Please fill out the fields below so we can help you better. Note: you must provide your domain name to get help. Domain names for issued certificates are all made public in Certificate Transparency logs (e.g., so withholding your domain name here does not increase secrecy, but only makes it harder for us to provide help.

My domain is:

I ran this command:
I’m trying to renew certificate via acme4j client

It produced this output:
Caused by: org.shredzone.acme4j.exception.AcmeRateLimitExceededException: Error creating new cert :: too many certificates already issued for exact set of domains: see

This was probably caused by my fault in client implementation. I’ve fixed the issue now. Can you please reset the limit so I can renew the certificate?


Sorry but this is not possible…

You must wait until the limit expired.

(However, you can bypass this issue by adding a new SAN into this cert)

Thank you

The certificate has been renewed dozens of times – as many times as possible – starting February 19. Do you have any of the other certificates, or at least their private keys?

I have 2 questions then:

  1. Do you know when the limit expires?
  2. Do all requests that currently fail still count?

This is described at

The answer to your second question is no; those are covered by a separate rate limit that expires after one hour.

There are multiple limits that can be hit. The limit for exact set of domains is five per seven days. There’s also a limit of twenty per registered domain per seven days. Failed attempts do not count against this, but they do count against the limit of five failures per hostname per hour.

I have rerun the request to get the certificate and I still hit the limit:

Error creating new cert :: too many certificates already issued for exact set of domains

I think there were no requests made for couple of days.

Do you have an information what limit that that domain reaches and when it expires?

The last 5 duplicate certificates were issued in the afternoon of March 28. You have to wait 1 week from that time to get more duplicates.

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