thru the zener, at the zener breakdown voltage.

That method causes decay of the solenoid field to occur much more slowly. For many print head applications, the additional decay time would hinder performance.

It's just a safe operating area consideration.

Interesting that the designer chose a faster driver than was recommended in the datasheet. Either faster action was sought, or the datasheet is a bit misleading.

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--Larry Brasfield
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Reply to
Larry Brasfield

Probably thinking of Sziklai connection, which is a bit different. The base current in this case comes from the supply rather than through the load, which means it uses more current from the supply and (potentially) has a far lower "on" voltage.

Definitely. A resistor is required.

The zener connection is not uncommon, especially in chips. The SOA of the BJT has to be about good enough for this service to safely switch the inductive load in the first place, so it's trading off a small

3-cent glass zener diode against a larger power zener.

Best regards, Spehro Pefhany

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Reply to
Spehro Pefhany

Ha, I had a feeling that was coming. ;-)

Sure, and the specs can be tighter on some high-volume parts- for example the ubiquitous 2SA1015(Y) part has hFE in the range of 120-240 (there are 3 beta bins for the PNP part and 4 for the NPN complement).

So, with 100uA base drive, you might expect 12-24mA at the base of driver (18mA +/-33%), which is not too bad. Of course that's at 25°C, it will drop at lower temperatures, just when the output drive transistor needs more juice, and increase with increasing temperature, just when you don't need it to drive the output transistor, and when the extra heating is undesired. So, maybe 10-40mA over a wide temperature range and from unit-to-unit, sloppy, but not crazy bad if you can afford to burn the extra current. IC designers had to deal with *resistors* just about that bad in the seventies, IIRC.

Best regards, Spehro Pefhany

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Reply to
Spehro Pefhany

Hello Philip,

The first transistor is unlikely able to pass enough current to blow the zener. The 4.7k limits base current and beta will most likely be below

300. Anyways, the 2nd transistor's base won't go much above 700mV and when that is the case its collector will reach saturation voltage. Depending on the size of that transistor that could be around 150mV or so. IOW there will be a bit less than a diode drop cross the zener while it is "in reverse".

It's a bit unusual but probably the goal was to run the 2nd transistor a bit into conduction when the voltage at its collector wants to rise above the zener voltage upon switch-off. IOW, to protect the 2nd transistor from exceeding that voltage. This is unusual because a strong solenoid can generate enough of a spike to fry the base region or the zener, or both. But maybe it's a small solenoid in your case.

Regards, Joerg

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Reply to
Joerg

Hi, I've just been reverse engineering a printer controller PCB and I've come across a rather unusual (to me anyway) transistor driver circuit in the printhead solenoid driver. Forgive my naff ASCII art, but here's the schematic: +V +V --- --- | | | 3 Solenoid | 3 e | 3 b | / PNP +--|>Z---*

-----[ 4k7 ]------|< | | c

| \\ | b | / +-----*-----|< NPN | \\ | e | --- GND

To me, that looks like a pretty normal complementary Darlington circuit. What I'm most interested in is the 20V Zener diode on the NPN - ZD.cathode to NPN.collector, ZD.anode to NPN.base. AFAICT, when the PNP switches on, the ZD will get put straight over the solenoid, with the anode more positive than the cathode, causing current to flow A-K to ground. Surely that would blow the Zener to bits (it's a BZX84C20 - 300mW SOT23)? If it was oriented cathode to +V, anode to NPN.collector, I'd have guessed it would have worked as a rather expensive spark-killer/back-EMF suppressor. In this configuration I'm not so sure...

Thanks.

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Reply to
Philip Pemberton

...

True. My supposition was that the NPN would be driven hard. As others have pointed out, the current from the PNP is limited enough that such concern is unwarranted. Of course, with the OP's later statement that there is a limiting resistor, the concern is irrelevant.

That's more or less implied in what I stated.

Maybe things have changed, but I once learned, (very expensively), to not avalanche MOSFETs. So I agree.

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Reply to
Larry Brasfield

ISTR it's also known as a "Sziklai" pair. You use an NPN and a PNP (or vice versa) instead of a pair of NPNs or a pair of PNPs. In this case, the pair acts like a PNP. There's another variant that acts like an NPN (like Darlingtons - you can have one made of PNPs that acts like a PNP, or one made of NPNs that acts like an NPN).

[scans board again] Yep. A 150-ohm 1206 SMD between the PNP's collector and +V. The solenoid is limited only by its own resistance. The 150R is shared between six solenoid drivers, but only two can ever be energised at once.

A huge NPN - an FZT649 (3A). The PNP is a BCW70.

Later.

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Reply to
Philip Pemberton

So it makes the transistor ground the coil when the back-EMF hits it? Ick. I'd have used a 1N4001 over the solenoid, but that's just me.

It's a printhead solenoid in a dot-matrix impact printer. Not a very big printer either - an Epson M-183 42-column receipt printer. Epson's recommended circuit (from the datasheet) is a Darlington NPN pair with a spark-killer diode over the solenoid (K to +V, A to drive transistor collector).

Later.

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Reply to
Philip Pemberton

Hello Spehro,

I have seen "designs" where they relied on the beta to be in a certain range to save a resistor. Mostly in disposable stuff though.

Regards, Joerg

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Reply to
Joerg

The datasheet is in some form of Engrish/Jengrish hybrid. Basically, English with the sentence structure of Japanese. Great fun to decipher. I looked at a Seiko thermal printer datasheet and the Epson M-183 datasheet. The Seiko datasheet is written in nice, clear, understandable English; the Epson datasheet makes you try and guess what you need to do to make the damn thing work.

I've looked up the solenoid driver ckt - it uses a 1S2075K small-signal diode (DO35 pacakge). I haven't run it through Quickcross yet, but going by the datasheet, it looks like the 2075 is similar to the 1N4148.

The tacho-generator circuit in the datasheet is even stranger. Two transistors, two resistors, two diodes, a capacitor and the tach coil. The one on the board I reverse-engineered is similar, only with one less transistor and one less diode.

Later.

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Reply to
Philip Pemberton

Actually, for most modern MOSFET designs (made in the last 15 years), that's not true.

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Reply to
Winfield Hill

won't the zener in series with the NPNs Vce be in parallel with the NPNs Vbe, i.e. The zener will never be on in that direction unless the NPN is driven really hard?

Its needs to dissipate the power somewhere, its either that or the zener...

On lots protected mosfets it's done with two back to back zeners, the mosfet can handle more energy "self clamped" like that than just letting it go in to avalance.

-Lasse

Reply to
Lasse Langwadt Christensen

[quoting somebody else]

Are you stating that most modern MOSFETs can handle the same pulse power in avalanche as when biased to keep vds just below breakdown? I don't recall seeing this in the datasheets of the MOSFETs I've used recently. I would love to see some data on this because it is often otherwise necessary to use higher voltage parts just to preclude the possibility of brief breakdown.

If you are questioning my learning experience, that was when small power MOSFETs were $100 a pop and many scopes too slow to fully capture the turn-off spikes made possible by their switching speed. (1980 or so)

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Larry Brasfield

The primary purpose is most certainly fast turn-off of the solenoid and the C-B connection is a zener power rating multiplying effect in that the zener peak current is Isolenoid/beta because the transistor is out of saturation whenever Vce=Vzener. Note that the solenoid inductance is not linear and a maximum when pulled in, so that turn-on is confronted with minimum inductance and initially small L/R time constant, whereas the turn-off is confronted with maximum inductance and initially large L/R. The Vz-V+ reverse turn-off drive compensates for that asymmetry. A further advantage is elimination of the abrupt di/dt transient on the supply line because the V+ to GND current varies continuously and is smoothed by the solenoid inductance at turn-on and turn-off.

Reply to
Fred Bloggs

Not only that, but the Zener doubles as an anti-saturation diode by absorbing the excess base current from the sloppily driven first transistor, where the designer relies on current gain to "get enough" current. Thus no need for base resistor on 2nd device.

All-in-All quite clever and saves some USD 0.0001's worth of components per printer;

Reply to
Frithiof Andreas Jensen
[snip]

And VCEsat=??

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
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Jim Thompson

[snip]

Just making sure you really understood the drops. Of course these configurations are butt-slow on turn-off.

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
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Reply to
Jim Thompson

Barfington ?:-)

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.
Reply to
Jim Thompson

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