Reverse transformer?

[Malcolm G6ANZ]

I need to generate a -170 voltage at a lowish current. I have tried running a mains to 6.3v trans former 'backwards' i.e. feeding in 6.3 and using the mains winding as the secondary. This works up to a point. When bridge rectified it gives out about -200 unloaded but this drops to about -80 with the load connected.
Is the poor regulation due to running the transformer backward or should I be looking for faults elsewhere ?
The apparatus is the Sun engine analyser and there are no circuits anywhere on the net
I got a circuit of the PSU with the analyser so I know which voltages I need to generate but can only guess what they do.

Malcolm

[Sideband]

When you say 'lowish current', exactly what is 'lowish'? I would think the current available from the primary (used as a secondary) would probably be just a few tens of milliamps depending of course on the rating of the transformer. What sort of transformer is it?

What value smoothing capacitors have you used?

[GrimJosef]

Depends what you mean by 'lowish' current (microamps ? milliamps ?). With such a large voltage difference a low current on the HV secondary can (obviously) correspond to a pretty substantial one on the LV primary. And small voltage drops in, say, the LV feed cables matter quite a lot when the overall primary voltage is so low anyway. I'd start by measuring the voltage across the 6.3V winding on and off load and seeing if it drops due to losses in the supply circuitry.

The other thing to worry about is the current waveform. If your bridge is feeding an enormous capacitor then the secondary current will consist of a 100Hz train of very short, very high value current spikes. The transformer might be OK for the average current but might saturate before it gets anywhere near being able to deliver these spikes. You could try doing the sums and/or reducing the capacitance and/or splitting the capacitance into two parallel sections (smaller one fed by the rectifier, larger one separated from the smaller by a choke - so-called 'pi filter') to see if any of those approaches helps.

Cheers,

GJ

[PETERg0rsq]

The transformer has a voltage ratio of about 38:1

If the transformer was perfect, then you would require 38X the current you need, feeding it!

To get 240V @ 100ma you need 6.3V @ 3.8 Amps feeding it, assuming the transformer is at least 24VA.

The figures above would also be AC RMS sinewave.

As mentioned the transformer is operating far from perfect, as it would have been optimized for step down operation.

Check to see what power the primary is taking and the power it is delivering to get an idea of its efficiency in the configuration you want to use it.

Also check the primary (6.3VAC) voltage, for a voltage drop, which would indicate the 6.3V supply can not supply enough current.

Peter

[Malcolm G6ANZ]

Thanks all for the replies. The measured current is approx 40mA. but this with the voltage dropping to -80. I think from what has been said 'it aint gonna work'. Therefore I shall hunt down a small transformer to give somewhere near the right voltage, preferably too high so it can be reduced.

Still it was worth a try even as an exercise that failed.

Malcolm

40mA is quite high for this method, there are many 55-0-55 transformers available, should suite your application well.

[Anthony Thomas]

I have just found the circuit diagram for my SUN99 Motortester and I believe that the circuit shares many similar traits to your QDT89 Malcom.

In the power supply there are two "negative" 170 Volt supply points to take into consideration, A and B both minus 170 Volt.

The mains transformer uses the 160-0-160 windings which feed into the positive of two BY127 diodes to supply C136.

A has a pair of reversed BY127 that feed into a 32uf electrolytic (C136) and smoothed by 2K7 0.5W resistor and this feeds to the vertical bias trim pot (R138) in the cathode circuit of V6 which is a 12AX7 Vert output to CRT deflector plates pins 10 and 11.

B is also fed from C136 but smoothed by an 8K2 0.5 W resistor smoothed by C137 (32uf) and then into the cathode circuits of : "V1 (2D21), V2 (6BL8) which is the sweep generator, and Horizontal amp V3 (12AX7) (horizontal bias trim pot (P113). It is also connected to the trigger level trimpot. Valve V3 is connected to plates 7 and 8 on the CRT.

Since the scope part uses + 700 Volt on the horizontal and vertical deflectors this 170 Volts negative is probably needed for to obtain suffient deflection voltages on the CRT.

PM me your post adress Malcom and I will send you a copy.

best wishes
Tony.

[Malcolm G6ANZ]

thank you all for the information. Anthony, a PM is on its way soon

Malcolm

[Anthony Thomas]

Thanks Malcom, if you pm me your terestrial adress its in the post in a heartbeat my scanner isn't cooperating today

[Anthony Thomas]

This my first attempt at loading up a pdf file so hope that it is done correctly.

The schematic has been copied from copies so a bit unclear and taped over.

Perhaps you can print out and tape together to get a better over view, I see the second image is upside down now.

Will post the copies anyway.

Tony

[Malcolm G6ANZ]

Anthony, those PDF's look to be the same as my analyser. The circuit of the PSU area and tube supplies are identical. The rest should be similar enough to get it going.
Many thanks

Malcolm

[Anthony Thomas]

Great, glad that it helped