Help Needed SCR Protection Using Free wheeling Diode

Thread Starter

Turbopower

Joined Jun 11, 2022
5
Hi Everyone

I have a high voltage pulse circuit that has a free wheeling diode to protect the SCR

The only problem I have is the diode is wasting energy and lowering the output voltage from the transformer

Would the 1UF capacitor on its own be enough to protect the SCR and remove the diode or should I use a diode and zener snubber ?

The SCR is a BT151-800R

Any advice appreciated
 

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crutschow

Joined Mar 14, 2008
30,100
How is the capacitor being charged and with how much voltage?

The SCR will take 800V in the reverse direction so why do you need the diode?
 

Thread Starter

Turbopower

Joined Jun 11, 2022
5
I've attached diagram to show more details

The diode was added as I thought it was needed to protect the SCR from spikes, but it seems to be having a negative effect on the performance
 

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MisterBill2

Joined Jan 23, 2018
12,399
The capacitor across the transformer (C2) will certainly reduce the output quite a bit.
To protect the SCR from the reverse voltage spike put a diode in series with the SCR, but make sure that the diode is rated for enough current. This will cause the spike voltage to be divided between the two diodes. THAT will provide some protection.
D2in it;s present location will certainly damp any ringing in the transformer, which will be good unless you want the maximum possible voltage out. That is why II suggest the series connection.
 

Thread Starter

Turbopower

Joined Jun 11, 2022
5
Hi

I have tested as above and with C2 removed and the output from the transformer is much lower

With a 1N5408 diode in series with the SCR there is no output from the transformer

Its only when D2 is removed the output is better

I need maximum voltage out
 
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MisterBill2

Joined Jan 23, 2018
12,399
The diode in series with the SCR must be connected to allow current to flow when the SCR conducts. It may need to have a 47K ohm resistor in parallel with the diode, so that the SCR can trigger.
 

kaindub

Joined Oct 28, 2019
97
I think you have failed to realise how an SCR works.
Lets set an initial condition where C5 is fully charged. When the ASCR fires, the current flows through the transformer and the SCR. The SCR only turns off when the current is zero. Since you have a source of power (T1, C1 and D1) its not possible or unlikely that the current through the SCR will fall to zero. So the SCR keeps onducting.
What you do need is a capacitor connected in series with the SCR. This blocks any DC path and hence the current through the SCR will fall to zero in the cycle.
There is no free wheeling current in your circuit as there is no inductor (the inductance of the pulse transformer is negligible. But you will need an RC snubber acros the SCR just to ensure when it cuts off, that the voltage rise across its terminals is slowed so it does not retrigger unexpectedly.
 

Ramussons

Joined May 3, 2013
1,273
Hi Everyone

I have a high voltage pulse circuit that has a free wheeling diode to protect the SCR

The only problem I have is the diode is wasting energy and lowering the output voltage from the transformer

Would the 1UF capacitor on its own be enough to protect the SCR and remove the diode or should I use a diode and zener snubber ?

The SCR is a BT151-800R

Any advice appreciated
That freewheeling diode - Across the SCR - does not seem to play any role .........
 
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MisterBill2

Joined Jan 23, 2018
12,399
Some CDI circuits have a signal that stops the oscillator momentarily when the SCR is triggered. AND just like already stated, the SCR does not switch off until the current drops below the maintaining voltage for the SCR. Thus there is not that second spike to suppress. The inductive spike that happens is the whole purpose of the circuit.

The impedance of the charging circuit portion must be great enough to allow the voltage to drop below the maintaining voltage of the SCR, which means that any ringing of the transformer may be useful in switching it off . So a better location for D2, shown in the original circuit, will be in parallel with C1 . C2 is in the wrong place in of the circuit.
Certainly the TS needs to examine the circuits of some CD ignition systems and understand how they function.
 

MrAl

Joined Jun 17, 2014
9,155
Hi Everyone

I have a high voltage pulse circuit that has a free wheeling diode to protect the SCR

The only problem I have is the diode is wasting energy and lowering the output voltage from the transformer

Would the 1UF capacitor on its own be enough to protect the SCR and remove the diode or should I use a diode and zener snubber ?

The SCR is a BT151-800R

Any advice appreciated

Hello there,

In applications where you have a high voltage spike you dont use a lone capacitor you use a snubber circuit.
A snubber circuit is a simple circuit with a diode, capacitor, and bleeder resistor. The capacitor works the same way you would think that a lone capacitor would work, but the diode controls the charging direction and the bleeder controls the discharge of the capacitor so that the circuit is ready for the next spike that comes along. These circuits are found in converters that use H bridges to protect the transistors form a fast rising spike.

The bleeder components have to be selected based on the energy in the spike which can be hard to calculate without knowing the characteristics of the other parts such as the transformer inductance, so a bit of trial and error is required to find out the best values. But of course first you need a schematic of the snubber. These can be found on the web, but it's nothing more than a capacitor with a diode to charge it and prevent discharging through the main part of the circuit, and a bleeder resistor to make sure the cap discharges enough to be ready for the next cycle.
So in place of the capacitor you added later, you just put a bleeder resistor across it with a diode in series with it such that when the spike appears the diode conducts and when the spike goes away when the diode does not conduct.

A little background on the original circuit and with the added snubber.
When the SCR is turned 'on' the transformer is energized. The inductance will cause some ringing which will turn off the SCR eventually as long as the gate turn on signal is removed. When the SCR turns off the transformer still rings, and because the impedance is a lot higher with the SCR off the current from the primary will generate a huge spike which can exceed the reverse voltage rating of the SCR. With a snubber in place, the spike will be reduced to an acceptable level thus protecting the SCR. As time progresses, the bleeder resistor discharges the capacitor somewhat getting it ready for the next turn on turn off cycle.
 

MisterBill2

Joined Jan 23, 2018
12,399
That ringing mentioned in post #11 is often damped by the load taking energy out of the circuit. We have no hint about the application of the circuit, but I am guessing that the intended load is not presently connected.
And if it happens to be a spark ignition system for an engine, damping the spark is the wrong direction to go.
 

Thread Starter

Turbopower

Joined Jun 11, 2022
5
The circuit is for electric fence, the capaictor is charged and released into the pulse transformer every 1.5 seconds

D2 seems to make the ringing worse and lowers the output voltage
 

MisterBill2

Joined Jan 23, 2018
12,399
For an electric fence charger that ringing is good because it makes the shock sting more while not adding any seriously greater amount of power. So the solution is to use a more highly rated SCR device. The concept does have a lot in common with the CD ignition system, which also uses a capacitor discharge thru an SCR.
 
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MrAl

Joined Jun 17, 2014
9,155
The circuit is for electric fence, the capaictor is charged and released into the pulse transformer every 1.5 seconds

D2 seems to make the ringing worse and lowers the output voltage
The hole idea with these problems is control. you have to be able to control what is happening, how high the voltage is allowed to go. This kind of control comes in different forms depending a lot on the application. For transistors in a bridge snubbers are the answer, but for this project maybe the best way to handle it is to just add a resistive load of the right value resistance and the right power rating in order to keep the peak of the wave lower than the rating of the SCR or whatever solid state device is being used.
Because these problems depend a lot on how much energy is in the pulse, it's hard to nail down the exact peak behavior you often just have to make measurements under controlled conditions and then add whatever you need as you go, usually increasing the bleed off current by lowering the resistor value until you get it to an acceptable level.

To do this you have to be able to monitor the peak and that usually involves an oscilloscope, and possibly a variac or adjustable power supply, and in this case a power resistor of some value with a decent power rating.
What you do is power the electrical input (assuming line voltage AC for now) with a variac, then turn it up to maybe half way, then look at the waveform trying to capture or just observe the peak. You add some resistance to the load then look again, and if it looks low then turn up the variac and repeat lowering the resistance each time. As you get up to full line voltage you should be able to find a value that loads it just enough to prevent the peak from going too high.
OF course this also assumes that your pulse device can accept a variable input ro can be temporarily changed into a mode that allows it.

So it would take some trial and error to get right but you also need the oscilloscope with the right high voltage probe too.
 
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