# What is the purpose of transistor and zener in this circuit?

simulate this circuit – Schematic created using CircuitLab (simulation will not be correct because the diodes are added just by its name.)

Hello, I am a student who is not really familiar with electronics, and these days I start studying several circuits which is being used in our experiments.

I have a circuit we insert between room ground and the circuit with power supplies. I made a diagram from the photo of the circuit that I took.

As far as I know, to avoid too high floating voltage between the circuit with power supplies and the ground, the circuit is to make negative voltage ~ MAX -50V relative to ground. (Zener diode 1N5347B are actually 5 in series, not one. I omitted for the simplicity in the diagram)

I sort of understood by Googling and learning myself that the capacitor and the diode is a safe circuit for sudden voltage peaks (am I right?) But, the center part in the circuit diagram, still no idea after several days study. Only figured out that 1N5347B Zener diode 10V 5W is to endure max -50 V.

Here are my questions..

1. What would this circuit be called, if it has a name?
2. How may the capacitor size 1 uF and resistor size 10 ohms be chosen?
3. Why and in what role are those resistors (cement resistors) and transistor connected additionally to the zeners? I thought to have max -50V, only the 5 zener diodes may be sufficient.
4. Is it sort of isolation circuit or floating regulation circuit?

----------------------added at 20:52 pm, 6th March, 2016 ----------

• Thank you for all who had commented.

• The following figure is the edited schematics. U1 - U3 are some user equipment boxes powered by DC power supplies 1, 2, 3 sharing Vcg with their '-' port. (PSs have '+, -, GND' ports, and'GND' ports in PSs are not connected to its own '-' ports.) So, the right side of Vcg are grounded, not to earth, but to the node Vcg. The circuit in the photo is the left side of Vcg.

simulate this circuit

So far what I learnt about the circuit (left side of the node, Vcg) are,

1. to keep 'Vcg' negative potential relative to earth. (which doesn't really have its own power supply), but up to max ~ -50V.
2. thanks to @transistor, the roles of the capacitor, diode, and shunt regulator.

Here, additional questions came up to me.

1. What role does the circuit itself have when it doesn't have a power supply to itself? would it be just being connected between Vcg and earth, and would start work when Vcg tends to go further negative than -50V?
2. In what case (I guess, when what happens in the right side circuit for operation of equipment), Vcg would tend to go more negative?
3. and let's say when it goes -51V, without any power supplies on the circuit, the current through R2 to turn on Q1 is going to come through the earth?
4. What kind of improvement would I be able to make for this circuit?

(from the previous question) 5. How may C1 1 uF and R2&R3 10 ohms have been chosen? (got to know about R1 to limits the current through D2, thanks to @transistor)

p.s. Am I asking too many questions in one? is putting additional questions not quite the thing?

-------------10:34 am 7th March 2016, updated--------

I think @transistor answer makes sense, and I understood its role. But my schematics may have error still, and the figure in p. 338 doesn't show clearly how the circuit will be closed during the thruster operation, I am trying to give as much information as possible. @transistor mentioned about 'closing circuit', and I added a drawing for that. Generated plasma near anode (inside in U1 in the drawing) accelerate to the vacuum chamber wall which is earth ground, so I expect this is the mechanism closing the circuit. In the drawing, the big square grounded is the vacuum chamber. I hope this help to comprehend the circuit for the additional questions.

• Consider re-drawing... ground at the bottom and the supplies at the top... it makes much more sense that way. Mar 6, 2016 at 8:14
• I've redrawn it with ground at the top since it's a negative voltage regulator. See my answer. Mar 6, 2016 at 10:28
• Show all the diodes, all the circuit connections and all the correct resistor values. Do not skimp on the circuit just because you believe it is unimportant. If power sources are AC then make sure this is clear. Mar 6, 2016 at 10:31
• Thank you all for the comments. I was lack of basic conventions. I will try to be as clear as possible.
– Hwi
Mar 6, 2016 at 10:56
• Thanks for the update, @hwi. The circuit doesn't quite make sense because there is only one connection between the regulator and the loads. There must be a second connection to complete the circuit. What sort of equipment is this connected to? Additional questions are fine. Mar 6, 2016 at 14:35

simulate this circuit – Schematic created using CircuitLab

Figure 1. Redrawn circuit with negative rail at bottom.

I'd say your circuit analysis is pretty good and your explanation of circuit use is a great help. Convention on circuit schematics is that current generally flows from top to bottom and since most of us reads left to right we generally put inputs (logic, analog, power) on the left and outputs on the right. What that in mind, I've redrawn your schematic.

What you have described is a "shunt regulator". If the voltage across the regulator rises above the preset voltage the regulator turns on and "shunts" current away from the power source to hold the voltage in control. They are not all that common because they waste power in the shunt circuit - hence the large heatsink and ceramic resistor in your circuit. Also, the shunt circuit relies on the source power-supply having some internal resistance otherwise a huge current would flow through the shunt as the power supplies try to keep the voltage up and the shunt tries to keep it down.

This circuit is a rather unusual, however, because there is no complete power circuit through the shunt circuit. Based on your Hall thruster link (page 338) your circuit schematic regarding the non-closed loop is correct.

So, if all the above is correct, what's happening is as follows:

• The circuit is designed to maintain the V1, 2 and 3 negative terminals $V_{CLAMP}$ somewhere between 0 V (actually + 0.7 allowing for the diode forward-voltage drop) and -50 V DC.
• C1 is a local bypass capacitor and will shunt any high-frequency noise away from the regulator. It won't have any effect on the DC.
• If $V_{CLAMP}$ rises above GND then D1 will be forward biased and prevent it from rising any futher.
• If $V_{CLAMP}$ falls below -50 V then D2 Zener diode will break down feeding current into the base of Q1. This will allow current to flow through R3 and Q1 to "pull $V_{CLAMP}$ back up" to -50 V.
• If the voltage is between 0 and -50 V the shunt circuit will take no action.

Please let me know if you think this is inconsistent with any other information you have. I know nothing about thrusters.

• This answer illustrates perfectly how important the layout of the schematic is. The circuit was incomprehensible at a glance from the original jumble of wires, but when redrawn, its purpose is immediately clear and recognizable.
– pipe
Mar 6, 2016 at 10:28
• Thank you so much for the redrawn circuit and the comment on the convention on circuit schematics. As many other commented, I was lack of even basics.
– Hwi
Mar 6, 2016 at 10:39
• and for the answer, the explanation was so clear that I could understand your answer well. One thing I think I didn't mention and show clearly caused a bit difference between the circuit I was trying to say and the redrawn circuit. One experimental equipment(not drawn in my original schematics) are powered by '+' side of each power supplies (PSs), sharing '-' of PSs. And '-' sides of PSs are connected to Vout in the redrawn one. I meant the 'room ground' in the original schematics as 'earth' (my fault for unclear wording).
– Hwi
Mar 6, 2016 at 10:54
• I will refer your redrawn and other's comments and try to re-make the clear layout.
– Hwi
Mar 6, 2016 at 10:59
• Thank you. There must be a risk of the whole circuit picking up a charge from the charged particle beams. This circuit prevents that rising too high. Mar 7, 2016 at 7:27