Here are some equations for current sense resistor:

$$R_{sense}=\frac{1}{I_{trip}}$$

$$P_{sense}=I_{trip}^2R_{sense}$$

$$V_{max}=\sqrt{R_{sense}P_{sense}}$$

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• 1x dc blocking capacitor(10nF to 10uF) C3(2.2uF Axiel Film Capacitor)
• 2x 100nF Capacitors C1 and C2
• 2x 1k ohm resistors R1 and R2
• 2x 4.7k ohm resistors R3 and R4
• 2x General Purpose NPN low power transistors Q1 and Q2(I used PN2222A)
• A 1.2 to 14v dc power source
• A alligator ground clip

This signal injector can come in handy for repairing radios, amplifers and old TVs and also square wave generator in a probe. Later on I will make a version that uses C1 and C2 as 470nF and R3 and R4 as 1 Megaohm and C3 as 47uF(Since the frequency is ultralow).

The specs

Where R=R2 or R3, C=C1 or C2 and \(C_{3}\)=C3

Frequency: \(f=\frac{1}{1.38RC}=1.5kHz\)

Output AC Resistance: \(X_{C3}=\frac{1}{2\pi*fC_{3}}\)

PCB Files

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Once the parts come in a YouTube video will be uploaded

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The following equation is used for figuring out the unknown capacitor(C1) when balanced. I used yacas to rewrite the equation for the other 3 variables.

$$C_{1}=C_{2}\frac{R_{2}}{R_{1}}$$

The equation will show up once MathJax loads

Notes about this project:

• This circuit requires a high impedance headphone(like ones designed for crystal radios) I tried it with headphones and it was quite loud for it to be balanced. I think this was because that it is sensitive to weaker signals.
• The frequency will need to be high, so it will not work with AC Step down transformers.
• This trick probably work best on a oscilloscope. With a oscilloscope you can use frequencies higher than sound frequencies so. Plus at lower capacitance the low current will not throw off the balance point.
• It is probably best ti use a resistor substitution box instead of a pot. This is because with a single turn pot the resistance is hard to find and if the resistance goes below a certain level it can look like you found the correct resistance even when it is not.

With the bridge being balanced you can see it is really close to zero.

Here it is again but this time unbalanced

For this project we will use reference capacitors values 1nF 10nF 100nF and 1µF. R2 should be high as possible.

Capacitor Symbol icon icon by Icons8

program desauty;
uses sysutils;
var c1,c2,r1,r2,cr,rr,ar:extended;
s:string;
label calculate;
begin
	calculate:
	c1:=0;c2:=0;r1:=0;r2:=0;
	write('Enter C1:');
	readln(s);
	if length(s)>0 then c1:=strtofloat(s);
	write('Enter C2:');
	readln(s);
	if length(s)>0 then c2:=strtofloat(s);
	write('Enter R1:');
	readln(s);
	if length(s)>0 then r1:=strtofloat(s);
	write('Enter R2:');
	readln(s);
	if length(s)>0 then r2:=strtofloat(s);
	if c1=0 then c1:=c2*r2/r1;
	if c2=0 then c2:=(c1*r1)/r2;
	if r1=0 then r1:=(c2*r2)/c1;
	if r2=0 then r2:=(c1*r1)/c2;
	cr:=c2/c1;
	rr:=r1/r2;
	ar:=cr-rr;
	Writeln('C1:',floattostr(c1));
	writeln('C2:',floattostr(c2));
	writeln('R1:',floattostr(r1));
	writeln('R2:',floattostr(r2));
	writeln('CRatio:',floattostr(cr));
	writeln('RRatio:',floattostr(rr));
	Writeln('ARatio:',floattostr(ar));
	if abs(ar)<2 then writeln('Balanced')
	else writeln('Unbalanced');
	write('Press enter to calculate again...');
	readln;
	goto calculate;
end.

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Here’s what you need:

1x 120VAC Output transformer

1x outlet connector

1x power cord

You can add a 1/2A fuse that goes on the output of the transformer

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