Electrical – How to add a capacitor and switch correctly to piezoelectric circuit

I assume AC symbol top right = motor?

Doable but some (not too great) risk. - see below.

A small external 2:1 stepdown transformer is safer.
What is Wattage on rating plate?

Warnings:

• C in series with 100R is a suppression capacitor. If it is not 230 VAC working rated it may tell you so loudly.

  • If the TRIAc is not rated for 230 VAC ... as aboive. It should have a part number on it.

  • The bridge rectifier suggests that the circuit is working with full wave rectified AC. The TRIAC may actually be an SCR.

  • The timing capacitor should never see full mains voltage but being 230 VAC rated would "be wise".

Your diagram is hard on the brain because it is "upside down". The TRIAC control voltage is referenced to the TOP of your circuit (usually = terminal VT1) and not the bottom. It is std practice to draw this reference connection at the bottom as "ground of sorts" :-).
As shown the implication is that the voltage that turrns on the TRIAC is between gate and the bottom of the circuit as shown. This is not the case.

Here is a more conventionally drawn circuit. Capacitor C is charged by R until it reaches a voltage high enough to trigger the DIAC and feed an energy pulse to the gate. The speed is slowed down by taking longer per cycle to charge the capacitor to DIAC firing level. This is achieved by increasing the minimum and maximum values of R. Rmin sets fastest charging time so earliest time in main cycle to trigger so maximum speed. Rmax sets longest time to charge so slowest speed.

Methods / approaches:

The aim throughout is to slow down the time taken to charge the timing cap to DIAC trigger voltage.

(1) It is possible that your Dremel is "designed" to be either 110 VAC or 230VAC and that the 50 ohm resistor across the pot reduces Rmax and Rmin enough so that it operates on 110 VAC. By removing that resistor you MAY increase Rmax enough to slow the motor down enough. But you won't affect Rmin.

(2) A simple method that will have a reasonable chance of working is to double all resistors. It MAY be enough to about double the 620 ohm to 1200 ohm - or bit more or a bit less.

(3) If you have room them doubling the value of the timing cap should work well. Or connect an indentical one electrically in parallel with it.

This circuit may look familiar. Compare to diagram and text above. Ask questions.

Timing cap size:

As noted, doubling the timing cap should APPROXIMATELY halve the effective motor voltages. Very approximate because of the waveform shape / timing issues.

You can try putting misc caps in parallel with existing cap and seeing what happens.

How big is the existing cap?
Cap must be charged to DIAC voltage in a half cycle worst case. Any longer is ineffective.
t = RC.
R seems to be about 620 R + (50R in // with pot)
Those values are sustpect.
Measuring 620R WITH POWER OFF AND PLUG OUT may be useful.

Say Rmax = 1k. T = 1/2 cycle at 60 Hz say = 1/120h second ~= 8 mS.
T = RC.
C = T/R = 0.008/1k = 0.000008 F = 8 uF. Is it an electrolytic? Seems too high but a few experiments with mains rated caps there will show. eg 0.1 uF in parallel should have minimal effect IF C = ~~ 8 uF
If 0.1 uF swamps existing cap and makes Dremel run really slow then try 0.01 and 0.001 etc.

BUT

Iwould be tempted to

• Read rating plate for max power.

  • Use a small isolating transformer.

An auto transformer may do.

Apply mains to a centre tapped 240 VAC winding. Connect Dremel across one half.

or

Take two identical transformers rated at 110:X or 240:X
Connect primaries in series = mains in.
Connect secondaries in parallel with correct phase = inter unit coupling.
Connect Dremel across one primary winding.

Here X V winding can be any voltage as long as transformers are identical.

Either can work correctly if designed properly. If you have a dumb rectifier supply feeding a 7805, then all the rectifier part needs to do is guarantee the minimum input voltage to the 7805 is met.

The problem is that such a power supply only charges up the input cap at the line cycle peaks, then the 7805 will drain it between the peaks. This means the cap needs to be big enough to still supply the minimum 7805 input voltage at the worst case current drain for the maximum time between the peaks.

The advantage of a full wave rectifier is that both the positive and negative peaks are used. This means the cap is charged up twice as often. Since the maximum time since the last peak is less, the cap can be less to support the same maximum current draw. The downside of a full wave rectifier is that it takes 4 diodes instead of 1, and one more diode drop of voltage is lost. Diodes are cheap and small, so most of the time a full wave rectifier makes more sense. Another way to make a full wave rectifier is with a center tapped transformer secondary. The center is connected to ground and there is one diode from each end to the raw positive supply. This full wave rectifies with only one diode drop in the path, but requires a heavier and more expensive transformer.

A advantage of a half wave rectifier is that one side of the AC input can be directly connected to the same ground as the DC output. That doesn't matter when the AC input is a transformer secondary, but it can be a issue if the AC is already ground-referenced.