The conventional Joule Thief uses a single cell at 1.5V for power.\u00a0 It also typically uses a coil that has both windings the same number of turns.\u00a0 This means that the voltage across the primary, which is also the voltage across the LED, is the same as the voltage across the feedback winding.\u00a0 Almost all LEDs have a forward voltage of 3.3 volts down to 1.8V for red, so a single cell will not light the LED.\u00a0 This is the purpose of the Joule Thief – to light the LED with a lower voltage.<\/p>\n
We can put two or more LEDs in series on the Joule Thief, which will have a forward voltage of more than 3.3 volts: from 3.6 volts for two red LEDs up to 6.6 volts for two white or blue LEDs.\u00a0 But with two white or blue LEDs, the voltage across the primary winding will be at least 6.6 volts, and this is reflected back to the feedback winding, since it has the same number of turns.\u00a0 The typical transistor has an absolute maximum reverse voltage across the emitter to base junction of 5 to 6 volts, and the peak voltage from the feedback winding will exceed this.\u00a0 So what do we do?\u00a0 Reduce the number of turns of the feedback winding, so the voltage stay be\u00a0 able s below the maximum.\u00a0 If the feedback winding has half the turns of the primary winding, then the voltage will be half, and 6.6 volts on the primary will be 3.3 volts on the feedback winding.<\/p>\n
Note: QS reminded me that the primary voltage has 1.5V or whatever the supply voltage is already added to it, so this voltage must be subtracted from the feedback voltage.\u00a0 With a 1.5V cell, the 3.3V LED should have 1.5V from the cell plus the rest of the voltage from the primary winding, which should be about 1.8V peak.\u00a0 But I’ve seen the voltage across\u00a0 the LED climb to 4.5 volts peak, so I would guess that the actual voltage the primary\u00a0 adds is about 3 volts.<\/p>\n
This will allow us to put two LEDs in series on the primary.\u00a0 Since the LED forward voltage is higher, it will also allow us to use two cells in series for a 3V supply.\u00a0 Normally, with a single cell, the starting voltage would be 1.5V, and the ending voltage would be 0.6V, which is about 0.9 V total.\u00a0 With two cells in series, the starting voltage would be 3.0 V, and the ending voltage 0.6 V, for a total of 2.4 V, a much greater change overall.\u00a0 Each cell should be able to run down to 0.3 V, half the typical JT value.<\/p>\n
One other benefit is that since the supply voltage is higher, the current can be lower for the same power to the LEDs.\u00a0 This also puts less demand on the transistor and the losses will be lower.\u00a0 More than\u00a0 two LEDs could be connected in series across the transistor, as long as the ratio of the coil windings is changed to keep the maximum voltage across the emitter to base junction below 5V, and the maximum voltage across the emitter to collector less than the transistor’s maximum rating, which might be 40V for a PN2222A or 2N4401, but could be as high as three hundred volts for a high voltage transistor such as the MPSA42.\u00a0 The coil windings might have a 5 or 10 to 1 ratio.<\/p>\n
Another way to boost a low voltage without using a specially wound coil is to use a two transistor circuit like the one in my previous blog<\/a>.\u00a0 The coil is a choke or inductor with a single winding.\u00a0 A small amount of the current is fed back through a small capacitor to the first transistor to keep the oscillations going.\u00a0 But the voltage is no longer limited by the first transistor, it is limited by the second transistor’s maximum collector voltage and the load, which might be several LEDs in series.\u00a0 One important point I should make is that the demands on this inductor are greater with a greater ratio between the input voltage, which was 3 volts in our earlier example, to the output voltage, which might be 20 or 30 volts.<\/p>\n","protected":false},"excerpt":{"rendered":" The conventional Joule Thief uses a single cell at 1.5V for power.\u00a0 It also typically uses a coil that has both windings the same number of turns.\u00a0 This means that the voltage across the primary, which is also the voltage across the LED, is the same as the voltage across the feedback winding.\u00a0 Almost all <\/p>\n (Read More…)<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[4],"tags":[],"class_list":["post-7101","post","type-post","status-publish","format-standard","hentry","category-joule-thief-smps-dc-dc"],"_links":{"self":[{"href":"https:\/\/rustybolt.info\/wordpress\/index.php?rest_route=\/wp\/v2\/posts\/7101","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/rustybolt.info\/wordpress\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/rustybolt.info\/wordpress\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/rustybolt.info\/wordpress\/index.php?rest_route=\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/rustybolt.info\/wordpress\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=7101"}],"version-history":[{"count":12,"href":"https:\/\/rustybolt.info\/wordpress\/index.php?rest_route=\/wp\/v2\/posts\/7101\/revisions"}],"predecessor-version":[{"id":7103,"href":"https:\/\/rustybolt.info\/wordpress\/index.php?rest_route=\/wp\/v2\/posts\/7101\/revisions\/7103"}],"wp:attachment":[{"href":"https:\/\/rustybolt.info\/wordpress\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=7101"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/rustybolt.info\/wordpress\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=7101"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/rustybolt.info\/wordpress\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=7101"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}