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I am debugging an ancient little impact printer and in my probing have found an unexpected voltage level. I'd like to understand this part of the circuit, why it's constructed as it is, and whether what I'm looking at is part of what's wrong, or as expected:

enter image description here (Detail from the schematic here)

The printer has a number of little solenoids and one of these driver sections for each one, that switches on the 40VDC rail to the coil as shown. The input from the left is a TTL level signal indicating whether the solenoid should be switched on or off. The buffer shown is a 7407.

The transistor is a TIP127, and the diode is a 15V zener. (Resistor is 2.7K fwiw).

When the printer has power, the 40VDC rail is at about 41V (ok). But when I measure the voltage at the bottom of that resistor, it's about 16V. (If I pull the 7407 out it's still 16V, which I assume is because the '07 was in high-z state.)

My questions are:

  1. What is the purpose of the zener diode in this configuration? My first thought is that it's for some sort of back flow protection of the TTL circuit, but then I don't expect it to do anything (except drop .7V) in normal operation.
  2. Why am I seeing 16V at the bottom of that resistor? I'm not assuming something is "wrong" here but I don't follow how the PNP transistor plus the zener is resulting in this voltage sitting there when the coil is doing nothing-- my sense would be that a TTL level voltage (only) should be turning the transistor on and off as a switch for the coil circuit. Since there are seven of these little compositions, and they all seem to show similar values, I'd be surprised if something's wrong with all of them, but I guess not impossible.

I'm trying to understand here more than I'm trying to debug as such, so thank you for explanations of what this is trying to achieve!

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2 Answers 2

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The 7407 is only rated for 30V, so the zener diode allows the transistor to turn off when the output is high. It can be expected to drop at least 10V when the transistor is 'on', likely much more for safety margin.

The 16V you are reading is an 'off' voltage. Something that that is not unexpected. The exact voltage you read when off is not an accurate indication of the zener voltage because it's dependent on leakages and meter loading.

If that is one of those Selectric conversion boxes, they were kind of a pos, tbh.

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    \$\begingroup\$ 7407 is non-inverting so low input = low output, yes. The ~40V voltage is dropped across the B-E junction (about 0.7V), the zener (say 24V), the output (maybe 0.1V) and the resistor. The resistor controls the current to an appropriate range to turn the transistor on and drive the solenoid. Current would be about 15V/R in this example. When the input is high or open the output goes high-Z so it will rise to about 40V minus the zener voltage at low current. When you put your meter there you are actually drawing a bit of current through the zener and the transistor base turning it on slightly. \$\endgroup\$ Commented Mar 14 at 19:31
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    \$\begingroup\$ Minor design critique: It would be preferred to put a resistor from base to emitter so that leakage in the 7407 doesn't slightly turn the transistor on, this kind of thing is particularly important for robust designs that have to work over a wide temperature range, but it does increase parts count. Something that would drain a few % of the 'on' base current away would be appropriate so maybe the same value as R1~R7 since it has 5% of the voltage across it when on (0.7V vs. 15V). \$\endgroup\$ Commented Mar 14 at 19:36
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    \$\begingroup\$ P.S. voltage you are measuring is not the zener voltage, it is ~40V minus the B-E junction at low current (maybe 300mV) minus the zener voltage at low current (in this case 24V) so you read about 16V. \$\endgroup\$ Commented Mar 14 at 19:38
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    \$\begingroup\$ OK, thank you! I will double check that but I think for my purposes here I have understood what this is doing when it's on and why I was measuring nonzero voltage there when it is off (essentially the meter itself was allowing some current to flow). Thanks \$\endgroup\$
    – BZo
    Commented Mar 14 at 20:10
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    \$\begingroup\$ Just to close the loop on this, you were right: the actual zener part in the unit I have is a 4750A, which looks to be a 27V zener. So your math checks out. Thanks! \$\endgroup\$
    – BZo
    Commented Mar 14 at 22:13
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Check the 7407 datasheet: it is an open-collector buffer with high-voltage output, rated up to 30V.

The supply shown is 40V, however. This would violate the rating.

By connecting a >10V zener in series, the logic-high rating can be respected, while still driving the node solidly low. The logic output can saturate to <1V when low, here putting about 5.5mA through the resistor; in fact, VOL for 16mA is max. 0.4V so we can be sure it's at least this well saturated.

Note a discrepancy: a zener diode only has significant voltage drop when there's some current flowing; while this can be quite small currents (~µA), it's decidedly nonzero. Does this cause the transistor to turn on when it shouldn't? Fortunately, the TIP127 includes base-emitter resistors, ensuring that a small output-high leakage doesn't turn it on.

That said, the IOH maximum of 0.25mA well exceeds the TIP127's input sink capacity of, say, 0.6V / 8k = 75µA; I would rather see an external pull-up resistor here. Presumably, the real IOH doesn't actually get that high, or not at modest temperatures anyway, but that's the limit given in the datasheet and they are allowed to sell you a part with performance that bad; it would be annoying to say the least, to discover a shipment of such outliers in production.

They could've also implemented the level-shifting capability by adding a cascode transistor (right):

schematic

simulate this circuit – Schematic created using CircuitLab

or as an inverting stage (left), both of which receive a logic-level input, no buffering required.

These options may've been overlooked, or ignored for various reasons -- perhaps transistors were more expensive (component plus assembly costs) at the time. They're pretty close in cost these days, and with SMTs being pervasive, either solution would be as compact as it would be effective.

Nowadays, we might also consider an IC for the task: high-side load switches are common for automotive use, with robust capabilities, and often handy features like current sensing. Not that these features would've been necessary for a printer, of course, but the electrical requirements of automotive fuel injectors for example are very similar to those of impact pin/hammer type printers back in the day.

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  • \$\begingroup\$ Thanks so much for taking the time to diagram and offer the alternative design schemes. The printer was DIY customer kit build (!), so maybe in addition to parts cost as you suggest it would have been easier to assemble with fewer transistor parts. Thanks \$\endgroup\$
    – BZo
    Commented Mar 14 at 20:13

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