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| minimum load circuit for lab psu |
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| nemail2:
--- Quote from: xavier60 on May 28, 2019, 06:53:24 am --- --- Quote from: nemail2 on May 28, 2019, 06:16:45 am ---that does look good, doesn't it? 300mV is certainly better than what I'm having now. also more stable. and still cheaper/less parts than the opamp solution. residual voltage at the output of the PSU itself? because of the drain? or because of the lack of a drain when set to 0mV where the load circuit isn't working? sorry for the stupid question... --- End quote --- With mine, the small Base bias current causes a small voltage drop across the Emitter resistor while the regulator is off. This small voltage is transferred to the output via the transistor. --- End quote --- let's say I have a short on the output, could that blow the ass out of the transistor? what about a diode to prevent current to flow in the wrong direction? |
| xavier60:
--- Quote from: nemail2 on May 28, 2019, 07:18:12 am --- --- Quote from: xavier60 on May 28, 2019, 06:53:24 am --- --- Quote from: nemail2 on May 28, 2019, 06:16:45 am ---that does look good, doesn't it? 300mV is certainly better than what I'm having now. also more stable. and still cheaper/less parts than the opamp solution. residual voltage at the output of the PSU itself? because of the drain? or because of the lack of a drain when set to 0mV where the load circuit isn't working? sorry for the stupid question... --- End quote --- With mine, the small Base bias current causes a small voltage drop across the Emitter resistor while the regulator is off. This small voltage is transferred to the output via the transistor. --- End quote --- let's say I have a short on the output, could that blow the ass out of the transistor? what about a diode to prevent current to flow in the wrong direction? --- End quote --- A short on the output shouldn't cause a problem. Because I sometimes charge batteries, the last two bench supplies that I built are designed to briefly tolerate applied reverse polarity. I use a diode and a polyswitch just to protect the preload circuit. Not all regulator topologies can be made to tolerate applied reverse polarity such as those with voltage follower output stages. |
| nemail2:
what concerns me a bit after thinking a while about it (sorry for the late response, didn't have time to hobby around recently) is the negative current when the emitter resistors are in place which is shown in your diagram at voltages below 0.1V and also my spice simulation says the same thing. also, I was thinking about using the BC847ALT1G for this, instead of the DMMT5551-7-F to reduce my BOM (and I have already bought the BC847ALT1G in high quantity (at least in regards of hobbyists use)). they should work as well, right? maximumoutput voltage is 20V. what about backfeed voltage of let's say motors or magnetic coils (relais) would they be an issue for the Vceo=45V of the BC847ALT1G? the DMMT5551-7-F has a much higher Vceo of 160V. not sure if that would even help though, when dealing with backfed voltage like that. i guess I'd need some other protection to mitigate this issue but that's for another topic... thanks! |
| nemail2:
ok so i have built this up with the DMMT5551-7-F matched pair and without the emitter resistors. what should i say... the magic smoke escaped from the DMMT5551-7-F at about 7-8V output set voltage. so i bodged in the emitter resistors (48 Ohms) and replaced the DMMT5551-7-F. The magic smoke did not escape again but now i can't get my outout below 120mV. before those emitter ristors my output went down to 3mV. Why is it so? LTSpice shows some negative current at the emittors, is it something to do with that? As soon as I disconnect the current mirror, the voltage drops to normal (3-5mV). What is going on there? Thanks! |
| duak:
I think the output voltage no longer goes to zero because the current mirror is injecting current into +V_ADJ when the output voltage is << 600 mV. The diode connected transistor T4a will develop a certain collector/base voltage as well as the voltage developed across its emitter resistor. This will forward bias T4b's base emitter junction. When +V_ADJ is lower than T4b's base voltage by a few hundred mV, T4b's base-collector junction will also start to conduct and allow current to flow into +V_ADJ. Did you say that T4 was damaged without the emitter resistors? What I think may have happened was that the thermal dissipation of T4b was exceeded due to thermal runaway. Even though these devices are matched fairly well (the spec sheet says they are adjacent dice) T4b will be hotter than T4a when +V_ADJ is increased and when this happens its collector current will be greater than the reference current. This increases its temperature which increases its collector current which increases its dissipation and so on... Do you have the recomended heat sink area on the PCB? I suggest reducing the value of the emitter resistor for T4a and measure T4b's emitter voltage to determine its emitter current and calculating the sinking current and its power dissipation at various output voltages. You may also find that by reducing the emitter resistor value the current mirror will not inject current into the output. |
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