2
\$\begingroup\$

I am in the process of adapting an old phone to be used as an Bluetooth phone. I am currently testing the ringer circuit, which I build using the schematics provided by Sparkfun.

I wired up the stuff as follows:

schematic as built

Almost exactly as they have done it, apart from a few minor changes (I use a different H-bridge and only had a 150uH inductor lying around).

For the H-bridge I use the SN75441ONE, and the DC-DC converter is the MC34063.

RING1 and RING2 are connected to my Arduino, and the 1000ohm ringer is from the old phone I have lying around.

Now, when I hook up just the right part of the schematic to a 3.5V source (Vcc), I get a nice 45V on the output (with a steady current draw of about 130mA). But as soon as I connect this to the ringer part, I measure an output voltage of only about 7 volts. Increasing Vcc keeps the '45V'-voltage at about twice the Vcc.

I have to turn up Vcc to about 9V to get the '45V'-voltage to about 18V, and then phone rings if I set RING1 and RING2 alternately to high (alternating at roughly 20Hz). It is not very loud, but it does work.

My question: why do I measure a lower voltage at the 45V output when I connect the ringer circuit? Why does this drop to a lot lower values? I don't want to need to provide 10V to the schematic, for what I know it should run on 3.5V.

(I hope I have provided enough information. Here is a picture of the actual implementation:)

board

\$\endgroup\$

2 Answers 2

5
\$\begingroup\$

If this is your 1000 pF timing cap you will be court-martialled.

enter image description here


Otherwise:

Temporarily short U1 drain to source - it MAY be current limiting for reasons best known to itself. (And/or measure voltage drop across it during operation. Should be close to zero).

Set supply you are using to current limit and short out R3 (0.25 Ohms (0.27 on your board)). This is a current limiting resistor which can cause problems if set wrong. Looks OK though.

Place a small capacitor (0.001 probably OK, maybe 0.01 uF) across R14/47k. This improves transient response and noise response and can help heaps on occasion.

R6 = 180R should be fine if that as they have specified. I do not know what the load is but R6 provides drive current to the internal Darlington second stage. At Vring ~= 3.8V then drive ~~ 3V/180R =~~ 16 mA. I don't know Beta of the 2nd transistor in the Darlington but say = 25 gives Vout max = 400 mA which should be OK. (About 1 Watt ringing).

Check all pins during operation to be sure they are at voltages you'd expect.
1 = 0/Vring/55V approx.
2 = hard ground.

Advise results.

Tell Sparkfun that the world rotates the other way (outputs are meant to be at right of circuit) and that pin numbers on circuit diagrams are obligatory.

\$\endgroup\$
8
  • \$\begingroup\$ let me start with your first suggestion: it is indeed a wrong capacitor, I was under the (shamefully wrong) assumption that 1000 picofarad = 1 micro farad (and I am really ashamed as I just graduated as a master in Aerospace Engineering. What can I say? We usually don't work with such small units?) \$\endgroup\$
    – Arvest
    Sep 24, 2012 at 11:05
  • \$\begingroup\$ @Arvest - I found that last - or I would have not have added the rest. It's about 103.5% certain that that is your problem. When you get back from the court martial tell me if it fixes things. Some of the other material is of possible future use anyway :-). \$\endgroup\$
    – Russell McMahon
    Sep 24, 2012 at 11:20
  • \$\begingroup\$ I very much appreciate the extra information. As a electronics-newbie I hardly get why half of the circuit is why it is as it is. Too bad the shop isn't open here today, I'll post the result tomorrow. \$\endgroup\$
    – Arvest
    Sep 24, 2012 at 11:33
  • \$\begingroup\$ I changed the capacitor to the correct value (1000 pF) and it works perfectly. I get a very loud ring, with a nice 55V at the output of the op-amp. One thing though is that the circuit draws about 350mA at 3.8V, and during ringing this rises up to 440mA. I thought that in steady state, once the capacitors are charged, the current draw should be a lot lower? I measured all the pins, and cannot find any odd voltages. \$\endgroup\$
    – Arvest
    Sep 25, 2012 at 11:06
  • \$\begingroup\$ @Arvest - Ensure all caps are correctly rated for voltages applied and that polarity is correct. Remove 50V from circuitry other than inverter and see if current changes. Lower timing cap to say 220 pF - may osc freq is about 200 kHz. A larger inductor MAY help. \$\endgroup\$
    – Russell McMahon
    Sep 25, 2012 at 13:07
1
\$\begingroup\$

The 130-mA draw at 3.5V (455mW input power) with no load on the regulator other than its own voltage divider (45mW output power) is a clue that it's running very inefficiently.

Looking at your photographs, I see that TC, the timing capacitor, which should be 1000 pF (1 nF), appears to be an electrolytic can, which suggests that it's way too large (1 µF would be 1000× the intended value). This causes the regulator to operate far too slowly, which means that it's wasting a lot of power (probably saturating the inductor) and can't deliver the intended output power.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.