I'm getting ~ 12VAC pretty stable without load with this circuit

However when I try to convert this signal into 6VDC up to 500mA or 1A with AC to DC conversion and a regulator; As seen here

I get huge voltage drops on Timer circuit. AC voltage oscillates around 7 8 Volts, converted DC voltage going into regulator drops down to 1 V. So I don't expect to get anything from output of the regulator. Any thoughts or explanations would help me a lot since I just started my career and willing to learn more and more! Thanks!

  • \$\begingroup\$ Just for clarification, are you trying to pull anything near an amp through \$R_9\$? (I'm assuming that the output of the top circuit is feeding the input of the bottom one. Maybe I'm just not understanding you, though.) \$\endgroup\$ – jonk Mar 17 '17 at 19:26
  • \$\begingroup\$ @jonk yes that assumption was totally correct. \$\endgroup\$ – Alian4life Mar 17 '17 at 19:31
  • \$\begingroup\$ Have you considered the voltage drop across \$R_9\$, then? It seems almost surprising to me that you'd see even one volt left over. Probably the diode, \$D_1\$ in the second circuit, is the reason you see anything at all. \$\endgroup\$ – jonk Mar 17 '17 at 19:35
  • \$\begingroup\$ @jonk Actually I tested the regulator circuit with 12VAC 1kHz from a reliable source and I got 6V and 1A on 6R load perfectly. But when I try using the timer circuit above, which supplies the same 12VAC 1kHz(without load), voltage drops occur and screw up everything. D1 diode is just blocking negative half cycle, I can't get rid of it, can I? \$\endgroup\$ – Alian4life Mar 17 '17 at 19:44
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    \$\begingroup\$ If you attempt to draw 0.5 Amp through R9, you will have a 165 volt drop across that resistor!! You can't get more than 33 mA through R9, even if you ground the output end of R9. \$\endgroup\$ – Peter Bennett Mar 17 '17 at 19:53

I tested the regulator circuit with 12VAC 1kHz from a reliable source and I got 6V and 1A on 6R load perfectly.

I believe you. If your reliable source happens to have sufficient current compliance then I don't see why it wouldn't work.

But when I try using the timer circuit above, which supplies the same 12VAC 1kHz(without load), voltage drops occur and screw up everything.

Yeah. Because of \$R_9\$ in your timer circuit. As long as you are not pulling any current through \$R_9\$, it won't drop any voltage and you will see an unloaded output voltage. The problem comes in trying to supply a load through \$R_9\$.

D1 diode is just blocking negative half cycle, I can't get rid of it, can I?

No, you need it or something to provide that kind of service for you. I wasn't suggesting that you remove it. I was trying to guess why you see anything at all when you load your \$6\:\textrm{V}\$ output with anything much.

Let's back off for a moment and focus on your LM317 regulator circuit. Without the intended load of hundreds of milliamps, and just as you show the circuit itself without a load present, you are already loading the input source to it. You are, because you have a voltage-setting resistor pair present: \$R_1\$ and \$R_2\$. The current in these is:

$$I=\frac{1.25\:\textrm{V}}{180\:\Omega}\approx 7\:\textrm{mA}$$

This is an average current load that exists without any external load on your regulator. And it must be supplied by your input source supply to it. This comes from your capacitor, \$C_1\$. Let's assume for a moment that you need about \$3\:\textrm{V}\$ above the output voltage in order to operate the LM317 properly, for now. This means the capacitor minimum voltage should be about \$9\:\textrm{V}\$. Let's work out the impact on your top circuit.

Since we know the average current required just to maintain the output voltage of the LM317 and have an estimate of the input voltage needed, as well, we can compute the energy (U is sometimes used, so is W, depending on various conventions in use and the field -- I'll use U here) per cycle (at \$1\:\textrm{kHz}\$) is:

$$U = \frac{9\:\textrm{V}\:\cdot\: 7\:\textrm{mA}}{1\:\textrm{kHz}}=63\:\mu\textrm{J}$$

This energy comes from the capacitor and must be replaced each cycle. We can work out the voltage difference needed on the capacitor in the following way:

$$\begin{align*} U &=\frac{1}{2}\: C\: V^2\\\\ \textrm{d} U &= C\: V\:\textrm{d}V\\\\ \textrm{d}V&=\frac{\textrm{d} U}{C\: V}=\frac{63\:\mu\textrm{J}}{470\:\mu\textrm{F}\:\cdot\:9\:\textrm{V}}\approx 15\:\textrm{mV} \end{align*}$$

Now we need to compute the current required to achieve that change. Taking into account the fact that \$D_1\$ limits the period, even with a square wave input [and would be much less time with a sine wave], to less than \$500\:\mu\textrm{s}\$, this is just:

$$\begin{align*} I_C &= C\:\cdot\frac{\textrm{d} V}{\textrm{d} t}\gt 470\:\mu\textrm{F}\:\cdot\frac{15\:\textrm{mV}}{500\:\mu\textrm{s}}\approx 15\:\textrm{mA} \end{align*}$$

Let's round this up to about \$20\:\textrm{mA}\$. Across \$R_9\$ this leads to a voltage drop of \$330\:\Omega\:\cdot\: 20\:\textrm{mA}=6.6\:\textrm{V}\$! The diode drop across \$D_1\$ would mean that this is more than \$7\:\textrm{V}\$. And if the capacitor were really able to be at \$9\:\textrm{V}\$ (which it obviously could not be), then the diode would be forced into reverse bias (which assumes that it was forward biased in order to allow the current in the first place.)

In short, it's not going to work. Not even to just maintain the output voltage without an external load. Let alone trying to do that with an external load.

  • \$\begingroup\$ Now I got the point, thanks a lot for this amazing explanation! \$\endgroup\$ – Alian4life Mar 17 '17 at 22:33

As jonk says, R9 is probably your main issue, just quickly thinking about ohms law 1A across 330 ohms would require a 115 volt drop. Although the amperage required out of the top circuit and into the bottom one may be less than 1A because of stepping down and AC-DC conversion with a 100% efficient AC-DC conversion, you're "wasting" half your cycle since the square wave only supplies energy while high, so 1A is probably a better estimate for your AC draw. This rough estimate is clearly several orders of magnitude beyond what it should be. I would recommend replacing R9 with a short and having Q3 and D1 the only place where voltage may be dropped between 12 volt supply and input to the regulator.

I don't know if the parts numbers in the diagram are accurate as to what you are using for the circuit, but there are a couple of issues with those too that you may experience, make sure you look over all their maximum ratings (mainly maximum current capabilities) to make sure they can supply the power you need.

Looking at the transistors you specified, they only have an absolute maximum current rating of 100mA, which would definitely not be good enough for the circuit. I would go with something closer to a 2A rating for Q3 just to be on the safe side. Q1 you can completely do without (although it does no harm to your circuit by actually having it there), it will only act to sink current which can never occur since current cannot flow backwards into your output because of diode D1. I would recommend simply taking it out.

The LM317 requires at least a 3 volt drop between the input and the output to function properly, so you will require at least 9V at the input of the LM317 after both the voltage drop over the driving transistors and and over D1. If you continue to have issues check to see if maybe you have less than the required voltage at your regulator input. If so, you may need to get a "Low drop out" regulator, which have much better minimum input to output voltage drop requirements, generally less than a volt. This circuit even with well picked components may still have around a 2V drop from the supply to the input to the regulator (over the diode and Q3), so it is likely you will end up having to go with a low drop out regulator.

I would get rid of R7 as Q3 likely requires a fair amount of current into it's base to drive 1A from it's output. Although it depends on the current gain of the transistor, generally I like to assume that the transistor will want about a 20th of it's output current driving it through it's base (most transistors generally have a current gain of 50 at absolute worst so this is generally quite a conservative estimate, although I recommend checking the data sheets of whatever transistors you use just to check it out). If you kept R7, using the assumption of requiring a 20th of the output current to drive the base the op-amp would need to be outputting around 50mA over R7 which would require at a 34 volt drop over it (according to ohms law).

Make sure the op-amp you're using can supply this 50mA, if not you may need to replace your single NPN follower output with a darlington pair NPN follower (essentially 2 NPNs in series, with a current gain equal to the product of their gains), just to guarantee that the current out of the op-amp can in fact drive the transistors to supply 1A.

Lastly, Double check as well that D1 can also survive at 1A, I again recommend getting a component with a voltage rating closer to 2A just to be on the safe side.

  • \$\begingroup\$ I'm not even sure why the upper circuit exists. Given there is already a \$+12\:\textrm{V}\$ rail. The LM317 can be directly driven from that. The upper circuit is a disaster of design, agreed. But what's the point of it, anyway, assuming it was done right? (It requires a negative rail, as well, which is just that much more to worry about.) I held short discussing how to repair the upper design, because that is like helping redesign a Rube Goldberg digging contraction added to the end of a shovel, when the shovel by itself would serve better still. \$\endgroup\$ – jonk Mar 17 '17 at 21:20

Here is the equivalent circuit of the output stage of your AC generator:-


simulate this circuit – Schematic created using CircuitLab

Consider what happens when the output is positive (SW1 - representing the output transistor inside the LM311 comparator - is off). Q2 receives Base current through R2. The BC547 ('A' version) has a typical current gain of 150, so if the load draws 30mA the Base of Q2 needs 30mA/150 = 200uA. R2 must then drop 200uA * 10k = 2V, and the Base-Emitter junction drops another ~0.6V, resulting in 12V - 2.6V = 9.4V at the Emitter.

So far it's not too bad - of the 12V supplied you have 9.4V left. But the current still has to go through R3. 30mA * 330Ω = 9.9V dropped across R3. But at 30mA only 9.4V is available, so the output voltage will collapse to zero before you can even get 30mA out of it!

The situation is better when the output is driven negative (SW1 on) because R1 can deliver ~10 times more current to the Base of Q3. If it wasn't for R3 you could get over 100mA at 12V during negative half cycles. However with R3 included the voltage still collapses under load (and since your rectifier only uses positive half cycles anyway...).

This circuit could barely power an LM317 with no load, and has no chance of delivering 0.5~1A. You could increase the generator's maximum output current by making R1, R2, and R3 much smaller (eg. 1K, 68Ω, and 33Ω) but to get 0.5~1A you would need higher current transistors and a better driver circuit.


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