[sdiy] Output impedance check...
phil.macphail at liivatera.com
Fri Feb 8 14:48:21 CET 2013
R8 should be scaled to prevent this from happening, although the variation in beta would make this quite crude. With 100k for both resistors the transistor will almost certainly be saturated in normal operation, which is why I suggested removing it - that would be far more predictable and free of what you describe,
On 8 Feb 2013, at 15:22, ChristianH <chris at chrismusic.de> wrote:
> I'm not so sure about R10 being redundant - what if the 072 drives the
> transistor into saturation? What would protect the LED from being zapped
> with 12V?
> Ok, 100k may be little high, but without the base current being
> controllable, I wouldn't leave it out completely.
> On Fri, 8 Feb 2013 14:45:10 +0200 Phil Macphail
> <phil.macphail at liivatera.com> wrote:
>> Hi Justin,
>> R10 is redundant in this circuit, unless you wan to limit the current through the LED to around 100uA, which seems a little low… And the LED won't light at all until the signal exceeds Vbe plus Vforward of the LED, which is probably 2.5-3 volts total. For negative signal swings Q3 and the LED are reverse-biased which is probably OK for small signals, but you need to check the breakdown voltage for the transistor.
>> So to answer your questions, the output impedance will still be 1k but it would be much better to use an op-amp if the output signal isn't digital. If the signal is digital, just drive the LED via a resistor from the op-amp without the transistor.
>> On 8 Feb 2013, at 14:30, Justin Owen <juzowen at gmail.com> wrote:
>>> Was hoping someone could take a quick look at this to double check my assumptions?
>>> Standard op-amp buffer as output. Input is from 'in-circuit', output is to the outside world - yes, where dragons be.
>>> Is the impedance of the jack output still 1K even though there is an LED driver also hanging off the same op-amp output?
>>> Any real reason why not to do this under 'normal' use?
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