0
\$\begingroup\$

Recently, I have a project that needs to heat up a metal filament with a single pulse voltage. I built a circuit with 555 timer and transistors to heat up the metal filament(0.2 ohm). The circuit diagram is as below:

schematic

simulate this circuit – Schematic created using CircuitLab

The circuit works pretty fine when R1C1=3s; 12V with 1.1A going through the load. However, when the time begins decreasing to 0.01s, the current reads from the output voltage source is very low (300 mA at 20V) and could not heat up the filament at all.

Normally, when directly connecting to voltage supply sources, the filament would light up at 2V with 1.5A. I wonder the way to light up the filament in such low pulse second. Could anyone help me with this? how to design?

Update】 Thanks for comments. I forgot to put common ground. The D1 and R2, R3 is for trigger purposes. I click a button to trigger that circuit, but sometimes trigger time would longer than pulse time. So I put those components there.

I tried to build this circuit with one BJT before, but it encountered the same problem. It worked pretty well ~3s but it could not light up the filament in just~ms. That's why I tried to put another BJT to increase current(maybe it's a bad idea).

Update】 Thanks for answer. Right now the filament could light up via the circuit below.

schematic

simulate this circuit

Still, there are some tiny detailed problems I encountered. First, the light intensity which comes out from the tungsten filament is not always the same. At mono-pulse 20 ms mode, it originally could not light up at 12V. But after pulse it again and again(small time gap(~1s) between each pulse), it becomes lighter and lighter until saturation. But after waiting a longer time gap, it returned back to its initial state which is darker or even could not light up. Though pulsing two to three pulse voltage(20 ms), it could finally light up again with saturation intensity. I am not sure why this happened. Maybe some parasitic inductor? Could anyone comment on this problem?

\$\endgroup\$
  • 1
    \$\begingroup\$ Are those the real components? I can't see how it will work with D2 being a 1N4148. What is D2 for? Have you tried putting the load where D2 is and connecting Q5 emitter to ground? Q4 probably isn't necessary. D3 and R3 probably are redundant too. Your two supplies also need a common ground. Your reset pin needs connecting to the supply too. \$\endgroup\$ – HandyHowie Sep 24 at 6:54
  • 1
    \$\begingroup\$ Ground connection missing on the 12V supply? \$\endgroup\$ – winny Sep 24 at 7:40
  • 1
    \$\begingroup\$ You should fix the details of your schematic like mentioned in the comments. If the circuit was really like shown the 1N4148 diode would explode instantly when Q5 starts to conduct. \$\endgroup\$ – Bimpelrekkie Sep 24 at 8:13
  • \$\begingroup\$ @eganlee What are you using to measure the current? A DMM will not be able to respond quickly enough to measure a 10ms pulse, the the filament resistance likely changes as it heats up. Do you have an additional series resistor somewhere to measure current? \$\endgroup\$ – Caleb Reister Sep 24 at 18:35
  • \$\begingroup\$ It is fine to use two transistors, though you will probably be better off configuring them as a Darlington pair, or even better, buy one. \$\endgroup\$ – Caleb Reister Sep 24 at 18:38
0
\$\begingroup\$

Your filament has a 'lit up' resistance of 2V/1.5A = 1.3Ω, but a cold resistance of only 0.2Ω. If powered by a constant 2V it would initially draw 2V/0.2Ω = 10A. So for fast 'lighting up' your circuit should be able to provide at least 10A at close to 2V, while not exceeding 2V when the filament heats up.

That's why I tried to put another BJT tried to increase current(maybe it's a bad idea).

Your idea is good. The MJE3055 can (just) deliver 10A, but only has a current gain of ~15 at this Collector current, so it needs a Base current drive of ~10/15 = 0.7A. The 555 Timer IC can only source ~0.2A, so you do need the extra transistor to amplify the current.

You main problem is that R5 and R6 are limiting Q5's Base current to ~0.5A, so it is not capable of providing the required 10A output. Replacing R6 with a short will increase Base current to ~0.8A which may be enough. If that is still not enough then you can reduce the value of R5 until it is.

But then you have another problem. When the filament heats up and its resistance increases the voltage will rise above 2V, which will probably burn it out. You need something to regulate the voltage, so it stays relatively constant even as the filament current varies. In the circuit below this is achieved with an extra transistor acting as a voltage comparator, and some diodes providing a voltage reference.

schematic

simulate this circuit – Schematic created using CircuitLab

Diodes D1 and D2 drop ~0.6V each, providing a reference of 1.2V. Q3 compares that voltage to the output voltage, and begins to turn on when it gets ~0.6V higher (ie. at ~1.8V). It then shunts some of the current coming out of R4 to ground, reducing current supplied to Q1. As output voltage rises more Q3 turns on more and shunts more current away from Q1. In this way it regulates the output voltage to around 1.8~2.0V.

Notes:

R2 puts ~10mA through the diodes to ensure that they drop the required voltage even when Q3 is off.

R1 limits Q3's Base current in case a fault condition causes the output voltage to rise above 2V (eg. filament blown and/or Q2 shorted out).

I increased the value of R4 to reduce power consumption in Q3 and the 555.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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